AP Inter 1st Year

For systematic learning, students can include AP Inter 1st Year English Grammar with Answers and Parts of Speech in English Inter 1st Year in their daily study routine.

AP Inter 1st Year English Parts of Speech with Answers

There are eight parts of speech. They are

1. Noun,
2. Pronoun,
3. Verb,
4. Adjective,
5. Adverb,
6. Preposition,
7. Conjunction and
8. Interjection.

Noun

A noun is a word for a person, place, thing, or idea. Nouns are often used with an article (the, a, an), but not always. Proper nouns always start with a capital letter; common nouns do not. Nouns can be singular or plural, concrete or abstract. Nouns show possession by adding’s. Nouns can function in different roles within a sentence; for example, a noun can be a subject, direct object, indirect object, subject complement, or object of a preposition.

నామవాచకం అనేది ఒక వ్యక్తి, స్థలం, వస్తువు లేదా ఆలోచనను సూచించే పదం. నామవాచకాలు తరచుగా ఒక article (the, a, an)తో ఉపయోగించబడతాయి, కానీ ఎల్లప్పుడూ కాదు. సరైన నామవాచకాలు ఎల్లప్పుడూ capital అక్షరంతో ప్రారంభమవుతాయి. సాధారణ నామవాచకాలు అలా చేయవు. నామవాచకాలు ఏకవచనం లేదా బహువచనం, కాంక్రీటు లేదా వియుక్తంగా ఉండవచ్చు. నామవాచకాలు ‘s ని జోడించడం ద్వారా possession ని చూపుతాయి. నామవాచకాలు ఒక వాక్యంలో వేర్వేరు పాత్రలలో పనిచేయగలవు; ఉదాహరణకు, ఒక నామవాచకం ఒక కర్త, ప్రత్యక్ష వస్తువు, పరోక్ష వస్తువు, కర్త యొక్క వస్తువు కావచ్చు.

Common Nouns : man, mountain, state, ocean, country, building, cat, airline, etc.

Proper Nouns: Himalayas, Andhra Pradesh, Atlantic Ocean, Australia, Vijayawada, Moon, Sun, Ram, Raghu, Sita, etc. Abstract Nouns: temperature, wealth, happiness, pride, fear, religion, belief, history, communication, fever, headache, heat, etc.

Concrete Nouns: stone, water, wood, iron, glass, grass, bird, photograph, banana, hands, dog, suitcase, flowers, etc. Countable Nouns: coin, bed, cat, movie, train, country, book, phone, match, speaker, clock, pen, pencil, violin, etc.

Uncountable Nouns : money, milk, rice, snow, rain, water, food, music, oil, sugar, hair, etc.

Compound Nouns : headmaster, handkerchief, tablecloth, eyeglasses, seabird, classmate, daughter-in-law, pigtails, sunlight, snowfall, playground, doorbell, etc.

Collective Nouns: bunch, audience, flock, team, group, family, band, bundle, socks, trousers, herd, flock, group, procession, staff, jury, etc.

Pronoun

A pronoun is a word used in place of a noun. A pronoun is usually substituted for a specific noun. Pronouns are further defined by type: personal pronouns refer to specific persons or things; possessive pronouns indicate ownership; reflexive pronouns are used to emphasize another noun or pronoun; relative pronouns introduce a subordinate clause; and demonstrative pronouns identify, point to, or refer to nouns.

సర్వనామం అనేది నామవాచకం స్థానంలో ఉపయోగించే పదం. సర్వనామం సాధారణంగా ఒక నిర్దిష్ట నామవాచకానికి ప్రత్యామ్నాయంగా ఉంటుంది, దీనిని, ఆమె అనే సర్వనామం యొక్క పూర్వపదం అమ్మాయి. వ్యక్తిగత సర్వనామాలు నిర్దిష్ట వ్యక్తులు లేదా వస్తువులను సూచిస్తాయి.

Personal Pronouns:

Demonstrative Pronouns : This, that, these, those.

• This is a book.
• That is a book.
• These are my books.
• Those are your books.

Indefinite Pronouns : Anyone, something, all, some, both. Anyone can do it.

• Anybody can do it.
• Someone can do it.
• Somebody can do it.
• All can do it.
• Both are my sisters.

Interrogative Pronouns: What, whom, whose, who, which. What happened?

• With whom did you come ?
• Whose are these books?
• Who do you want?
• Which is your pen?

Distributive Pronouns : Each, everyone, either, neither. Each one of you will get a chance.

• Everyone of you will get a chance.
• Either of you can get a chance.
• Neither of you will get a chance.

Verb

The verb in a sentence expresses action or being. There is a main verb and sometimes one or more helping verbs. (“She can sing.” Sing is the main verb; can is the helping verb.) A verb must agree with its subject in number (both are singular or both are plural). Verbs also take different forms to express tense.

ఒక వాక్యంలోని క్రియ చర్య లేదా ఉనికిని వ్యక్తపరుస్తుంది. ఒక ప్రధాన క్రియ మరియు కొన్నిసార్లు ఒకటి లేదా అంతకంటే ఎక్కువ సహాయ క్రియలు ఉంటాయి. (“ఆమె పాడగలదు.” sing అనేది ప్రధాన క్రియ; can అనేది సహాయ క్రియ.) ఒక క్రియ దాని కర్తతో ఏకీభవించాలి (రెండూ ఏకవచనం లేదా రెండూ బహువచనం). క్రియలు కూడా కాలాన్ని వ్యక్తీకరించడానికి వేర్వేరు రూపాలను తీసుకుంటాయి.

• Action verbs : go, come, sit, stand, walk, jump, take, drive, do, drink, clean, cut, drive, eat, fly, go, live, make, play, read, run, shower, sleep, smile, stop, sweep, swim, think, throw, trip, wash, work, write etc. (There are thousands of these verbs.)
• Being verbs : am, is, are, was, were (be, been).
• Possessive verbs : have, has, had.
• Helping verbs : Am, is, are, was, were, have, has, had, will, would, shall, should, can, could, may, might, need, dare, ought, used, and must.

Adjective

An adjective is a word used to modify or describe a noun or a pronoun. It usually answers the question of which one, what kind, or how many. (Articles [a, an, the] are usually classified as adjectives.)
Examples : homeless, adorable, delightful, homely, adventurous, horrible, aggressive, determined, hungry, real, agreeable, etc.

విశేషణం అనేది నామవాచకం లేదా సర్వనామాన్ని సవరించడానికి లేదా వివరించడానికి ఉపయోగించే పదం. ఇది సాధారణంగా ఏది, ఏ రకం లేదా ఎన్ని అనే ప్రశ్నకు సమాధానం ఇస్తుంది. (articles [a, an, the] ను సాధారణంగా విశేషణాలుగా వర్గీకరిస్తారు.)

Kinds of adjectives:

a. Quantitative Adjectives : Quantitative adjectives provide information about the quantity of the nouns / pronouns. This type belongs to the question category of ‘how much’ and ‘how many.’
Examples:

I have just 200 rupees in my purse. (How much)
They have three children. (How many)
You should have completed the whole task. (How much)
There are a few students in the class. (How many)

b. Descriptive Adjectives : Descriptive adjectives are words which describe nouns and pronouns. Most of the adjectives belong to this type. These adjectives provide information and attribute to the nouns/pronouns they modify or describe. Descriptive adjectives are also called qualitative adjectives.

Examples:

• I have a fast car which can be used in races.
• I am hungry so let us go to any hotel.
• The mischievous children are making a lot of noise.
• I saw a flying Eagle.
• He bought an expensive old silver mirror.
• He gave me four fresh yellow roses.
• I want to buy a diamond ring.

c. Proper Adjectives: Proper adjectives are the adjective form of proper nouns. When proper nouns modify or describe other nouns/pronouns, they become proper adjectives. ‘Proper’ means ‘specific.’
Example:

• American cars are comparatively very strong.
• Chinese people are hard workers but their leaders are cunning.
• KFC restaurants are found only in big cities.
• Marxist philosophers don’t like capitalism.
• He belongs to the Communist party of India.
• Indian culture is getting diluted.

d. Demonstrative Adjectives: Demonstrative adjectives directly refer to something or someone. Demonstrative adjectives include the words: this, that, these, those.

Examples:

• These books are borrowed from our college library.
• This house is ours.
• That problem needs a very good solution.
• Those bridges were built by the British.

e. Possessive adjectives : Possessive adjectives indicate possession or ownership. It suggests the belongingness of something to someone/something. Some of the most used possessive adjectives are my, his, her, our, their, your.

Examples:

• Your cycle was stolen yesterday.
• We all liked his performance.
• The students of class submitted their assignments.
• I could not understand her intentions.
• My mind stops working when I think about the problem.
• Our intention is not to hurt anyone’s feelings.

f. Interrogative Adjectives : Interrogative adjectives ask a question. An interrogative adjective must be followed by a noun or a pronoun. The interrogative adjectives are: which, what, whose. These words will not be considered as adjectives if a noun does not follow right after them. ‘Whose’ also belongs to the possessive adjective type.

Examples:

• Whose book was that?
• Which pen do you like more?
• What books are you buying today?
• Which team scored the higher in the last match?
• What curry do you like on Sundays?
• Which song are you listening to?

g. Indefinite Adjectives : Indefinite adjective describes or modifies a noun unspecifically. They provide indefinite/ unspecific information about the noun. The common indefinite adjectives are few, many, much, most, all, any, each, every, either, nobody, several, some, etc.

Examples:

• I gave some ice cream to my friend.
• I want a few moments to take a decision.
• Several writers wrote about the recent incidents.
• Each student will have to submit homework tomorrow.

Adverb

An adverb describes or modifies a verb, an adjective, or another adverb, but never a noun. It usually answers the questions of when, where, how, why, under what conditions, or to what degree. Adverbs often end in ‘ly’.

ఒక క్రియా విశేషణం ఒక క్రియ, విశేషణం లేదా మరొక క్రియా విశేషణాన్ని వర్ణిస్తుంది లేదా మారుస్తుంది, కానీ ఎప్పుడూ నామవాచకంగా ఉండదు. ఇది సాధారణంగా ఎప్పుడు, ఎందుకు, ఎలా, ఎందుకు, ఏ పరిస్థులలో, లేదా ఎంతవరకు అనే ప్రశ్నలకు సమాధానం ఇస్తుంది. క్రియా విశేషణాలు తరుచుగా ly తో ముగుస్తాయి.

Adverbs of questions : How, when, where, why.
How did they do it ?
When did they do it ?
Where did they do it ?

Adverbs of Time : today, yesterday, later, now, then, last year, all day, for a while, sometimes,

Adverbs of Frequency: hourly, daily, weekly, monthly, yearly, every, second, once a minute, twice a year, once, twice, once or twice, three times, always, constantly, ever, frequently, generally, never, normally, occasionally, often, rarely, regularly, seldom, sometimes, usually, annually, daily, fortnightly, hourly, monthly, nightly, quarterly, weekly, yearly, yes, no, not, so, too, very, etc.

Preposition

A preposition is a word placed before a noun or pronoun to form a phrase modifying another word in the sentence. Therefore a preposition is always part of a prepositional phrase. The prepositional phrase almost always functions as an adjective or as an adverb.

సర్వనామం ముందు ఉంచిన పదం, వాక్యంలోని మరొక పదాన్ని సవరించి విభక్తి ప్రత్యయము ఒక పదబంధాన్ని ఏర్పరుస్తుంది. అందువల్ల, విభక్తి ప్రత్యయము రెండు పదాల మధ్య బంధాన్ని ఏర్పరుస్తుంది.

Examples: aboard, about, above, across, after, against, ahead of, along, amid, amidst, among, around, as, as far as, as of, aside from, at, athwart, atop barring, before, behind, below, beneath, beside, besides, between, beyond, but, by, despite, down, during, except, excluding, for, from, in, minus, near, next to, of, off, on, opposite, out, outside, over, since, through, till, to, toward, towards, under, underneath, unlike, until, up, upon, with, within, without, etc.

Conjunction

A conjunction joins words, phrases, or clauses, and indicates the relationship between the elements joined.

Coordinating conjunctions connect grammatically equal elements: and, but, or, nor, for, so, yet.

Subordinating conjunctions connect clauses that are not equal: because, although, while, since, etc.

There are other types of conjunctions as well.

ఒక సంయోగం పదాలు,పదబంధాలు లేదా నిబంధనలను కలుపుతుంది మరియు జతచేయబడిన elements మధ్య సంబంధాన్ని సూచిస్తుంది. సమన్వయ సంయోగాలు వ్యాకరణపరంగా సమాన అంశాలను కలుపుతాయి: and, but, or, nor, for, so, yet. సబార్డినేటింగ్ సంయోగాలు సమానం కానీ నిబంధనలు కలుపుతాయి: because, although, while, since, etc. ఇతర రకాల సంయోగాలు కూడా ఉన్నాయి.

Subordinating conjunctions: though, although, even though, even if, as, since, because, for, after, as if, as long as, as much as, as soon, as though, before, even if, if, lest, than, that, till, unless, until, when, where, how, what, why, whenever, wherever, while etc.

Coordinating conjunctions: but, yet, still, so, therefore, that’s why, and or, otherwise, then, either………… or, neither………… nor, not only……… but also etc.

Interjection

An interjection is a word used to express emotion. It is often followed by an exclamation point (!).

భావోద్వేగానికి వ్యక్తీకరించడానికి ఉపయోగించే పదం దాని తర్వాత తరచుగా ఆశ్చర్యార్థంకంగా గుర్తు (!) ఉంటుంది.

Example: Aha! Ahh! Ahoy! Alas! Boo! Bravo! Congratulations! Eh! Eureka! Goodbye! Goodness! Hey! Hurray! Oh! Oops! Ouch! Phew! uhhuh!

Wahoo! Wow! Yeah!, etc.

Examples

Question 1.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 2.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

(Note : In sentence 5, ‘out’ modifies the verb ‘went’ by indicating direction or movement away from a place, which is typical of an adverb of place.)

Exercises For Practice

Question 1.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 2.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 3.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 4.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 5.
Here’s a day from the life of Padmini. Can you identify at least 15 nouns from the given paragraph ?

Last weekend, Padmini visited the museum with her friends. They admired the paintings and marvelled at the sculptures in the art gallery. Later, they had lunch at a cozy cafe, where Padmini ordered a delicious sandwich and a cup of coffee. The group talked about their plans for the upcoming holiday and shared their ideas for a fun trip. Everyone enjoyed the afternoon, and they decided to visit the museum again next month.
Answers:

1. weekend
2. museum
3. friend
4. paintings
5. sculptures
6. gallery
7. lunch
8. cafe
9. sandwich
10. cup
11. coffee
12. group
13. plans
14. holiday
15. ideas
16. trip
17. afternoon
18. month

Question 6.
Fill in the blanks with correct plural form of the nouns given in the brackets.

1. The park was full of playful ___________. (child)
Answer:
children

2. We ran several ___________. (diagnosis)
Answer:
diagnoses

3. He explained the ___________ (focus) of the study.
Answer:
foci

4. My ___________ (foot) were sore after the run.
Answer:
feet

5. My dentist says I have healthy ___________. (tooth)
Answer:
teeth

6. A flock of ___________ (goose) flew across the sky.
Answer:
geese

7. All my ___________ (brother-in-law) are teachers.
Answer:
brothers-in-law

Question 7.
Fill in the blanks with appropriate Pronouns to complete the story.

One day, a young boy named Sam went to the park with ___________ (1) sister. ___________ (2) both loved playing on the swings and running around. While Sam was climbing a tree, ___________ (3)
saw a bird’s nest. ___________ (4) called out to ___________ (5) sister, “Look! There’s a nest up here!” ___________ (6) sister looked up and said. “Be careful, Sam. Don’t scare the birds!” Sam carefully climbed down and joined ___________ (7) on the bench. Together, ___________ (8) decided to draw the birds in ___________ (9) notebooks. When ___________ (10) finished, ___________ (11) showed the drawings to ___________ (12) parents, who were very impressed.
Answers:
1. his
2. They
3. he
4. He
5. his
6. His
7. her
8. they
9. their
10. they
11. they
12. their

Question 8.
Read each sentence carefully and underline or highlight the Adjectives.

1. The beautiful garden was filled with vibrant flowers.
Answer:
The beautiful garden was filled with vibrant flowers.

2. She wore a red dress and carried a small bag.
Answer:
She wore a red dress and carried a small bag.

3. The tall building cast a long shadow on the street.
Answer:
The tall building cast a long shadow on the street.

4. We had a delicious meal at the new restaurant.
Answer:
We had a delicious meal at the new restaurant.

5. His quick decision helped to avoid a serious mistake.
Answer:
His quick decision helped to avoid a serious mistake.

6. The young boy found a golden key in the ancient chest.
Answer:
The young boy found a golden key in the ancient chest.

7. It was a cold night with a bright moon in the sky.
Answer:
It was a cold night with a bright moon in the sky.

8. The teacher gave us a challenging yet interesting assignment.
Answer:
The teacher gave us a challenging yet interesting assignment.

9. I love my cozy, little home in the peaceful countryside.
Answer:
I love my cozy, little home in the peaceful countryside.

10. The angry crowd demanded immediate action from the authorities:
Answer:
The angry crowd demanded immediate action from the authorities.

Question 9.
Identify the Adverbs in the following sentences.

1. She sings melodiously during every performance.
Answer:
melodiously

2. The train arrived late because of the storm.
Answer:
late

3. He quickly finished his homework before heading out.
Answer:
quickly

4. The team played extremely well in the finals.
Answer:
extremely, well

5. They are always kind to their neighbours.
Answer:
always

6. The baby slept peacefully in her crib.
Answer:
peacefully

7. We will visit the museum tomorrow.
Answer:
tomorrow

8. He spoke very softly during the meeting.
Answer:
very, softly

9. The cat crept silently across the room.
Answer:
silently

10. I have never seen such a beautiful sunset.
Answer:
never

Question 10.
Underline the Conjunctions in the following sentences.

1. She wanted to go to the party, but she was feeling unwell.
Answer:
She wanted to go to the party, but she was feeling unwell.

2. I will call you when I reach home.
Answer:
I will call you when I reach home.

3. He likes both cats and dogs, but he cannot decide which to adopt.
Answer:
He likes both cats and dogs, but he cannot decide which to adopt.

4. They left early because they had a train to catch.
Answer:
They left early because they had a train to catch.

5. You can stay here or you can go with them.
Answer:
You can stay here or you can go with them.

Exercises

Question 1.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 2.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 3.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 4.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 5.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 6.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 7.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 8.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 9.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 10.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 11.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 12.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Referring to the AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series Exercise 8e Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sequences and Series Solutions Exercise 8e

I. Find the sum to n terms of each of the series in Exercises 1 to 7.

Question 1.
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
Solution:
The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … n terms
Hence, a_n = n(n + 1)
Therefore, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) a_k = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k(k + l) = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k² + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k

= \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{\mathrm{n}(\mathrm{n}+1)}{2}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\frac{2 \mathrm{n}+1}{3}+1\right)\)

= \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\frac{2 \mathrm{n}+4}{3}\right)\)

= \(\frac{\mathrm{n}}{3}\)(n + 1)(n + 2)

Question 2.
1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
Solution:
The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + … n terms
Hence, a_n = n(n + 1) (n + 2) = (n² + n) (n + 2) = n³ + 3n² + 2n
Therefore, S_n = \(\sum_{k=1}^n\) ak = \(\sum_{k=1}^n\)(k³ + 3k² + 2k) = \(\sum_{k=1}^n\) k³ + 3 \(\sum_{k=1}^n\) k² + 2 \(\sum_{k=1}^n\) k

Question 3.
3 × 1² + 5 × 2² + 7 × 3² + …
Solution:
The given series is 3 × 1² + 5 × 2² + 7 × 3² + …. n terms
Hence, a_n = (2n + 1) n² = 2n³ + n²
Therefore, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) a_k = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) (2k³ + k²) = 2\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k³ + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k²

= 2\(\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]^2\) + \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\)

= \(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{2}+\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\)

= \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\mathrm{n}(\mathrm{n}+1)+\frac{2 \mathrm{n}+1}{3}\right)\)

= \(\frac{n(n+1)}{2}\left[\frac{3 n^2+3 n+2 n+1}{3}\right]\)

= \(\frac{n(n+1)}{2}\left[\frac{3 n^2+5 n+1}{3}\right]\)

= \(\frac{n}{6}\)(n + 1)(3n + 5n + 1)

Question 4.
\(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}\) + ………..
Solution:
The given series is \(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}\) + ……….. n terms
Hence, a_n = \(\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\)
∴ a₁ = \(\frac{1}{1}-\frac{1}{2}\), a₂ = \(\frac{1}{2}-\frac{1}{3}\), a₃ = \(\frac{1}{3}-\frac{1}{4}\), …………….., a_n = \(\frac{1}{n}-\frac{1}{n+1}\),

Adding the above terms column wise, we obtain
a₂ + a₂ + …… + a_n = \(\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{\mathrm{n}}\right]-\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{\mathrm{n}+1}\right]\)

Thus, S_n = 1 – \(\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}\)

Question 5.
5² + 6² + 7² + ………. +20²
Solution:
The given series is 5² + 6² + 7² + ………. +20²
Hence, a_n = (n + 4)² = n² + 8n + 16

Therefore, S_n = \(\sum_{k=1}^n\)a_k = \(\sum_{k=1}^n\)(k² + 8k + 16)

S_n = \(\sum_{k=1}^n\)k² + 8\(\sum_{k=1}^n\)k + \(\sum_{k=1}^n\) 16

= \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{8 \mathrm{n}(\mathrm{n}+1)}{2}\) + 16n

Since, 16^th term is (16 + 4)² = (20)²

Then, S_n = \(\frac{16(16+1)(2(16)+1)}{6}+\frac{8(16)(16+1)}{2}\)+ 16(16)

= \(\frac{(16)(17)(33)}{6}+\frac{(128)(17)}{2}\) + 256

= 1496 + 1088 + 256 = 2840

Thus, 5² + 6² + 7² + ………. +20² = 2840

Question 6.
3 × 8 + 6 × 11 + 9 × 14 + …………..
Solution:
The given series is 3 × 8 + 6 × 11 + 9 × 14 + ………….. n terms
Hence, a_n = (n^th term of 3, 6, 9, ………..) × (n^th term of 8, 11, 14 ……)
= (3n)(3n + 5)
= 9n² + 15n

Therefore, S_n = \(\sum_{k=1}^n\) ak = \(\sum_{k=1}^n\) (9k² + 15k)

= 9 \(\sum_{k=1}^n\) k² + 15\(\sum_{k=1}^n\)k

= 9 × \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\) + 15 × \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

= \(\frac{3 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{2}+\frac{15 \mathrm{n}(\mathrm{n}+1)}{2}\)

= \(\frac{3 \mathrm{n}(\mathrm{n}+1)}{2}\)(2n + 1 + 5)

= \(\frac{3 \mathrm{n}(\mathrm{n}+1)}{2}\)(2n + 6)

= 3n(n + 1)(n + 3)

Question 7.
1² + (1² + 2²) + (1² + 2² + 3²) + ………..
Solution:
The given series is 1² + (1² + 2²) + (1² + 2² + 3²) + ……….. n terms

Hence, a_n = (1² + 2² + 3² + …… + n²) = \(\frac{n}{2}\)(n + 1)(2n + 1)
= \(\frac{n}{6}\) (2n² + 3n + 1)
= \(\frac{1}{3}\)n³ + \(\frac{1}{2}\)n² + \(\frac{1}{6}\)n

Therefore,

II. Find the sum to n terms of the series in Exercises 1 to 3 whose nth terms is given by

Question 1.
n(n + 1) (n + 4).
Solution:
The given n* term is a_n = n(n + 1) (n + 4)
Hence, a_n = n(n + 1) (n + 4) = n(n² + 5n + 4) = n³ + 5n² + 4n
Therefore, S_n = \(\sum_{k=1}^n\) a_k = \(\sum_{k=1}^n\)k³ + 5 \(\sum_{k=1}^n\) k² + 4 \(\sum_{k=1}^n\) k

Question 2.
n² + 2n
Solution:
The given n^th term is an = n² + 2ⁿ
Hence, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k² + 2^k

= \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k² + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)2^k …………..(1)
Consider \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 2^k = 2¹ + 2² + 2³ +…

The above series 2¹ + 2² + 2³ + … is a G.P with both the first term and common ratio equal to 2.
\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 2^k = \(\frac{1}{2}\) = 2(2ⁿ – 1) …………..(2)

From (1) and (2) we obtain
S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k² + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 2^k
= \(\frac{n}{6}\)(n + 1)(2n + 1) + 2(2ⁿ – 1)

Question 3.
(2n – 1)²
Solution:
The given n^th term is a_n = (2n – 1)²
Hence, a_n = (2n – 1)² = 4n² – 4n + 1

Therefore, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) a = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) (4k² – 4k + 1)

= 4\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k2 – 4\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 1

= \(\frac{4 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\frac{4 \mathrm{n}(\mathrm{n}+1)}{2}\) + n

= \(\frac{2 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{3}\) – 2n(n + 1) + n

= n[\(\frac{2\left(2 n^2+3 n+1\right)}{3}\) – 2(n + 1) + 1]

= n\(\left[\frac{4 n^2+6 n+2-6 n-6+3}{3}\right]\) = n\(\left[\frac{4 \mathrm{n}^2-1}{3}\right]\)

= \(\frac{\mathrm{n}}{3}\) (2n + 1)(2n – 1)

Referring to the AP Inter 1st Year Maths Study Material Chapter 7 Binomial Theorem Exercise 7a Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Binomial Theorem Solutions Exercise 7a

I. Expand each of the expressions in Exercises 1 to 8.

Question 1.
(1 – 2x)⁵
Solution:
By using Binomial theorem, the expression (1 – 2x)⁵ can be expanded as
⁵C₀(1)⁵(-2x)⁰ + ⁵C₁(1)⁴(-2x)¹ + ⁵C₂ (1)³ (-2x)² + ⁵C₃(1)²(-2x)³ + ⁵C₄(1)¹ (-2x)⁴ + ⁵C₅(1)° (-2x)⁵
= 1 – 5(2x) + 10(4x²) – 10(8x³) + 5(16x⁴) – (32x⁵)
= 1 – 10x + 40x² – 80x³ + 80x⁴ – 32x⁵.

Question 2.
\(\left(\frac{2}{x}-\frac{x}{2}\right)^5\)
Solution:
By using Binomial theorem, the expression \(\left(\frac{2}{x}-\frac{x}{2}\right)^5\) can be expanded as

Question 3.
(2x – 3)⁶
Solution:
By using Binomial theorem, the expression (2x – 3)⁶ can be expanded as ⁶C₀ (2x)⁶ (-3)° + ⁶C₁ (2x)⁵ (-3)¹ + ⁶C₂ (2x)⁴ (-3)² + ⁶C₃ (2x)³ (-3)³ + ⁶C₄ (2x)² (-3)⁴ + ⁶C₅ (2x)¹ (-3)⁵ + ⁶C₆ (2x)⁰ (-3)⁶
= 64x⁶ – 6(32x⁵) (3) + 15(16x⁴)(9) – 20(8x³)(27) + 15(4x²)(81) – 6(2x)(243) + 729
= 64x⁶ – 576x⁵ + 2160x⁴ – 4320x³ + 4860x² – 2916x + 729

Question 4.
\(\left(\frac{x}{3}+\frac{1}{x}\right)^5\)
Solution:
By using Binomial theorem, the expression \(\left(\frac{x}{3}+\frac{1}{x}\right)^5\) can be expanded as

Question 5.
\(\left(x+\frac{1}{x}\right)^6\)
Solution:
By using Binomial theorem, the expression \(\left(x+\frac{1}{x}\right)^6\) can be expanded as

Question 6.
(4x + 5y)⁷
Solution:
By using Binomial theorem,
(4x + 5y)⁷ = ⁷C₀ (4x)⁷ (5y)⁰ + ⁷C₁ (4x)⁶ (5y)¹ + ⁷C₂ (4x)⁵ (5y)² + ⁷C₃ (4x)⁴ (5y)³ + ⁷C₄ (4x)³ (5y)⁴ + ⁷C₅ (4x)² (5y)⁵ + ⁷C₆ (4x)¹ (5y)⁶ + ⁷C₇ (4x)° (5y)⁷
= \(\sum_{r=0}^7\) ⁷C_r (4x)^7-r (5y)^r

Question 7.
\(\left(\frac{2}{3} x+\frac{7}{4} y\right)^5\)
Solution:
By using Binomial theorem, we have

Question 8.
\(\left(\frac{2 p}{5}-\frac{3 q}{7}\right)^6\)
Solution:

Question 9.
Write down and simplify.
i) 6^th term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\)
Solution:
Given that \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\)
The general term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\) is
T_r+1 = ⁹C_r \(\left(\frac{2 x}{3}\right)^{9-r}\left(\frac{3 y}{2}\right)^r\)
If r = 5
T₆ = ⁹C₅\(\frac{1}{2}\)
= ⁹C₅\(\frac{1}{2}\)
= ⁹C₄\(\frac{1}{2}\) x⁴y⁵
= \(\frac{1}{2}\). x⁴y⁵
= 189 x⁴y⁵

ii) 7^th term in (3x – 4y)¹⁰.
Solution:
Given that (3x – 4y)¹⁰
The general term in (3x – 4y)¹⁰ is T_r+1 = (-1)^{r 10}C_r (3x)^10-r (4y)^r
If r = 6 ⇒ T₇ = (-1)^{6 10}C₆ (3x)⁴ (4y)⁶
= ₂C₄3₄4₆x₄y₆
= \(\frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4}\). 3₄.4₆. x₄. y₆
= 280 (12)₅x₄y₆ (∵ _nC_r = _nC_n-r)

iii) 10^th term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\)
Solution:
Given that \(\left(\frac{3 p}{4}-5 q\right)^{14}\)
The general term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\) is T_r+1 = (-1)^{r 14}C_r \(\left(\frac{3 \mathrm{p}}{4}\right)^{14-\mathrm{r}}\) (5q)^r
If r = 9
T₁₀ = (-1)^{9 14}C₉\(\left(\frac{3 p}{4}\right)^5\) (5q)⁹
= – ¹⁴C₅\(\left(\frac{3}{4}\right)^5\) 5⁹.p⁵.q⁹ (∵ _nC_r = _nC_n-r)
= \(-\frac{14 \times 13 \times 12 \times 11 \times 10}{1 \times 2 \times 3 \times 4 \times 5} \cdot \frac{3^5}{4^5}\).5⁹.p⁵.q⁹
= \(\frac{-(2002) 3^5 \cdot 5^9}{4^5}\)p⁵.q⁹

iv) r^th term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) (1 ≤ r ≤ 8)
Solution:
Given that \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\)
The general term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) is T_r+1 = ⁸C_r \(\left(\frac{3 a}{5}\right)^{8-r}\left(\frac{5 b}{7}\right)^r\)
Replace r by r – 1, then
T_r = ⁸C_(r-1)\(\left(\frac{3 a}{5}\right)^{8-r+1}\left(\frac{5 b}{7}\right)^{r-1}\)
= ⁸C_(r-1)\(\left(\frac{3 a}{5}\right)^{9-r}\left(\frac{5 b}{7}\right)^{r-1}\) (1 ≤ r ≤ 8)

v) Middle term in \(\left(3-\frac{x^3}{6}\right)^7\)
Solution:
Given that \(\left(3-\frac{x^3}{6}\right)^7\)

The expression contains 8 terms. Thus T₄ and T₅ are middle terms.
T₄ = T₃₊₁ = ⁷C₃ (3)^7-3\(\left(-\frac{x^3}{6}\right)^3=\frac{35 \times 3^4(-x)^9}{6^3}=-\frac{105}{8}\)x⁹.

T₅ = T₄₊₁ = ⁷C₄ (3)^7-4\(\left(-\frac{x^3}{6}\right)^4=\frac{35 \times 3^3 x^{12}}{6^4}=\frac{35}{48}\)x¹²

Thus, the middle terms in the expansion of \(\left(3-\frac{x^3}{6}\right)^7\) are \(-\frac{105}{8}\) x⁹ and \(\frac{35}{48}\)x¹²

vi) Middle term in \(\left(\frac{x}{3}+9 y\right)^{10}\)
Solution:
Given that \(\left(\frac{x}{3}+9 y\right)^{10}\)
The number of terms in the given expansion is 11. Thus T₆ is the middle term.
T₆ = T₅₊₁ = ¹⁰C₅ \(\frac{1}{2}\) (9y)⁵
= \(\frac{1}{2}\) 3¹⁰.x⁵y⁵
= 252 × 3⁵ x⁵.y⁵
= 61236 x⁵y⁵

vii) 4th term in (x – 2y)¹²
Solution:
Given that (x – 2y)¹²
General term in the expansion of (x + y)ⁿ is Tr+ x = nCr Xn_r (Y)r
Here, X = x, Y = -2y, and n = 12; T_r+1 = ¹²C_r (x)^12-r (-2y)^r
Putting r = 3, ,
T₃₊₁ = ¹²C₃x^12-3 (-2y)³
T₄ = \(\frac{12 \times 11 \times 10}{6}\) x⁹(-2)³y³
= 2 × 11 × 10x⁹(-8)y³
= 220 × (-8)x⁹y³
= -1760 x⁹y³

viii) 13^th term in \(\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}\), x ≠ 0
Solution:
The general term in the expansion of (X + Y)ⁿ is T_r+1 = ⁿC_r X^n-rY^r

Here X = 9x, Y = –\(\frac{1}{3 \sqrt{x}}\), and n = 18

The General term in the expansion of \(\frac{1}{2}\) is T_r = ¹⁸C_r (9x)^18-r\(\left(\frac{-1}{3 \sqrt{x}}\right)^r\)
Putting r = 12
T₁₂₊₁ = ¹⁸C₁₂ (9x)⁶ \(\left(\frac{-1}{3 \sqrt{x}}\right)^{12}\)
= ¹⁸C₁₂ 9⁶ × x⁶\(\frac{1}{3^{12}(\sqrt{x})^{12}}\) (∵ _nC_r = _nC_n-r)
= \(\frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \times \frac{3^{12} \times x^6}{3^{12} \times x^6}\)
= 18564

Question 10.
Find the number of terms in the expansion of
i) \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\)
Solution:
The number of terms in (x + a)ⁿ is (n + 1), where n is a positive integer.
Hence number of terms in \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\) are 9 + 1 = 10

ii) (3p + 4q)¹⁴
Solution:
The number of terms in (x + a)ⁿ is (n + 1), where n is a positive integer.
Hence number of terms in (3p + 4q)¹⁴ are 14 + 1 = 15

II. Using binomial theorem, evaluate each of the following.

Question 1.
(96)³
Solution:
96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial theorem can be applied, it can be written that, 96 = 100 – 4
Therefore, (96)³ = (100 – 4)³
= ³C₀ (100)³ (-4)⁰ + ³C₁ (100)² (-4)¹ + ³C₂ (100)¹ (-4)² + ³C₃ (100)° (-4)³
= (100)³ – 3(100)²(4) + 3(100)(4)² – (4)³
= 1000000 – 120000 + 4800 – 64
= 884736

Question 2.
(102)⁵
Solution:
102 can be expressed as the sum or difference of two numbers whose powers are
easier to calculate and then, Binomial theorem can be applied.
It can be written that, 102 = 100 + 2.
Therefore, (102)⁵ = (100 + 2)⁵
= ⁵C₀ (100)⁵ (2)° + ⁵C₁ (100)⁴ (2)¹ + ⁵C₂ (100)³ (2)² + ⁵C₃ (100)² (2)³ + ⁵C₄ (100)¹ (2)⁴ + ⁵C₅ (100)° (2)⁵
= (100)⁵ + 5(100)⁴ (2) + 10(100)³(2)² + 10(100)²(2)³ + 5(100) × (2)⁴ + (2)⁵
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032

Question 3.
(101)⁴
Solution:
101 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial theorem can be applied.
It can be written that, 101 = 100+ 1.
(101)⁴ = (100 + 1)⁴ = ⁴C₀ (100)⁴(1)°+ ⁴C₁ (100)³ (1)¹ + ⁴C₂ (100)² (1)² + ⁴C₃ (100)¹ (1)³ + ⁴C₄ (100)° (1)⁴
= (100)⁴ + 4(100)³ + 6(100)² + 4(100)¹ + 1
= 100000000 + 4000000 + 60000 + 400 + 1
= 104060401

Question 4.
(99)⁵
Solution:
99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial theorem can be applied.
It can be written that 99 = 100 – 1
=> (99)⁵ = (100 – 1)⁵
= ⁵C₀(100)⁵(-1)° + ⁵C₁(100)⁴(-1)¹ + ⁵C₂(100)³(-1)² + ⁵C₃(100)²(-1)³ + ⁵C₄(100)¹(-1)⁴ + ⁵C₅(100)°(-1)⁵
= (100)⁵ – 5(100)⁴ + 10(100)³ – 10(100)² + 5(100) – 1
= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1
= 10010000500 – 500100001
= 9509900499

Question 5.
Using Binomial theorem, indicate which number is larger (1.1)¹⁰⁰⁰⁰ 0r 1000.
Solution:
By splitting 1.1 and then applying Binomial theorem, the first few terms of (1.1)10000 can be obtained as (1.1)¹⁰⁰⁰⁰
= (1 + 0.1)¹⁰⁰⁰⁰
= ¹⁰⁰⁰⁰C₀ + ¹⁰⁰⁰⁰C₁(0.1)¹ + other positive terms
= 1 + 10000 × 0.1 + other positive terms
= 1001 + other positive terms > 1000
Hence, (1.1)¹⁰⁰⁰⁰ > 1000

Question 6.
Find (a + b)⁴ – (a – b)⁴, hence evaluate (√3 + √2)⁴ – (√3 – √2)⁴
Solution:
Using Binomial theorem, the expression, (a + b)4 – (a – b)4, can be expanded as (a + b)⁴ – (a – b)⁴
= ⁴C₀a⁴b° + ⁴C₁b¹ + ⁴C₂a²b² + ⁴C₃b³ + ⁴C₄a°b⁴ – [⁴C₀a⁴b° – ⁴C₁a³b¹ + ⁴C₂a²b² – ⁴C₃a¹b³ + ⁴C₄a°b⁴]
= 2[⁴C₁a³b¹ + ⁴C₃a¹b³]
= 2[4a³b + 4ab³]
= 8ab(a² + b²)

Now putting a = √3 and b = √2 , we have
(√3 + √2)⁴ – (√3 – √2)⁴ = 8(√3)(√2) [(√3)² +(√2)²]
= 8√6(3 + 2)
= 40√6

Question 7.
Find (x + 1)⁶ + (x – 1)⁶. Hence or otherwise evaluate (√2 + 1) + (√2 – 1)
Solution:
Using Binomial theorem, the expression, (x + 1)⁶ + (x – 1)⁶, can be expanded as (x + 1)⁶ + (x – 1)⁶
= ⁶C₀x⁶1⁰ + ⁶C₁⁵1¹ + ⁶C₂x⁴1² + ⁶C₃x³1³ + ⁶C₄x²1⁴ + ⁶C₅x¹1⁵ + ⁶C₆x°1⁶ + ⁶C₀x⁶1°- ⁶C₁x⁵1¹ + ⁶C₂x⁴1² – ⁶C₃x³1³ + ⁶C₄x²1⁴ – ⁶C₅x¹1⁵ + ⁶C₆x°1⁶
= 2[⁶C₀x⁶ + ⁶C₂x⁴ + ⁶C₄x² + ⁶C₆x⁰]
= 2(x⁶ + 15x⁴ + 15x² + 1)
Now putting x = √2 , we have (√2)⁶ + (√2)⁶
= 2[(√2)⁶ + 15(√2)⁴ + 15(√2)² + 1]
= 2(8 + 60 + 30 + 1)
= 198

Question 8.
Show that 9n +1 – 8n – 9 is divisible by 64, whenever n is a positive integer.
Solution:
Using Binomial theorem,
(1 + x)ⁿ⁺¹ = ⁿ⁺¹C₀ x° + ⁿ⁺¹C₁ x¹ + ⁿ⁺¹C₂ x² + …………. + ⁿ⁺¹C_n+1 xⁿ⁺¹
Now putting x = 8, we have
(1+8)ⁿ⁺¹ = ⁿ⁺¹C₀ 8° + ⁿ⁺¹C₁ 8¹ + ⁿ⁺¹C₂ 8² + ⁿ⁺¹C_n+1 8ⁿ⁺¹
⇒ 9ⁿ⁺¹ = 1 + (n + 1)(8) + 8²[ⁿ⁺¹C₂ + …….. + ⁿ⁺¹C_n+1 8^n-1]
⇒ 9ⁿ⁺¹ – 1 – 8(n+1) = 64[ⁿ⁺¹C₂ + + ⁿ⁺¹C_n+1 8^n-1]
⇒ 9ⁿ⁺¹ – 8n – 9 = 64 [ⁿ⁺¹C₂ + …. + ⁿ⁺¹C_n+1 8^n-1]
Now, RHS is divisible by 64 for all positive integers n. So, LHS is also divisible by 64. Hence, 9ⁿ⁺¹ – 8n – 9 is divisible by 64, whenever n is a positive interger.

Question 9.
Prove that \(\sum_{r=0}^n\)3^rnC_r = 4ⁿ.
Solution:
LHS = \(\sum_{r=0}^n\)3^rnC_r = 3⁰ⁿC₀ + 3¹ⁿC₁ + 3²ⁿC₂ + …………… 3^{n-1 n}C_n-1 + ⁿC_n+1. 3ⁿ

= ⁿC₀ 3° + ⁿC₁ 3¹ + ⁿC₂ 3² + …. + 3^n-1 + ⁿC_n.3ⁿ
= (1 + 3)ⁿ = 4ⁿ = RHS

Question 10.
Using the binomial theorem, prove that 50” – 49n – 1 is divisible by 492 for all positive integers n.
Solution:
Given 50ⁿ – 49n – 1 = (49 + 1)ⁿ – 49n – 1
= ⁿC₀(49)ⁿ + ⁿC₁(49)^n-1 + ⁿC₂(49)^n-2 + … + ⁿC_n 2(49)² + ⁿC_n-1(49) + ⁿC_n(49)° – 49n – 1
= (49)ⁿ + ⁿC₁(49)^n-1 + ⁿC₂(49)^n-2 + … + ⁿC_n-2(49)² + (n) (49) + 1 – 49n – 1
= (49)² [(49)^n-2 + ⁿC₂ (49)^n-3 + ⁿC₂(49)^n-4 + … + ⁿC_n-2]
= (49)² [a positive integer]
∴ 50ⁿ – 49n – 1 is divisible by 492 for all positive integers of n.

Practice AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Sequences and Series MCQ

Question 1.
Which of the following is finite sequence ?
1) 48, 24, 12, ………..
2) 1, 2, 3, …….
3) 2, 4, 6, 8, 10 ………..
4) 2, 3, 5, 7, 11, 13, …….
Answer:
3) 2, 4, 6, 8, 10 ………..

Explanation:
2, 4, 6, 8, 10 is a finite sequence.

Question 2.
Which of the following relation gives Fibonacci sequence ?
1) a_n = a_n-1 + a_n-2, n > 2
2) a_n-1 = a_n + a_n-2, n > 2
3) a_n-2 = a_n + a_n-1, n> 2
4) a_n = a_n+1 + a_n-2, n > 2
Answer:
1) a_n = a_n-1 + a_n-2, n > 2

Explanation:
Relation of fibonacci sequence is a_n = a_n-1 + a_n-2, n > 2

Question 3.
The first term of Fibonacci sequence is
1) 0
2) 1
3) 2
4) 3
Answer:
2) 1

Explanation:
The first term of fibonacci sequence is “1”.

Question 4.
If a_n = 4n + 6, then 15^th term of the sequence is
1) 6
2) 10
3) 60
4) 66
Answer:
4) 66

Explanation:
a_n = 4n + 6
a₁₅ = 4(15) + 6
= 60 + 6 = 66

Question 5.
A series can also be denoted by the symbol …………
1) πa_n
2) Σa_n
3) Φa_n
4) θa_n
Answer:
2) Σa_n

Explanation:
Series denoted by Σa_n

Question 6.
The sum of first five terms of series 2 + 4 + 6 + ….. is
1) 14
2) 16
3) 20
4) 30
Answer:
4) 30

Explanation:
Given series 2 + + 6 + ………..
Sum of even number S_n = n(n + 1)
S₅ = 5(5 + 1) = 5.(6) = 30

Question 7.
\(\sum_{n=1}^4\) 2n + 3 = ……….
1) 5
2) 12
3) 21
4) 32
Answer:
4) 32

Explanation:
\(\sum_{n=1}^4\) 2n + 3, If n = 1, 2, 3, 4; Series is 5, 7, 9, 11 …… ; S₄ = 5 + 7 + 9 + 11 = 32

Question 8.
If a_n+1 = a_n. r then the sequence is called
1) Arithmetic progression
2) Geometric progression
3) Harmonic progression
4) Special progression
Answer:
2) Geometric progression

Explanation:
a_n+1 = a_n. r is G.P.

Question 9.
The nth term of G.P. is ( 3 )
1) a_n = a + (n – 1) d
2) a_n = a + nd
3) a_n = a . r^n-1
4) a_n = arⁿ
Answer:
3) a_n = a . r^n-1

Explanation:
n^th term of G.P. is a_n = a . r^n-1

Question 10.
If r = 1 in G.P. then the sum of first n terms is
1) na
2) a/n
3) (n – 1) a
4) (n + 1) a
Answer:
1) na

Explanation:
If r = 1, the sum of first n terms = na

Question 11.
If 4, 8, 16, 32 are in G.P. then the sum upto 5th term is
1) 16
2) 64
3) 128
4) 124
Answer:
4) 124

Explanation:
4, 8, 16, 32, are in G.P; a = 4, r = 2
S₅ = \(\frac{4\left(2^5-1\right)}{2-1}=\frac{4(32-1)}{1}\) = 4(31) = 124

Question 12.
The Sum of series 1 + \(\frac{1}{2}+\frac{1}{4}\) + …….. upto 6 terms is
1) \(\frac{63}{32}\)
2) \(\frac{32}{63}\)
3) \(\frac{26}{53}\)
4) \(\frac{53}{26}\)
Answer:
1) \(\frac{63}{32}\)

Explanation:
Series 1 + \(\frac{1}{2}+\frac{1}{4}\) + …….. upto 6 terms
Sum of 6 terms = \(\frac{\left(1-\left(\frac{1}{2}\right)^6\right)}{1-\frac{1}{2}}\)
= 2(1 – \(\frac{1}{64}\))
= 2 – \(\frac{1}{32}=\frac{64-1}{32}=\frac{63}{32}\)

Question 13.
The geometric mean of 3 and 12 is ………….
1) 4
2) 6
3) 9
4) 12
Answer:
2) 6

Explanation:
Geometric mean of 3 and 12 = \(\sqrt{3 \times 12}=\sqrt{36}\)
= 6

Question 14.
If A.M of two numbers is \(\frac{15}{2}\) and their G.M is 6 then the two numbers are
1) 6 and 8
2) 12 and 3
3) 24 and 6
4) 27 and 3
Answer:
2) 12 and 3

Explanation:
Let the two numbers be and b
A.M = \(\frac{15}{2} \Rightarrow \frac{a+b}{2}=\frac{15}{2}\)
⇒ a + b = 15 ……….(1)

G. M = 6 ⇒ \(\sqrt{\mathrm{ab}}\)
⇒ ab = 36

a – b = \(\sqrt{(a+b)^2-4 a b}=\sqrt{225-144}=\sqrt{81}\)
a – b = 9 …(2)
(1) + (2) ⇒ a = 12, b = 3

Question 15.
If A means arithmetic mean and G means geometric mean then which of the following is true ?
1) A > G
2) A ≥ G
3) G > A
4) G ≥ A
Answer:
2) A ≥ G

Explanation:
A ≥ G

Question 16.
The ratio of A.M. and G.M of two positive numbers a and b is 5 : 3. Then the ratio of a to b is ( 1 )
1) 9 : 1
2) 3 : 5
3) 1 : 9
4) 3 : 1
Answer:
\(\frac{A \cdot M}{G \cdot M}=\frac{5}{3} \Rightarrow \frac{a+b}{2 \sqrt{a b}}=\frac{5}{3}\)
⇒ 9(a + b)² = 25 (4ab)
⇒ 9a² + 18ab + 9b² = 100 ab
⇒ 9a² – 82ab + 9b² = 0
⇒ (9a – b) (a – 9b) = 0
9a-b = 0
9a = b
\(\frac{a}{b}=\frac{1}{9}\)

a – 9b = 0
a = 9b
\(\frac{a}{b}=\frac{9}{1}\)
∴ a : b = 9 : 1 (∵ a > b)

Referring to the AP Inter 1st Year Maths Study Material Chapter 7 Binomial Theorem Exercise 7b Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Binomial Theorem Solutions Exercise 7b

I.

Question 1.
If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer. [Hint : Write aⁿ = (a – b + b)ⁿ and expand]
Solution:
Here, aⁿ = (a – b + b)ⁿ = [(a – b) + b)ⁿ
By using Binomial theorem, the expression [(a – b) + b]ⁿ can be expanded as
ⁿC₀(a- b)ⁿ b⁰ + ⁿC₁(a- b)^n-1 b¹ + ⁿC₂(a- b)^n-2 b² + …. + ⁿC_n-1(a – b)b^n-1 + ⁿC_n(a- b)⁰bⁿ
= (a – b)ⁿ + ⁿC₁(a – b)^n-1 b + ⁿC₂(a – b)^n-2 bⁿ + ………….. + ⁿC_n-1(a – b)b^n-1] + bⁿ
⇒ aⁿ = (a – b)[(a – b)^n-1 + ⁿC₁(a – b)^n-2 b + ⁿC₂(a – b)^n-3 b² + ⁿC_n-1b^n-1] + bⁿ
⇒ aⁿ – bⁿ = (a – b)[(a – b)^n-1 + ⁿC₁(a – b)^n-2 b + ⁿC₂(a – b)^n-3 b² + ……. + ⁿC_n-1b^n-1]
now, a – b is a factor of R.H.S, so, a – b is also a factor of L.H.S hence, a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.

Question 2.
Evaluate (√3 + √2)⁶ – (√3 – √2)⁶
Solution:
(√3 + √2)⁶ – (√3 – √2)⁶
= ⁶C₀(√3)⁶ + ⁶C₁(√3)⁵(√2) + ⁶C₂(√3)⁴(√2)² + ⁶C₃(√3)³(√2)³ + ⁶C₄(√3)²(√2)⁴
+ ⁶C₅(√3)¹(√2)⁵ + ⁶C₆(√2)⁶ – [⁶C₀(√3)6 – ⁶C₁(√3 )⁵(√2) + ⁶C₂(√3)⁴(√2)² – ⁶C₃(√3)³(√2)³ + ⁶C₄(√3)²(√2)⁴ – ⁶C₅(√3)¹(√2 )⁵ + ⁶C₆(√2)⁶]
= 2[⁶C₁(√3)⁵(√2) + ⁶C₃(√3)³(√2)³ + ⁶C₂(√3)¹(√2)⁵]
= 2[6 × 9√6 + 20 × 6√6 + 6 × 4√5]
= 2[54 + 120 + 24]√6
= 396√6

Question 3.
Find the value of (a² + \(\sqrt{a^2-1}\))⁴ + (a² – \(\sqrt{a^2-1}\))⁴
Solution:

= 2(a⁸ + 6a⁴(a² – 1) + (a² – 1)²]
= 2[a⁸ + 6a⁶ – 6a⁴ + a⁴ + 1 – 2a²]
= 2[a⁸ + 6a⁶ – 5a⁴ – 2a² + 1]
= 2a⁸ + 12a⁶ – 10a⁴ – 4a² + 2

II.

Question 1.
Find an approximation of (0.99)⁵ using the first three terms of its expansion.
Solution:
0.99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then Binomial theorem can be applied. It can be written that, 0.99 = 1 – 0.01
Therefore, (0.99)⁵ = (1 – 0.01)⁵
= ⁵C₀(-0.01)° + ⁵C₁(-0.01)¹ + ⁵C₂ (-0.01)² + ………………..
= 1 – 5(0.01) + 10(0.0001) + …………
= 1 – 0.05 + 0.001
= 0.951

Question 2.
Expand (1 + \(\frac{x}{2}-\frac{2}{x}\))⁴ using Binomial theorem,
Solution:
Here (1 + \(\frac{x}{2}-\frac{2}{x}\))⁴ = [(1 + \(\frac{x}{2}\)) – \(\frac{2}{x}\)]⁴
Using Binomial theorem, the expression (1 + \(\frac{x}{2}-\frac{2}{x}\))⁴ can be expressed as

Again applying Binomial theorem, we have

Now putting the value of (1 + \(\frac{x}{2}\))⁴ and (1 + \(\frac{x}{3}\))³ in (1) we have

Question 3.
Find the expansion of (3x² – 2ax + 3a²)³ using binomial theorem.
Solution:
Here, (3x² – 2ax + 3a²)³ = ((3x² – 2ax) + 3a²)³
Using Binomial theorem, the expression [(3x² – 2ax) + 3a²)³ can be expressed as
³C0(3x² – 2ax)³ + ³Cj(3x² – 2ax)² (3a²) + ³C2(3x² – 2ax)¹ (3a²)² + ³C3(3a²)³ = (3x² – 2ax)³ + 3(9x⁴ + 4a²x² – 12ax³)(3a²) + 3(3x² – 2ax) (9a⁴) + (27a⁶)
= (3x² – 2ax)³ + 81a²x⁴ + 36a⁴x² – 108a³x³ + 81a⁴x² – 54a⁵x + 27a⁶

Again applying Binomial theorem, we have
(3x² – 2ax)³ = ³C₀(3x²)³ + ³C₁(3x²)² (-2ax)¹ + ³C₂(3x²) (-2ax)² + ³C₃(-2ax)³
= 27x⁶ – 3 × 18ax⁵ + 3 × 12a²x⁴ – 8a³x³
= 27x⁶ – 54ax⁵ + 36a²x⁴ – 8a³x³

Now putting the value of (3x² – 2ax)³, we have
= 27x⁶ – 54ax⁵ + 36a²x⁴ – 8a³x³ + 81a²x⁴ + 36a⁴x² – 108a³x³ + 81a⁴x² – 54a⁵x + 27a⁶
= 27x⁶ – 54ax⁵ + 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶

Question 4.
Prove that C₀ + 3.C₁ + 5.C₂ + … + (2n + 1). C_n = (2n + 2). 2^n-1
Solution:
Let S = C₀ + 3.C₁ + 5.C₂ + … + (2n + 1).C_n ……….. (1)
By writing the terms in (1) in the reverse order we get
S = (2n + 1). C_n + (2n – 1) . C_n-1 + (2n – 3).C_n-2 + …….. + 3.C₁ + 1.C₀
= (2n + 1).C₀ + (2n – 1).C₁ + (2n – 3).C₂ + ……… + 3.C_n-1 + 1.C_n …………. (2)
On adding (1) and (2), we get
2S = (2n + 2).C₀ + (2n + 2).C₁ + … + (2n + 2).C_n-1 + (2n + 2) . C_n
= (2n + 2) (C₀ + C₁ + C₂ + … + C_n) = (2n + 2).2n Therefore, S = (2n + 2) 2^n-1

Question 5.
Prove that, for any real numbers a, d,
a.C₀ + (a + d) . C₂ + (a + 2d) . C₂ + … + (a + nd) . C_n = (2a + nd) . 2^n-1
Solution:
Let S = a.C₀ + (a + d) . C₂ + (a + 2d) . C₂ + … + (a + nd) . C_n ….. (1)
∴ S = (a + nd) . C_n + [a + (n – 1) d] . C_n-1 + … aC₀
⇒ S = (a + nd) . C₀ + [a + (n – 1)d]. + [a + (n – 2)d].C₂ + … + a . C_n ……….. (2)
(1) + (2) => 2S = (2a + nd) C₀ + (2a + nd) C₁ + (2a + nd) C₂ + … + (2a + nd) C_n
= (2a + nd) (C₀ + C₁ + C₂ + … + C_n)
= (2a + nd) 2n ⇒ S = (2a + nd) 2_n – 1.

Question 6.
If n is a positive integer, then prove that
C₀ + \(\frac{C_1}{2}+\frac{C_2}{3}\) + …… + \(\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\)
Solution:
Let S = C₀ + \(\frac{C_1}{2}+\frac{C_2}{3}\) + …… + \(\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\)
Now multiplying on both sides with (n + 1), we get
(n + 1) S = (n + 1)ⁿC₀ + \(\frac{1}{2}\) (n + 1)ⁿC₁ + —(n + 1)ⁿC₂ + … + ⁿC_n
= (n +1) + \(\frac{(\mathrm{n}+1) \mathrm{n}}{1 \cdot 2}+\frac{(\mathrm{n}+1) \mathrm{n}(\mathrm{n}-1)}{1 \cdot 2 \cdot 3}\) + … + 1
= ⁽ⁿ⁺¹⁾C₁ + ⁽ⁿ⁺¹⁾C₂ + ⁽ⁿ⁺¹⁾C₃ + ……… + ⁽ⁿ⁺¹⁾C_n+1
= ⁽ⁿ⁺¹⁾C₁ + ⁽ⁿ⁺¹⁾C₁ + ⁽ⁿ⁺¹⁾C₂ + ⁽ⁿ⁺²⁾C₃ ……… + ⁽ⁿ⁺¹⁾C_n+1 – ⁽ⁿ⁺¹⁾C₀
= 2ⁿ⁺¹ -1
⇒ S = \(\frac{2^{n+1}-1}{n+1}\)
C₀ + \(\frac{C_1}{2}+\frac{C_2}{3}\) + …… + \(\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\)

Question 7.
If n is a positive integer and x is any non-zero real number, then prove that
C₀ + C₁.\(\frac{x}{2}\) + C₂.\(\frac{x^2}{3}\) + C₃.\(\frac{x^3}{4}\) + ……….. + C_n.\(\frac{x^n}{n+1}=\frac{(1+x)^{n+1}-1}{(n+1) x}\)
Solution:
Let S ≡ C₀ + C₁.\(\frac{x}{2}\) + C₂.\(\frac{x^2}{3}\) + C₃.\(\frac{x^3}{4}\) + ……….. + C_n.\(\frac{x^n}{n+1}\)

Multiplying both sides with be ‘x’

Question 8.
Prove that 2.C₀ + 5.C₁ + 8.C₂ + ……….. + (3n + 2) C_n = (3n + 4)2^n-1
Solution:
Let S = 2.C₀ + 5.C₁ + 8.C₂ + ……….. + (3n – 1) C_n-1 + (3n + 2) C_n ……(1)
S = (3n + 2) C₀ + (3n – 1)C₁ + (3n – 4) C₂ + + 5C_n-1 + 2.C_n … (2)
(∵ C_n = C₀.C_n-1 = Ci…)
Adding (1) 86 (2) we get 2S = (3n + 4)C0 + (3n + 4)Cj + (3n + 4)C2 + … + (3n + 4) Cn = (3n + 4) (C0 + Cj + C2 + … + Cn) = (3n + 4)2n
S = (3n + 4) . 2^n-1

Question 9.
Prove tnat C₀ +2C₁ + 4.C₂ + 8.C₃ + ……….. + 2ⁿ.C_n = 3ⁿ
Solution:
We know that,
(1 + x)ⁿ = C₀ + C₁x + C₂x² + C₃x² + ………… + C_nxⁿ
Put x = 2 in equation (1) we get,
(1 + 2)ⁿ = C₀ + C₁(2) + C₂(2)² + C₃(2)³ + ………. + C_n(2)ⁿ
∴ C₀ +2C₁ + 4.C₂ + 8.C₃ + ……….. + 2ⁿ.C_n = 3ⁿ

Question 10.
If (1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ………… + a_2nx²ⁿ, then prove that
i) a₀ + a₁ + a₂ + ………… + a_2n = 3ⁿ
Solution:
(1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ………… + a_2nx²ⁿ
Put x = 1 ⇒ a₀ + a₁ + a₂ + ………… + a_2n = (1 + 1 + 1)ⁿ = 3ⁿ …….(1)
Put x = -1 ⇒ a₀ – a₁ + a₂ + ………. + a_2n = (1 – 1 + 1)ⁿ = 1 ………. (2)

a₀ + a₁ + a₂ + ………… + a_2n = 3ⁿ

ii) a₀ + a₂ + a₄ + ………… + a_2n = \(\frac{3^n+1}{2}\)
Solution:
(1) + (2) ⇒ 2(a₀ + a₂ + a₄ + ………… + a_2n) = 3ⁿ + 1
⇒ ∴ a₀ + a₂ + a₄ + ………… + a_2n = \(\frac{3^n+1}{2}\)

iii) a₁ + a₃ + a₅ + ….. + a_2n-1 = \(\frac{3^n-1}{2}\)
Solution:
(1) – (2) ⇒ 2(a₁ + a₃ + a₅ + ….. + a_2n-1) = 3ⁿ – 1
⇒ ∴ a₁ + a₃ + a₅ + ….. + a_2n-1 = \(\frac{3^n-1}{2}\)

iv) a₀ + a₃ + a₆ + a₉ + ………… + a_2n = 3^n-1
Solution:
Put x = 1
⇒ a₀ + a₃ + a₆ + a₉ + ………… + a_2n = 3ⁿ …(a)
Put x = ω
a₀ + a₁ω + a₂ω² + a₃ω³ + ………… + a_2nω²ⁿ = 0
Hint: 1 + ω + ω² = 0; ω³ = 1 ……………. (b)
Put x = ω²
⇒ a₀ + a₁ω² + a₂ω⁴ + a₃ω⁶ + ………… + a_2nω⁴ⁿ = 0 …..(c)
Adding (a), (b), (c)
∴ 3₀ + a₁(1 + ω + ω²) + a₂(1 + ω³ + ω⁶) + ……… + a_2n(1 + ω²ⁿ + ω⁴ⁿ) = 3ⁿ
⇒ 3a₀ + a₁(0) + a₂(0) + 3a₃ + ……. = 3ⁿ
∴ a₀ + a₃ + a₆ + ……… = \(\frac{3^{\mathrm{n}}}{3}\) = 3^n-1

Question 11.
If P and Q Me the sum of odd terms and the sum of even terms respectively in the expansion of (x + a)ⁿ then prove that
i) P² – Q² = (x² – a²)ⁿ
ii) 4PQ = (x + a)²ⁿ – (x – a)²ⁿ
Solution:
We know that (x + a)ⁿ = ⁿC₀xⁿ + ⁿC₁x^n-1a + ⁿC₂ x^n-2a² + ………………
(x + a)ⁿ = T₁ + T₂ + T₃ + T₄ + T₅ + T₆ + …………….
Given that P = (T₁ + T₃ + T₅ + T₇ + …) and Q = (T₂ + T₄ + T₆ + ………….)
⇒ (x + a)ⁿ = P + Q …………(1) and (x – a)ⁿ = P – Q ………….(2)

i) P² – Q² = (P + Q) (P – Q) = (x + a)ⁿ (x – a)ⁿ = [(x + a) (x – a)]ⁿ = (x² – a²)ⁿ

ii) 4PQ = (P + Q)² – (P – Q)²
= [(x + a)ⁿ]² – [(x – a)ⁿ]²
= (x + a)²ⁿ – (x – a)²ⁿ

Question 12.
If (1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + … + a₂₀x²⁰, then prove that
i) a₀ + a₁ + a₂ + … + a₂₀ = 2¹⁰
ii) a₀ – a₁ + a₂ – a₃ + … + a₂₀ = 4¹⁰
Solution:
(1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + … + a₂₀x²⁰ ………….. (1)

i) Put x = 1 in eq. (1)
(1 + 3 – 2)¹⁰ = a₀ + a₁ + a₂ + … + a₂₀
∴ a₀ + a₁ + a₂ + … + a₂₀ = 2¹⁰

ii) Put x = -1 in eq (1)
(1 – 3 – 2)¹⁰ = a₀ – a₁ + a₂ – a₃ + … + a₂₀
∴ a₀ – a₁ + a₂ – a₃ + … + a₂₀ = (-4)¹⁰ = 4¹⁰

Question 13.
Show that the middle term in the expansion of (1 + x)²ⁿ is \(\frac{1 \cdot 3 \cdot 5 \ldots(2 n-1)}{n!}\) 2ⁿ . xⁿ, where n is a positive integer.
Solution:
As 2n is even, the middle term of the expansion (1 + x)²ⁿ is i.e., (n + 1)^th term
Now T_n+1 = ²ⁿC_n (1)^2n-n (x)ⁿ = ²ⁿC_n xⁿ = \(\frac{(2 \mathrm{n})!}{\mathrm{n}!\mathrm{n}!}\)xⁿ

Question 14.
The second, third and fourth terms in the binomial expansion (x + a)ⁿ are 240, 720, 1080 respectively. Find the value of x, a and n.
Solution:
Given expression is (a + x)ⁿ. General term T_r+1 = ⁿC_r a^n-r x^r
T₂ = T₁₊₁ = ⁿC₁ a^n-1 x¹ = 240 … (1)
T₃ = T₂₊₁ = ⁿC₂ a^n-2 x² = 720 … (2)
T₄ = T₃₊₁ = ⁿC₃ a^n-3 x³ = 1080 … (3)

⇒ 4n – 8 = 3n – 3
⇒ 4n – 3n = -3 + 8
∴ n = 5

Substitute n = 5 in (4) ⇒ 4x = 6a ⇒ x = 3a/2
Substitute x, n values in (1)
⇒ 5 a⁴\(\left(\frac{3 \mathrm{a}}{2}\right)\) = 240
⇒ a⁵ = \(\frac{240 \times 2}{15}\) = 32
⇒ a⁵ = 2⁵
⇒ a = 2

Substitute the values of n = 5 and a = 2 in (6)
⇒ x = \(\frac{3(2)}{2}\) = 3
∴ a = 2, x = 3, n = 5

Regular practice with AP Inter 1st Year Physics Study Material Chapter 2 Motion in a Straight Line Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Physics 2nd Lesson Motion in a Straight Line Questions and Answers

I. Multiple Choice Questions

Question 1.
The velocity of a body depends on time according to the equation V = 10 + 2t². The body undergoing
(1) uniform acceleration
(2) uniform retardation
(3) non-uniform acceleration
(4) zero acceleration
Answer:
(3) non-uniform acceleration

Question 2.
Which of the following is true?
(1) Speed is always equal to magnitude of velocity.
(2) Speed is always greater than the magnitude of velocity.
(3) Velocity is always greater than the speed.
(4) Speed in general is either equal to or greater than the magnitude of velocity.
Answer:
(1) Speed is always equal to magnitude of velocity.

Question 3.
A bullet is dropped from the same time another bullet is fired horizontally. They will hit the ground
(1) one after the other
(2) simultaneously
(3) depends on the observer
(4) depends on the velocity of the fired bullet
Answer:
(2) simultaneously

Question 4.
If a car is moving with a uniform velocity of 60 km/h. Then its acceleration is
(1) 0km / h²
(2) 10km / h²
(3) 20km / h²
(4) 30km / h²
Answer:
(1) 0km / h²

Question 5.
A ball is thrown vertically upward with an initial velocity of 50 m/s. Its maximum height is (take g = 10/s²)
(1) 125m
(2) 250 m
(c) 500 m
(d) 1000 m
Answer:
(1) 125m

Question 6.
A particle moves in a straight line with an initial velocity of 10m/s and a constant acceleration of 2m/s². Its displacement after 5 seconds will be
(1) 50 m
(2) 75 m
(3) 150 m
(4) 200 m
Answer:
A = ut + \(\frac{1}{2}\)at² (or) x = v₀t + \(\frac{1}{2}[latex]at², ∴ x = (10 × 5) + [latex]\frac{1}{2}\)(2) × 5² = 50 + 25 = 75 m

Question 7.
A body starts from rest and moves with a uniform acceleration of 2m / s². Its velocity after 10 seconds will be
(1) 5 m/s
(2) 10 m/s.
(3) 15 m/s
(4) 20 m/s
Answer:
(4) 20 m/s
V = u + at = 0 + (2 × 10) = 20m/s

Question 8.
For the motion with uniform velocity, the slope of the velocity-time graph is equal to
(1) 1 m/s
(2) zero
(3) initial velocity
(4) final velocity
Answer:
(2) zero

Question 9.
If a cyclist takes two minutes to complete half revolution on a circular path of 120 m radius, what is the average speed?
(1) 1 m/s
(2) 2 m/s
(3) 3.14 m/s
(4) 4.14 m/s
Answer:
(3) 3.14 m/s
Average Speed = \(\frac{\text { Total distance }}{\text { Total Time }}=\frac{\pi \mathrm{r}}{2 \times 60}=\frac{3.14 \times 120}{2 \times 60}\) = 3.14 m/s

Question 10.
In a particular time, the net displacement travelled by the body is equal to
(1) the area which v -1 graph encloses with displacement axis
(2) the area which x -1 graph encloses with the time axis
(3) the area which v -1 graph encloses with time axis
(4) the area which a -1 graph encloses with displacement axis
Answer:
(3) the area which v -1 graph encloses with time axis

Question 11.
A particle moves along a straight line. At a time t, the distance x of the particle from the origin given by x = 20 + 12t – t³ where x is in meters and t in sec, the time taken for the particle fo coming to rest is
(1) 1 s
(2) 2 s
(3) 4 s
(4) 6 s
Answer:
(2) 2 s
x = 20 + 12t – t³, \(\frac{d x}{d t}\) = 12 – 3t² = V, 12 – 3t² = 0 ⇒ 3t² = 12 ⇒ t² = 4, t = 2s

Question 12.
A car travels half of the distance with speed v and the remaining distance with speed 2v. Its average speed is
(1) \(\frac{4 v}{3}\)
(2) \(\frac{3 v}{3}\)
(3) \(\frac{v}{3}\)
(4) \(\frac{2 v}{3}\)
Answer:
(1) \(\frac{4 v}{3}\)
Average Speed = \(\frac{2 u v}{u+v}=\frac{2 \times(v)(2 v)}{v+2 v}=\frac{2 \times v \times 2}{3}=\frac{4}{3} v\)

Question 13.
A car runs at constant speed on a circular track of radius 100 m, taking 62.8 seconds for every circular lap. The average velocity and average speed of each circular lap respectively is
(1) 10 m/s, 0 m/s
(2) 0m/s. 0 m/s
(3) 0 m/s, 10 m/s
(4) 10 m/s, 10 m/s
Answer:
(3) 0 m/s, 10 m/s
Hint: Average Velocity = 0
Average Speed = \(\frac{2 \pi \mathrm{r}}{62.8}=\frac{2 \times 3.14 \times 100}{62.8}=\frac{6.28 \times 100}{62.8}\) = 10m/s

Question 14.
The motion of the particle is given by the equation S = (2t³ + 5t² + 16t + 8)m. The value of acceleration of the particle at t = 1 sec is
(1) 20 m/s²
(2) 22 m/s²
(3) 24 m/s²
(4) 26 m/s²
Answer:
(2) 22 m/s²
S = 2t³ + 5t² + 16t + 8, v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 6t² + 10t + 16; a = \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 12t + 10
at t = 1, a = 12 + 10 = 22m/s²

Question 15.
The ratio of distances travelled by a freely falling body in the 1^st, 2^nd, 3^rd, 4^th second
(1) 1:2:3:4
(2) 1:4:9:16
(3) 1:3:5:7
(4) 1:1:1:1
Answer:
(3) 1:3:5:7
S_n = u + \(\frac{\mathrm{a}}{2}\)(2n – 1), S₁ = \(\frac{\mathrm{g}}{2}\)(2(1) – 1) = \(\)
S₂ = \(\frac{\mathrm{g}}{2}\)(2(2)- 1) = \(\frac{\mathrm{3g}}{2}\), S₃ = \(\frac{\mathrm{g}}{2}\)(2(3) – 1) = \(\frac{\mathrm{5g}}{2}\) ⇒ S₁ : S₂: S₃ : S₄ = 1 : 3 : 5 : 7

Question 16.
A stone falls freely under gravity. It covers distances h₁, h₂ and h₃ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h₁, h₂ and h₃ is
(1) h₂ = 3h₁ and h₃ = 3h₂
(2) h₁ = h₂ = h₃
(3) h₁ = 2h₂ = 3 h₃
(4) h₁ = \(\frac{\mathrm{h}_2}{3}=\frac{\mathrm{h}_3}{5}\)
Answer:
(4) h₁ = \(\frac{\mathrm{h}_2}{3}=\frac{\mathrm{h}_3}{5}\)
h₁ = \(\frac{1}{2}\)g(5²) = \(\frac{1}{2}\)g(25), x₁ = \(\frac{1}{2}\)g(10²) = \(\frac{1}{2}\)g(100)
h₂ = x₁ – h₁ = \(\frac{1}{2}\)g (100 – 25) = \(\frac{1}{2}\)g(75), x₂ = \(\frac{1}{2}\)g (15²) = \(\frac{1}{2}\)g(225)
h₃ = x₂ – x₁ = \(\frac{1}{2}\)g(225 – 100) = \(\frac{1}{2}\)g(125), h, = \(\frac{\mathrm{h}_2}{3}=\frac{\mathrm{h}_3}{5}\) (or) 1:3:5:7

II. Fill in the Blanks

Question 1.
________ deals with the study of motion of objects without taking into account, factors which cause the motion.
Answer:
Kinematics

Question 2.
The shortest path travelled by a moving object in a given time from initial to final position is called _______.
Answer:
displacement

Question 3.
The distance covered by a body along a semicircular path of radius r is _______ .
Answer:
πr

Question 4.
The slope of velocity-time graph gives _______.
Answer:
acceleration

Question 5.
The slope of displacement-time graph gives _______.
Answer:
velocity

Question 6.
The area under acceleration-time graph gives _____.
Answer:
change in velocity

Question 7.
The acceleration of a body moving with uniform velocity is ________.
Answer:
zero

Question 8.
The velocity of a body at any instant of time is called _______.
Answer:
instantaneous velocity

Question 9.
The value of an acceleration of a freely falling body is ________.
Answer:
g = 9.8 m/s²]

Question 10.
The ratio of distances travelled by a freely falling body from the rest in the first three seconds is ______.
Answer:
1 : 3 : 5

III. One Word Answer Questions

Question 1.
What is uniform motion ?
Answer:
If an object moving along a straight line covers equal displacement in equal intervals of time, said to be in uniform motion along a straight line.

Question 2.
What is acceleration ?
Answer:
The rate of change of velocity is called acceleration.

Question 3.
What is average speed ?
Answer:
The ratio of total distance travelled by a body to total time is called average speed.

Question 4.
What is instantaneous velocity ?
Answer:
The velocity of a particle at a particular instant is called instantaneous velocity.

Question 5.
Draw position – time graph for an object at rest.
Answer:

Question 6.
What represents the area under the cure of velocity – time graph ?
Answer:
Area = v × t = displacement.

Question 7.
What is the maximum height reached by a body when it is vertically projected up with an initial velocity ‘u’ ?
Answer:
H_max = \(\frac{u^2}{2 g}\)

Question 8.
Mention one similarity between speed and velocity.
Answer:
Having same units and dimensions.

Question 9.
Draw velocity-time graph for a particle moving with uniform velocity.
Answer:

Question 10.
What is the time taken by the body, when it is freely dropped from a height “h” ?
Answer:
h = \(\frac{1}{2}\)gt²; t² = \(\frac{2 h}{g} \Rightarrow t=\sqrt{\frac{2 h}{g}}\)

IV. Very Short Answer Questions

Question 1.
The states of motion and rest are relative. Explain. [dlarch – 2026]
Answer:
REST: If the position of a body does not change with respect to surroundings, it is said to be at “Rest”.

MOTION : If the position of a body changes with respect to surroundings, it is said to be in “Motion”. By definitions, rest and motion are relative with respect to surroundings.

Question 2.
Is it necessary that the direction of acceleration and velocity of a body should be the same? Give an example.
Answer:
No. For a decelerating body velocity is in forward direction and acceleration will be in opposite direction.

Question 3.
Is it possible for a body to have acceleration but zero velocity? Give an example.
Answer:
In case of vertically projected body at maximum height its velocity v = 0. But acceleration due to gravity ‘g’ is not zero.
So even though velocity v = 0 ⇒ acceleration is not zero.

Question 4.
Which is greater, a velocity of 72kmph or 25m/s ?
Answer:
V₁ = 72 × \(\frac{5}{18}\) = 20m/s; V₂ = 25m/s ∴ 25m/s is greater than 72kmph.

Question 5.
Give an example of one-dimensional motion where a particle moving along a straight line comes to rest and moves backward.
Answer:

1. A body projected vertically upwards.
2. When the length of pendulum is high and amplitude is less then its motion is along a straight line. The pendulum will come to a stop at extreme position and moves back in forward direction (‘x’ + ve) periodically.

Question 6.
A vehicle travels half the distance L with speed Vj and the other half with speed v₂. What is the average speed ?
Answer:
The average speed of a vehicle for the two equal parts \(\frac{\mathrm{L}}{2}, \frac{\mathrm{~L}}{2}\) is
V_avg = \(\frac{\text { Total distance }}{\text { Total time }}=\frac{\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}}{\mathrm{t}_1+\mathrm{t}_2}\)
V_avg = \(\frac{\mathrm{L}}{\left(\frac{\frac{\mathrm{~L}}{2}}{\mathrm{~V}_1}+\frac{\frac{\mathrm{L}}{2}}{\mathrm{~V}_2}\right)}=\frac{\mathrm{L}}{\frac{\mathrm{~L}}{2}\left(\frac{1}{\mathrm{~V}_1}+\frac{1}{\mathrm{~V}_2}\right)}=\frac{1}{\frac{1}{2}\left(\frac{\mathrm{~V}_1+\mathrm{V}_2}{\mathrm{~V}_1 \mathrm{~V}_2}\right)}\) (∵ t = \(\frac{\mathrm{L} / 2}{\mathrm{~V}}\)) ⇒ V_avg = \(\frac{2 V_1 V_2}{V_1+V_2}\)

Question 7.
A body covers a distance of 5m along a semicircular path from its one end to another end. What is the ratio of its distance covered to its displacement ?
Answer:
πr = 5; r = 5/π

\(\frac{\text { Distance }}{\text { Displacement }}=\frac{\pi \mathrm{r}}{2 \mathrm{r}}=\frac{\pi}{2}=\frac{22}{7 \times 2}=\frac{11}{7}\)

Question 8.
Displacement travelled by a body is given by S = 2t + 5t² m. What is the acceleration of that body ?
Answer:
S = 2t + 5t²; V = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 2 + 10t; a= \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 10m/s²

Question 9.
An object falling through a fluid is observed to have an acceleration given by a = g – bv where g is the gravitational acceleration and b is a constant. After a long time it is observed to fall with a constant velocity. What would be the value of this constant velocity ?
Answer:
Acceleration, a- = g – bv when moving with constant velocity, a = 0 ⇒ 0 = g – bv
∴ Constant velocity, v = \(\frac{\mathrm{g}}{\mathrm{~b}}\) m/sec.

Question 10.
Under what circumstances is the relationship s = vt valid ?
Answer:
Body moving with uniform velocity (a = 0).

V. Short Answer Questions

Question 1.
Can the equations of kinematics be used when the acceleration varies with time ? If not, what form would these equations take ?
Answer:
The equations of motion are 1) v =u + at 2) s = ut + 1/2at² and 3) v² – u² = 2as. All these three equations applicable to a body moves with uniform acceleration ‘a’.

No, the equations are not applicable when the acceleration varies with time.

Question 2.
A particle moves in a straight line with uniform acceleration. Its velocity at time t = 0 is V₁ and at time t = t is v₂. The average velocity of the particle in this time interval is (v₁ + v₂) / 2. Is this correct ? Substantiate your answer.
Answer:
t₁ = 0 ⇒ u = v₁ ; t₂ = t ⇒ v = v₂; Average velocity = \(\frac{u+v}{2}=\frac{v_1+v_2}{2}\)

Question 3.
Can the velocity of an object be in a direction other than the direction of acceleration of the object ? If so, give an example.
Answer:
Yes. Velocity of a body and its acceleration may be in different directions.
Ex :
1) In case of vertically projected body ⇒ velocity of body is in the upward direction and acceleration is in downward direction.
2) When brakes are applied the velocity of body before coming to rest is opposite to retarding acceleration.

Question 4.
Obtain the equation of motion x = ut + \(\frac{1}{2}\)at² for constant acceleration using graphical method.
Answer:
The velocity – time graph of a body moving with initial velocity ‘u’ and with uniform acceleration ‘a’ as shown. Let V be the velocity of the body after a time t.
In v – t graph, area of velocity-time graph = total displacement travelled by it. Area under velocity – time graph = area of OABCD
∴ Area of Rectangular part OACD = Area of OACD + Area of ABC.
A₁ = OA × OD = ut …..(1)
Area of triangle ABC = A₂
A₂ = \(\frac{1}{2}\) × Base × height = \(\frac{1}{2}\) × AC × (BC) = \(\frac{1}{2}\)t(u – v)
But v – u = at; ∴ A₂ = \(\frac{1}{2}\)t. at = \(\frac{1}{2}\)at²
∴ Total area under graph = s = A₁+ A₂; s(n) = ut + \(\frac{1}{2}\)at².
∴ s = ut + \(\frac{1}{2}\)at² is graphically proved.

Question 5.
Explain the terms the average velocity and instantaneous velocity. When are they equal?
Answer:
Average velocity : It is the ratio of total displacement to total time taken. It is independent of path of the body.
∴ Average velocity = \(\frac{\mathrm{s}_2-\mathrm{s}_1}{\mathrm{t}_2-\mathrm{t}_1}\) (Or) V_avg = \(\frac{\mathrm{ds}}{\mathrm{dt}}\)
Velocity of a particle at a particular instant of time is known as instantaneous velocity. Here time interval is very small. V = \(\frac{\mathrm{dx}}{\mathrm{dt}}\)
Only in uniform motion, instantaneous velocity = average velocity. For all other cases instantaneous velocity may differ from average velocity.

Problems

Question 1.
A man walks on a straight road from his home to a market 2.5 km away with a speed of km h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1. What is the (a) magnitude of average velocity mid (b) average speed of the man over the time interval 0 to 50 minutes ?
Solution:
Time taken by man to go from his home to market, t₁ = \(\frac{\text { distance }}{\text { speed }}=\frac{2.5}{5}=\frac{1}{2} \mathrm{~h}\)
Time taken by man to go from market to his home, t₁ = \(\frac{2.5}{7.5}=\frac{1}{3} h\)
∴ Total time taken = t₁ + t₂ = \(\frac{1}{2}+\frac{1}{3}=\frac{5}{6} h\) = 50min.
In time interval 0 to 50 min,
Total distance travelled = 2.5 + 2.5 = 5 km.; Total displacement = zero.

1. Average velocity = \(\frac{\text { displacement }}{\text { time }}\) = 0
2. Average speed = \(\frac{\text { distance }}{\text { time }}=\frac{5}{5 / 6}\) = 6 km/h

Question 2.
A car travels the first third of a distance with a speed of 10 kmph, the second third at 20 kmph and the last third at 60 kmph. What is its mean speed over the entire distance ?
Solution:
Total distance = s; distance travelled, S₁ = \(\frac{s}{3}\); velocity, v₁ = 10 kmph ; distance, s₂ = \(\frac{s}{3}\) velocity, v₂ = 20 kmph; distance, s₃ = \(\frac{s}{3}\); velocity, v₃ = 60 kmph
Average velocity = \(\frac{\text { total distance }}{\text { total time }}\)

Question 3.
A bullet moving with a speed of 150 m s^-1 strikes a tree and penetrates 3.5 cm before stopping. What is the magnitude of its retardation in the tree and the time taken for it to stop after striking the tree?
Solution:
Velocity of bullet, u = 150 m/s; Final velocity, v = 0 ; Distance travelled, s = 3.5 cm = 3.5 × 10^-2 m,
a) Acceleration, a = \(\frac{v^2-u^2}{2 s}=\frac{0^2-150^2}{2 \times 3.50 \times 10^{-2}}=\frac{-22500}{7 \times 10^{-2}}\)
= -3.214 × 10⁵ m/sec² (-ve sign for retardation)

b) Time taken to stop, t = \(\frac{v-u}{a}=\frac{-150}{-3.214 \times 10^5}\) = 4.67 × 10^-4 sec.

Question 4.
A motorist drives north for 30 min at 85 km/h and then stops for 15 min. He continues travelling north and covers 130 km in 2 hours. What is his total displacement and average velocity ?
Solution:
In first part:
Velocity, v₁ = 85 kmph ; Time, t₁ = 30 min ; Distance travelled, s₁ = v₁ t₁ = 85 × \(\frac{30}{60}\) = 42.5 km

In second part:
Distance travelled, s₂ = 0 ; Time, t₂ = 15.0 min.

In third part:
Distance travelled, s₃ = 130 km ; Time, t₃ = 2 hours = 120 min.
a) Total distance of the motorist, s = s₁ + s₂ + s₃ = 42.5 + 0 + 130 = 172.5 km

b) Total time travelled,
t = t₁ + t₂ + t₃ = 30 + 15 + 120 = 165 minutes = 2 hrs 45 minutes = 2\(\frac{3}{4}\) hrs. = \(\frac{11}{4}\) hrs.
∴ Average velocity, v_avg = \(\frac{\text { total displacement }}{\text { total time }}=\frac{172.5}{(11 / 4)}\) = 62.7 kmph.

Question 5.
A ball A is dropped from the top of a building and at the same time an identical ball B is thrown vertically upward from the ground. When the balls collide the speed of A is twice that of B. At what fraction of the height of the building did the collision occur ?
Solution:
Given at time of collision velocity of A = V_A = 2 × V_B (velocity of B)
Let the body be dropped from a height ‘h’. Let the two stones collide at x from the ground.
For the body dropped, s = h – x = \(\frac{1}{2}\)gt² → (1); For the body thrown up, x = ut – \(\frac{1}{2}\)gt² → (2)
For the body dropped, v = u + at ⇒ V_A = gt → (3)
For the body thrown up, v = u – gt ⇒ V_B = u – gt → (4)
Given V_A = 2V_B ⇒ gt = 2 (u – gt) or u = \(\frac{3gt}{2}\) → (5)
Divide equation (1) with equation (2)

∴ Fraction of height of collision = \(\frac{2}{3}\)

Question 6.
Drops of water fall at regular intervals from the roof of a building of height 16 m. The first drop strikes the ground at the same moment as the fifth drop leaves the roof. Find the distances between successive drops.
Solution:
Height of building, h = 16 m
∴ Time taken to reach the ground, t = \(\sqrt{\frac{2 h}{g}}\)
∴ Time taken to fall, t = \(\sqrt{\frac{2 \times 16}{9.8}}=\sqrt{\frac{32}{9.8}}=\sqrt{3.26}\) = 1.80

Number of drops, n = 5
∴ Number of intervals = n – 1 = 5 – 1 = 4
Time interval between drops = \(\frac{1.8}{4}\) = 0.45 sec
Time of travel of 1st drop, t₁ = 4 × 0.45 = 1.8
∴ Distance travelled by
1st drop, S₁ = \(\frac{1}{2}\)gt₁² = \(\frac{1}{2}\) × 9.8 × 1.8 × 1.8 = 16 m

For 2nd drop, t₂ = 3 × 0.45 = 1.35 sec., S₂ = \(\frac{1}{2}\)gt₂² ;
∴ S₂ = \(\frac{1}{2}\) × 9.8 × 1.35² = 4.9 × 1.822 = 8.9278 ≅ 9m

For 3rd drop, t₃ = 2 × 0.45 = 0.9 sec.
Distance, S₃ = \(\frac{1}{2}\)gt₃² = \(\frac{1}{2}\) × 9.8 × 0.9² = 3.97 ≅ 4m

For 4th drop, t₄ = 1 × 0.45 = 0.45 sec.
Distance travelled, S₄ = \(\frac{1}{2}\)gt₄² = \(\frac{1}{2}\) × 9.8 × (0.45)² ≅ 1

For 5 th drop, t₅ = 0 ⇒ S₅ = 0
Distance between 1^st and 2^nd drop S_{1, 2} = S₁ – S₂ = 16 – 9 = 7m
Distance between 2^nd and 3^rd drop S_{2, 3} = S₂ – S₃ = 9 – 4 = 5 m
Distance between 3^rd and 4^th drop S_{3, 4} = S₃ – S₄ = 4 – 1 = 3m
Distance between 4^th and 5^th drop S_{4, 5} = S₄ – S₅ = 1 – 0 = 1 m
∴ Distances between successive drops are 7m, 5m, 3m and 1m.

Question 7.
A food packet is dropped from an aeroplane, moving with a speed of 360 kmph in a horizontal direction, from a height of 500m. Find (i) its time of descent (ii) the horizontal distance between the point at which the food packet reaches the ground and the point above which it was dropped.
Solution:
Velocity of plane, V = 360 kmph = 360 × \(\frac{5}{18}\) = 100 m/s.
Height above the ground, h = 500 m; g = 10 m/s²

i) Time of descent, t = \(\) = 10sec.

ii) Horizontal distance between point of dropping and point where it reaches the ground = Range R.
∴ R = u \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\) = 100 \(\sqrt{\frac{2 \times 500}{10}}\) = 100 x 10 = 1000 m.

Motion in a Straight Line Class 11 NCERT Solutions

Question 1.
In which of the following examples of motion, can the body be considered approximately point object.
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on the top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
Answer:
(a) A railway carriage can be considered as a point object because the size of carriage is much smaller than the distance between the two stations.

(b) A monkey sitting on the top of the cyclist can be considered as a point object if the cyclist describes a circular path of radius which is sufficiently larger than the size of monkey.

(c) A spinning cricket ball cannot be considered as a point object because the size of ball is not negligible as compared to the distance through which the ball may turn on hitting the ground.
Remark: The spinning motion of cricket ball is due to its size. A point mass cannot have spin motion.

(d) A beaker slipping off the edge of a table cannot be considered as a point object because the size of beaker is not negligible as compared to the height of the table.

Question 2.
The position – time (x – t) graphs for two children A and B returning from their school O to their homes P and Q respectively as the shown in Fig.
Choose the correct entries in the brackets below :

(a) (A/B) lives closer to the school than (B/A).
(b) (A/B) starts from the school earlier than (B/A).
(c) (A/B) walks faster than (B/A).
(d) A and B reach home at the (same/different) time.
(e) (A/B) overtakes (B/A) on the road (once/twice).
Answer:
(a) As OP < OQ ; A lives closer to school than B.
(b) For A, x = 0 at t = 0 and for B, x = 0 at t = t₀ (let)
This implies that A starts from the school earlier than B.
(c) The speed = slope of x – t graph. As slope of B is greater than the slope of A, therefore, B walks faster than Answer:
(d) From graph, it is clear that the positions P and Q are attained at the same time, this implies that both A and B reach home at the same time.
(e) x – t graphs for A and B intersect each other only once, so B overtakes A on the road once.

Question 3.
A woman starts from her home at 9.00 am, walks with a speed of 5 kmh^-1 on a straight road upto her office 2.5 km away, stays at the office upto 5.00 pm and returns home by an auto with a speed of 25 kmh^-1. Choose suitable scales and plot the x-t graph of her motion.
Answer:
For motion from home to office :
Time to which woman leaves the office = 09.00 am
Speed = 5 km/h, distance traversed, s = 2.5 km
Time taken, t₁ = \(\frac{\text { Distance traversed }}{\text { Speed }}=\frac{2.5}{5}\) = 0.5 h = 30 min.
Time at which woman reaches office
= 9.00 am + 0.30 = 9.30 am
Stay at office : 9.30 am – 5.00 pm (position fixed)

For motion from office to home :
Time at which she leaves office = 5.00 pm
Speed of auto = 25 km/h, distance = 2.5 km
Time spent, t₂ = \(\frac{\text { Distance }}{\text { Speed }}=\frac{2.5}{25}\) = 0.1 h = 0.1 × 60 min = 6 min
Time at which she reaches home = 5.06 pm
The position-time (x – t) graph of motion of woman is as shown in the Fig.

Question 4.
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x – t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Answer:
The drunkard takes 5 steps forward and 3 steps backward. Each step is of 1 m length and takes 1 second. Therefore, the displacement of drunkard as a function of time is as shown below.

The position – time graph of motion of drunkard is as shown in the Fig.
From graph, it is clear that the drunkard will fall into the pit 37 seconds after the beginning.

Question 5.
A car moving along a straight highway with a speed of 126 kmir1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform); and how long does it take for the car to stop ?
Answer:
Initial speed of car, u = 126 km/h
1 km/h = \(\frac{1000 \mathrm{~m}}{3600 \mathrm{~s}}=\frac{5}{18}\) m/s ; u = 126 × \(\frac{5}{18}\) = 35 m/s
Final speed of car, v = 0, Distance traversed, s = 200 m
Using the formula v² = u² + 2as, we get
0 = (35)² + 2a × 200 ⇒ -400a = (35)² ⇒ a = \(\frac{(35)^2}{400}\) = -3.06 ms^-2
Retardation of Car = 3.06 ms^-2
If t is the time taken, then formula v = u + at gives
0 = 35 – 3.06 t ⇒ 3.06t = 35 ⇒ t = \(\frac{35}{3.06}\) = 11.4 s

Question 6.
A player throws a ball upwards with an initial speed of 29.4 ms1.
(a) What is the direction of acceleration during the upward motion of the ball ?
(b) What are the velocity and acceleration of the ball at the highest point of its motion ?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands ?[Take g = 9.8 ms^-2 and neglect air resistance].
Answer:
(a) The ball moves under gravity throughout its motion. The acceleration due to gravity is always directed vertically downward. So, the direction of acceleration during the upward motion of the ball is vertically downward.
(b) At the highest point of motion, the velocity of ball is zero and the acceleration is 9.8 ms^-2 directed vertically downward.
(c) As the ball remains between highest point and earth, so during upward and downward motion of ball, x > 0 (or x is positive).
During vertically upward direction v is negative (v < 0) and during vertically downward direction v is positive (v > 0).
Acceleration is always g = + 9.8 ms^-2 (positive).
(d) Using formula v² = u² + 2as, we have
a = + g = 9.8 ms^-2, s = – H, v = 0 (at highest point)
u = – 29.4 ms^-1 (given)
0 = (-29.4)² – 2 × 9.8 × H ⇒ H = \(\frac{(29.4)^2}{2 \times 9.8}\) m = 44.1 m
Also from relation, s = ut + \(\frac{1}{2}\)at²,
we have s = 0, u = -29.4 m/s, a = 98m/s²
0 = -29.4 t + \(\frac{1}{2}\) × 9.8t² or \(\frac{1}{2}\) 9.8t² = 29.4 t ⇒ t = \(\frac{29.4}{4.9}\) = 6s.
∴ Height, (H) = 44.1m and time taken (t) = 6s.

Question 7.
Read each statement below carefully and state with reasons and examples, if it is true or false.
A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant.
(b) with zero speed may have non-zero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.
Answer:
(a) True; at highest point of particle thrown in vertically upward direction, the speed is zero but acceleration is g = 9.8 ms^-2 vertically downward.
(b) False; because velocity = speed + direction. As speed is zero, velocity must be zero.
(c) True; as v = constant; acceleration a = \(\frac{d v}{d t}\) = 0
(d) False, when a body is thrown upward its speed decreases, but gravitational acceleration is downward (assumed positive direction).
However, if body continues to move along a straight line on a smooth horizontal surface, then positive acceleration will necessarily imply speeding up.

Question 8.
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Answer:
Let t₁ be the time taken by the ball to hit the floor.
Given, u = 0, h = 90 m, g = 9.8 m/s²
Formula, h = ut + \(\frac{1}{2}\)gt² gives
90 = 0 + \(\frac{1}{2}\)9.8t₁² ⇒ t₁² = \(\frac{90}{4.9}=\frac{900}{49}\) ⇒ t₁ = \(\frac{30}{7}\) = 4.3 s.
The speed of the ball when it hits the floor first time is given by
v₁ = u + gt₁ = 0 + 9.8 × \(\frac{30}{7}\) = 42 m/s
Now the ball bounces upward after collision. As the ball loses one-tenth of speed during collision, therefore, upward speed of ball just after collision
v₂ = \(\frac{9}{10}\) × 42 = 37.8 m/s
The ball now starts moving upward with retardation g = 9.8 m/s². The speed of ball at any instant t will be (u¹ = v₂ = 37.8 m/s).
∴ v = u¹ – gt = 37.8 – 9.8 t
The speed of ball becomes zero at highest point after time t₂¹ is given by
0 = 37.8 – 9.8 t₂¹ ⇒ t₂¹ = \(\frac{37.8}{9.8}\) = 3.85 s
Now the ball begins to fall downward and attains a speed of 37.8 m/s just before collision with the floor. As time of ascent = time of descent.

∴ Time taken by ball to hit the floor
t₂¹¹ = 3.85 s
Total time upto this instant
T = t₂ + t₂¹ + t₂¹¹ = 4.3 + 3.85 + 3.85 = 12 s
The speed of ball just after second collision
v₃ = \(\frac{9}{10}\) × 37.8 ≅ 34m/s
The graph of speed and time is as shown in the Fig.
Remark: Speed is never negative.

Question 9.
Explain clearly, with examples, the distinction between
a) magnitude of displacement (sometimes called distance) over an interval of time and the total length of the path covered by a particle over the same interval.
b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total paths length divided by the time interval].
Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only].
Answer:
For simplicity we suppose one-dimensional motion, where an object traverses from A to B and then returns back in total time t where AB = s.
(a) Displacement = 0 and total path length = AB + AB = 2s.
(b) Average velocity = \(\frac{\text { Displacement }}{\text { Time }}=\frac{0}{t}\) = 0
Average speed = \(\frac{\text { Total path length }}{\text { Time }}=\frac{2 \mathrm{~s}}{\mathrm{t}}\)

In both the cases (a) and (b), the second quantity (i.e., total path length and speed) turns out to be greater than the first (i.e., magnitude of displacement and velocity respectively). However, if the motion takes place along a straight line without change in direction, then equality sign will be true.
That is then, magnitude of displacement = Total path length
(i.e., motion along a straight line in one direction) and magnitude of velocity = Speed.

Question 10.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 kmh^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 kmh ‘. What is the
(a) Magnitude of average velocity, and
(b) The average speed of the man over the interval of time
(i) 0 to 30 min,
(ii) 0 to 50 min,
(iii) 0 to 40 min ?
Answer:
Distance of market from home = 2.5 km; Speed of man from home to market = 5 km/h
Time taken by man to reach market, t =Speed of man, while returning from the market = 7.5 km/h
∴ Time taken by man to return the home, t₁ = \(\frac{\text { Distance }}{\text { Speed }}=\frac{2.5}{5}=\frac{1}{2} \mathrm{~h}\) = 20 min
Speed of man, while returning from the market = 7.5 km/h
Time taken by man to return the home, t₂ = \(\frac{2.5}{7.5}=\frac{1}{3} h\) = 20 min
Total time of man returning to his home = 30 min + 20 min = 50 min

(i) Over the interval 0 to 30 min :
During this time, man goes from home to market; therefore, displacement, s = 2.5 km
Average velocity = \(\frac{\text { Displacement }}{\text { Time }}=\frac{2.5}{1 / 2}\) = 5km/h
Average speed = \(\frac{\text { Distance }}{\text { Time }}=\frac{2.5}{1 / 2}\) = 5 km/h
Clearly, during motion along a straight line (without turn), the average speed and average velocity are equal in magnitude.

(ii) Over the interval 0 to 50 min :
In 50 min the man goes from home to market and returns back to home.
Net displacement = zero
Average velocity = \(\frac{\text { Net displacement }}{\text { Total time }}=\frac{0}{50}\)
Average speed = \(\frac{\text { Total distance }}{\text { Total time }}=\frac{2.5 \mathrm{~km}+2.5 \mathrm{~km}}{(50 \mathrm{~min})}+\frac{5 \mathrm{~km}}{50 \times \frac{1}{60} \mathrm{~h}}\) = 6 km/h.

(iii) Over the interval 0 to 40 min :
During first 30 min man goes from home to market covering a distance of 2.5 km; in next 10 min the man is in path from market to home and covers a distance
= (75 km/h) × \(\left(\frac{10}{60} h\right)\) = 1.25 km
Displacement = 2.5 – 1.25 = 1.25 km
Distance = 2.5 + 1.25 = 3.75 km
Average velocity = \(\frac{\text { Displacement }}{\text { Time }}=\frac{1.25}{40 \times \frac{1}{60} \mathrm{~h}}\) = 1.875 km/h
Average speed = \(\frac{\text { Distance }}{\text { Time }}=\frac{3.75 \mathrm{~km}}{40 \times \frac{1}{60} \mathrm{~h}}\) = 5.625 km/h

Question 11.
In problems 9 and 10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why ?
Answer:
Instantaneous velocity, v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\)
Where ds is infinitesimal displacement in infinitesimal time interval.
For infinitesimal interval of time there is no change in direction so the magnitude of displacement is equal to the length of the path; hence magnitude of instantaneous velocity = instantaneous speed.

Question 12.
Look at the graphs (a) to (d) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Answer:
(a) It does not represent one-dimensional motion because according to this graph particle has two positions at the same time, which is impossible.

(b) It does not represent one dimensional motion because according to this graph the particle have equal positive and negative velocity at the same time, which is impossible.

(c) It cannot represent one-dimensional motion because according to graph speed is negative at certain instants, which is impossible (speed can never be negative).

(d) It cannot represent one-dimensional motion because according to this graph after a certain time the total path length begins to decrease with increase of time, which is impossible (total path length will always increase with increase of time).

Question 13.
Fig. shows the x – t graph of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0 ? If not suggest a suitable physical context for this graph.
Answer:
For t < 0, the particle is at rest; since, its position remains at x = 0. For t > 0, x – t graph is parabolic, it is possible if motion of particle is uniformly accelerated since x = ut + \(\frac{1}{2}\)at²
If u = 0, x = \(\frac{1}{2}\)at² or x ∝ t² (parabola)

Thus, it is not correct to say that the particle moves in a straight line for t < 0, because during t < 0, the particle does not move at all. However, for t > 0, the trajectory is parabolic which is possible if particle moves along a straight line with uniform acceleration. It is again not correct to say that particle moves on parabolic path for t > 0.

Physical context: The graph may represent the motion of a particle released from the top of a hill or tower.

Question 14.
Suggest a suitable physical situation for each of the following graphs.

Answer:
(a) The given x – t graph shows that initially x = 0, then ball gains constant velocity which then reverses in direction and finally its velocity becomes zero.
Physical situation : The ball is initially at rest on a smooth floor. It is kicked and given a constant velocity. Then, it strikes a wall, rebounds with reduced speed and strikes the opposite wall which stops it.

(b) The given velocity – time graph (v – t) shows that the velocity changes sign again and again with the passage of time and each time, it loses some speed.
Physical situation : A ball thrown up with some initial velocity rebounding from the floor with reduced speed after each collision.

(c) The given a – t graph shows that initially acceleration of object is zero. It gains acceleration for a very short duration and immediately it is decelerated and again, its acceleration is zero.
Physical situation: A cricket ball moving with a uniform speed is hit back with a bat for a very short duration.

Question 15.
Fig. gives the x – t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s and – 1.2 s.
Answer:
Position is positive when particle is above t – axis and negative if particle is below t – axis. Velocity is positive, when x increases with increase of time and velocity is negative when x decreases with increase of time. Acceleration is positive when the slope of x – t curve increases and negative when slope of x – t curve decreases.

At 0.3 s, x is negative, v is negative and a is positive.
At 1.2 s, x is positive, v is positive and a is negative.
At – 1.2 s, x is negative, v is positive and a is positive.

Question 16.
Fig. shows x – t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval.

Answer:
The average speed inx-t graph is equal to the slope of line joining the initial and final points in that interval. From the graph the slope is greatest in interval 3 and least in interval 2. Therefore, average speed is greatest in interval 3 and average speed is least in interval 2. Slope is positive in intervals 1 and 2 and negative in interval 3, so the average velocity is positive during intervals 1 and 2 and negative during interval 3.

Question 17.
Fig. shows a speed – time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which intervals is the average acceleration greatest in magnitude ? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ?

Answer:
The average acceleration is given by the slope of line joining the two points onv-f graph for given interval of time. From the graph, out of three intervals the slope is maximum in interval (2) but it is negative. Therefore, the magnitude of average acceleration is greatest in interval 2.
Average speed in a given interval = \(\frac{v_1+v_2}{2}\)
From the graph, it is clear that the magnitudes of v₁ and v₂ are maximum in interval 3, so average speed is maximum in interval 3.

Signs of v: The speed v is always positive. Therefore, it is positive in all the three intervals.
Signs of a: In interval T slope at v -1 graph is positive, therefore, acceleration is positive. In interval 2, the slope of v – t graph is negative, therefore, the acceleration is negative. In interval 3, the slope of v – t graph is positive, therefore, the acceleration is positive.

The instantaneous acceleration is equal to the slope of tangent on v – t graph at given instant. The tangents drawn on given v – t curve at points A, B, C and D are parallel to time axis, so slope is zero.
Hence, acceleration at points A, B, C and D are zero.

Question 18.
A police van moving on a highway with a speed of 30 km/h fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km/h, if the muzzle speed of the bullet is 150 m/s, with what speed does the bullet hit the thief’s car ?
(Note : Obtain that speed which is relevant for damaging thief’s car ?)
Answer:
Let v_p be the velocity of police van, v_T be the velocity of thief’s car and u_BP the muzzle velocity of bullet. Muzzle velocity means the velocity with respect to gun (or police van).
Given, v_P = 30 km/h = 30 × \(\frac{1100 \mathrm{~m}}{(60 \times 60) \mathrm{s}}=\frac{25}{3} \mathrm{~m} / \mathrm{s}\)
v_T = 192 km/h = 192 x \(\frac{1100 \mathrm{~m}}{(60 \times 60) \mathrm{s}}=\frac{160}{3} \mathrm{~m} / \mathrm{s}\)
u_B = 150 m/s
All velocities are in same direction, therefore, if u_B is the velocity of bullet with respect to earth, the velocity of bullet with respect to police van
u_BP = u_B + v_P = 150 + \(\frac{25}{3}=\frac{475}{3}\)m/s
Relative velocity of bullet with respect to thief’s car
u_BT = u_B – v_T = \(\frac{475}{3}-\frac{160}{3}=\frac{315}{3}\)= 105 m/s.

Regular practice with AP Inter 1st Year Chemistry Study Material Chapter 6 Equilibrium Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Chemistry 6th Lesson Equilibrium Questions and Answers

I. Multiple Choice Questions

Question 1.
The pH of a solution of hydrochloric acid is 4. The molarity of the solution is :
1) 4.0 M
2) 0.4 M
3) 0.0001 M
4) 0.04 M
Answer:
3) 0.0001 M
If [H⁺] = 0.0001M (Molarity), then
pH = -log₁₀ [H⁺] = -log 0.0001 = -log 10^-4 = 4.

Question 2.
The strong conjugate base in the following is
1) NO₃^–
2) Cl^–
3) SO₄^2-
4) CH₃COO^–
Answer:
4) CH₃COO^–
Strong conjugate base is CH₃COO^–, which is obtaining from weak acid.

Question 3.
When NH₄Cl is added to NH₄OH solution the dissociation of ammonium hydroxide is reduced. It is due to
1) common ion effect
2) hydrolysis
3) oxidation
4) reduction
Answer:
1) common ion effect
Due to common ion effect dissociation of NH₄OH is reduced.

Question 4.
The following reaction is at equilibrium.
Fe³⁺_(aq) + SCN^–_(aq) ⇌ [Fe(SCN)]²⁺_(aq)
In the above reaction, colour intensity of red colour can be increased by
l) addition of oxalic acid which reacts with Fe³⁺ ions.
2) addition of Hg²⁺ ions which reacts with SCN^– ions.
3) addition of KSCN.
4) red colour intensity cannot be changed.
Answer:
3) addition of KSCN.
\(\mathrm{Fe}_{(\mathrm{aq})}^{+3}+\mathrm{SCN}_{(\mathrm{aq})}^{-} \rightleftharpoons[\mathrm{Fe}(\mathrm{SCN})]_{(\mathrm{aq})}^{+2}\)
on adding KSCN colour intensity of red colour increased.

Question 5.
K_sp of a sparingly soluble salt AB₂ is 4 × 10^-12. The solubility of the salt is
1) 2 × 10^-6M
2) 4 × 10^-4M
3) 1 × 10^-12
4) 1 × 10^-4M
Solution:
4) 1 × 10^-4M
For AB₂ type of salt, K_sp = 4s³ ⇒ s² = \(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\)
⇒ s = \(\sqrt[3]{\frac{\mathrm{K}_{\mathrm{sp}}}{4}}=\sqrt[3]{\frac{4 \times 10^{-12}}{4}}\) = 10^-4m

Question 6.
An aqueous solution of ammonia consists of
1) H⁺ only
2) OH^– only
3) NH₄⁺ only
4) NH₄⁺ and OH^–
Answer:
Aqueous solution of ammonia contains NH₄⁺ and OH^– ions.
NH₄OH_(aq) → NH₄⁺ + OH^–

Question 7.
The pH of a 0.05 M of H₂SO₄ solution is
1) 0.05
2) 1
3) -1
4) 1.3010
Answer:
2) 1
For 0.05M of H₂SO₄
[H⁺] = 2 × 0.05(Di basic H₂SO₄) = 0.1
pH = -log₁₀[H⁺] = -log 0.1 = -log 10^-1 = 1

Question 8.
If the molar solubility of Zirconium phosphate Zr₃(PO₄)₄ is ‘S’. Solubility product K_sp of Zirconium phosphate is
1) 6912 (S)⁵
2) 6912 (S)⁷
3) 108 (S)⁷
4) 27 (S)⁵
Answer:
2) 6912 (S)⁷

Question 9.
For a reversible reaction, if the concentration of reactants is increased, the equilibrium constant of the reaction
1) Increases
2) Remains constant
3) Decreases
4) Depends on amount of reactant
Answer:
Equilibrium constant remains constant on increasing the concentration of reactants.

Question 10.
The pK_a of acetic acid and pK_b of ammonium hydroxide are 4.76 and 4.75 respectively.
1) 6.098
2) 7.005
3) 8
4) 5.0
Answer:
2) 7.005
CH₃COONH₄ is the salt formed by weak acid and weak base
∴ pH = 7 + — (PK_a – PK_b) = 7 + \(\frac{1}{2}\) (4.76 – 4.78) = 7 × \(\frac{1}{2}\) × 0.01 = 7.005

Question 11.
Which among the following salt solutions is basic in nature?
1) Ammonium chloride
2) Ammonium sulphate
3) Ammonium nitrate
4) Sodium acetate
Answer:
4) Sodium acetate
Sodium acetate is the salt obtained from sodium hydroxide (Strong base) and acetic acid (weak acid). So aq. CH₃COONa is in basic nature.

Question 12.
Observe the following equilibrium at T (Kelvin)
H_2(g) + I_2(g) ⇌ HI_(g)
Which one of the following does not disturb the above equilibrium at constant volume?
1) Addition of H_2(g)
2) Removal of HI_(g)
3) Addition of I_2(g)
4) Addition of He_(g)
Answer:
4) Addition of He_(g)
He is a Noble gas, it is chemically inert, so by adding inertgas the equilibrium does not disturbed.

Question 13.
Solution of 0.1 N NH₄OH and 0.1 N NH₄Cl has pH 9.25. Find out pK_b of NH₄OH.
1) 9.25
2) 4.75
3) 3.75
4) 8.25
Solution:
2) 4.75
P^OH = PK_b + log \(\frac{[\text { salt }]}{[\text { base }]}\) (basic buffer)
pH + P^OH = 14 ⇒ P^OH = 14 – 9.25 = 4.75
4.75 = PK_b + log \(\frac{0.1}{0.1}\) ⇒ PK_b = 4.75

Question 14.
For the following process, the expression for the K_c
Ni_(s) + 4CO_(g) ⇌ Ni(CO)_4(g)
1) K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]^2}{[\mathrm{Ni}][\mathrm{CO}]^4}\)
2) K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]}{[\mathrm{CO}]^4}\)
3) K_c = \(\frac{[\mathrm{Ni}][\mathrm{CO}]^4}{\left[\mathrm{Ni}(\mathrm{CO})_4\right]^{2-}}\)
4) K_c = \(\frac{[\mathrm{CO}]^4}{\left[\mathrm{Ni}(\mathrm{CO})_4\right]^2}\)
Answer:
2) K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]}{[\mathrm{CO}]^4}\)
K_c = \(\frac{\text { Molar concentrations of products }(\mathrm{g})}{\text { Molar concentrations of reactants }(\mathrm{g})}\) ⇒ K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]}{[\mathrm{CO}]^4}\)

Question 15.
K₁ and K₂ are equilibrium constants for reactions (i) & (ii). Then the relation between
K₁ and K₂
(i) H_2(g) + I_2(g) ⇌ 2HI(g)
(ii) HI_(g) ⇌ 1/2 H_2(g) + 1/2 I_2(g)
1) K₁ = K₂²
2) K₁ = \(\frac{1}{\mathrm{~K}_2}\)
3) K₁ = K₂⁰
4) K₁ = \(\frac{1}{K_2^2}\)
Solution:
4) K₁ = \(\frac{1}{K_2^2}\)
H_2(g) + I_2(g) ⇌ 2HI(g) …………… (1)
HI_(g) ⇌ \(\frac{1}{2}\) H_2(g) + \(\frac{1}{2}\) I_2(g) ………… (2)
K₁ = Equilibrium constant
K₂ = Equilibrium constant
K₂ = \(\frac{1}{\sqrt{\mathrm{~K}_1}}\) [eq.(2) is reverse reaction and halved]
Squaring on both sides, \(\mathrm{K}_2^2=\frac{1}{\mathrm{~K}_1}\)
⇒ K₁ = \(\frac{1}{K_2^2}\)

II. Fill in the Blanks

Question 1.
Acidity of BF₃ can be explained on the basis of __________ theory of acids and bases.
Answer:
Lewis acid base theory

Question 2.
The nature of aqueous solution of ammonium chloride (NH₄Cl) is ______.
Answer:
NH₄Cl is the salt of weak base (NH₄OH) and strong acid (HCl).
So aq. solution of NH₄Cl is acidic.

Question 3.
K_p for the following equation: ______.
CaCO_3(s) ⇌ CaO_(s) + CO_2(g)
Answer:
K_p = P_CO2

Question 4.
The molarity of pure water is ______.
Answer:
Molarity of pure water is = 55.5.M
M = \(\frac{\mathrm{W}}{\mathrm{GMW}} \times \frac{1000}{\mathrm{~V}(\mathrm{ml})}=\frac{1000}{18} \times \frac{1000}{1000}\) = 55.5 M

Question 5.
If K is in the range of ________ appreciable concentrations of both reactants and products are present.
Answer:
10^-3 to 10³

III. One Word Answer Questions

Question 1.
What is a bronsted base?
Answer:
The substance which can accept a proton is said to be bronsted base. Eg : Cl^–, HCO₃^–.

Question 2.
What is Lewis acid?
Answer:
Lewis acid is the substance which can accept a pair of electrons and form a coordinate covalent bond. Eg : BF₃, H⁺.

Question 3.
What is the value of K_w at 25°C? What are its units ?
Answer:
Ionic product of water K_w = 1 × 10^-14 ; Units are mole²/lit².

Question 4.
What is conjugate acid-base pair?
Answer:
An acid-base pair which differ by proton (H⁺) is called a conjugate acid base pair.
Eg : HCl and Cl^–.

Question 5.
Write the relation between Gibbs energy and equilibrium constant.
Answer:
∆G° : – 2.303 RT log K_c
R = gas constant ; T= temperature ; K_c = Equilibrium constant

IV. Very Short Answer Questions

Question 1.
State law of chemical equilibrium.
Answer:
The ratio of product of molar concentrations of products to the product of molar concentrations of reactants is a constant value. This constant is called equilibrium constant and this is called law of chemical equilibrium.
Eg : aA + bB ⇌ cc + dD (or) Equilibrium constant (K_c) = \(\frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{~A}]^{\mathrm{a}}[\mathrm{~B}]^{\mathrm{b}}}\)
Eg : H_2(g) + I_2(g) ⇌ (K_c) = \(\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}\)

Question 2.
What is homogeneous equilibrium? Give two examples.
Answer:
The equilibrium in which all the substances are present in the same phase is known as homogeneous equilibrium.
Eg: N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
CH₃COOH (l) + C₂H₅OH (l) ⇌ CH₃COO C₂H₅ (l) + H₂O (l)

Question 3.
What is heterogeneous equilibrium? Give two examples.
Answer:
The equilibrium in which the substances involved are present in different phases is called heterogeneous equilibrium.
CaCO₃ (s) ⇌ CO(s) + CO₂ (g) and H₂O (l) ⇌ H₂O (g)

Question 4.
What is meant by dynamic equilibrium?
Answer:
A state in a reversible reaction where the forward and backward reaction continue to takes place with same rate is known as dynamic equilibrium.

Question 5.
Give two chemical equilibrium reactions for which K_p > K_c
Answer:
In order that K_p > K_c, ∆n should be +ve
1) PCl₅ (g) ⇌ PCl₃ (g) + Cl₂(g); ∆n = (2 – 1)
2) N₂O₄(g) ⇌ 2NO₂ (g) ; ∆n = (2 – 1)

Question 6.
Give two chemical equilibrium reactions for which K_p <K_c.
Answer:
In order that K_p <K_c, ∆n should be -ve
1) N₂ (g) + 3H₂ (g) ⇌ 2NH₃(g) ; ∆n = (2 – 4)
2) 2SO₂ (g) + O₂ (g) ⇌ 2SO₃(g) ; ∆n = (2 – 3)

Question 7.
Write the equations for the conversion of K_c to K_p for the following reaction
CO_(g) + H₂O_(g) ⇌ CO_2(g) + H_2(g)
Answer:
CO_(g) + H₂O_(g) ⇌ CO_2(g) + H_2(g)
K_p = K_c (RT)^∆n
∆n = (1+1) – (1 + 1) = O
K_p = K_p (RT)
K_p = K_c

Question 8.
What are the factors which influence the chemical equilibrium?
Answer:
Molar concentration, temperature and pressure influence the equilibrium.

Question 9.
What is meant by ionic product of water?
Answer:
The product of concentration of H⁺ and OH^– ions in aqueous solution (or) in pure water is called ionic product of water.

Question 10.
Give two examples of salts whose aqueous solutions are acidic.
Answer:
NH₄Cl (Ammonium Chloride); Cu₂SO₄ (copper sulphate) and (NH₄)₂SO₄ (Ammonium sulphate) .
These aqueous solutions are acidic in nature.

V. Short Answer Questions

Question 1.
Derive the relation between K_p and K_c for the equilibrium reaction.
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
Answer:
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
K_c = \(=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}\) ; K_p = \(\frac{P_{\mathrm{NH}_3}^2}{P_{\mathrm{N}_2} \times \mathrm{P}_{\mathrm{H}_2}^3}\)
Partial pressures of P_N2, P_H2 and P_NH3 can be written as
P_N2 = \(\frac{\mathrm{n}_{\mathrm{N}_2} \mathrm{RT}}{\mathrm{~V}}\) = C_N2RT ; P_H2 = \(\frac{\mathrm{n}_{\mathrm{H}_2} \mathrm{RT}}{\mathrm{~V}}\) = C_H2RT and P_NH3 = \(\frac{\mathrm{n}_{\mathrm{NH}_3} \mathrm{RT}}{\mathrm{~V}}\) = C_NH3
Substituting these values in K_p equation, K_p = \(\frac{\left(\mathrm{C}_{\mathrm{NH}_3}\right)^2}{\left(\mathrm{C}_{\mathrm{N}_2}\right)\left(\mathrm{C}_{\mathrm{H}_2}\right)^3}\) × [RT]^-2
∴ K_p = K_c[RT]^-2 [∵ \(\frac{\mathrm{C}_{\mathrm{NH}_3}^2}{\left(\mathrm{C}_{\mathrm{N}_2}\right)\left(\mathrm{C}_{\mathrm{H}_2}\right)^3}\) = K_c]

Question 2.
State Le Chateliers’s Principle. Discuss the application of Le Chatelier’s principle for the industrial synthesis of Ammonia by Haber’s process.
Answer:
Le Chatelier’s principle : If a chemical reaction at equilibrium is subjected to a change in temperature, pressure or concentration that govern the chemical equilibrium, the equilibrium position shifts in the direction in which the change is reduced or nullified.
Application of Le Chatelier’s principle to by Haber’s process :
N_2(g) + 3H_2(g) ⇌ 2NH_3(g); ∆H = -92 kJ

Effect of pressure :
Increase of external pressure on the reaction at equilibrium favours the reaction in the direction in which the volume or the number of molecules decreases.
In the above reaction of formation of ammonia forward reaction involves decrease in volume or number of molecules. So high pressure is favourable for the formation of ammonia. Hence ammonia is manufactured at 200 atm pressure.

Effect of temperature :
In a reversible reaction, if the forward reaction is exothermic, the backward reaction will be endothermic.
High temperature favours the endothermic reaction while low temperature favours the exothermic reaction.

• Since the formation of ammonia is exothermic reaction, low temperature is favourable for the manufacture of ammonia.
• But at low temperature, the reaction is slow. So an optimum temperature of about 725 – 775 K is used.

Effect of catalyst:

• To increase the rate of formation of ammonia iron powder mixed with molybdenum powder is used as catalyst.
• Molybdenum acts as promoter to iron catalyst.

Effect of concentration :

• Increase in concentrations of reactants increases the rate of forward reaction.
• Removal of the products from reaction mixture decreases the rate of backward reaction.
• So more ammonia can be obtained by increasing the concentration of N₂ and H₂ and removing the NH₃ formed at regular intervals.
• Optimum conditions for the synthesis of ammonia are
  Pressure = 200 atm
  Temperature = 725 – 775 K
  Catalyst = Fe
  Promoter = Mo

Question 3.
State Le Chateliers’s Principle. Discuss the application of Le Chatelier’s principle for the manufacture of SO₃ by contact process.
Answer:
Le Chatelier’s principle : If a chemical reaction at equilibrium is subjected to a change in temperature, pressure or concentration that govern the chemical equilibrium, the equilibrium position shifts in the direction in which the change is reduced or nullified.

Application of Le Chatelier’s principle by contact process :
2SO_2(g) + O_2(g) ⇌ 2SO_3(g) + Heat; ∆H = -189 kJ

Effect of pressure :

• Increase of external pressure on the reaction at equilibrium favours the reaction in the direction in which the volume or the number of molecules decreases.
• Formation of SO₃ in the above reaction involves decrease in volume.
• According to Le Chatelier’s principle high pressure is favourable for the formation of SO₃.
• So, SO₃ is prepared at a pressure of 1.5 to 1.7 atmospheres.

Effect of temperature :

• In a reversible reaction if the forward reaction is exothermic, the backward reaction is endothermic.
• High temperatures are favourable for the endothermic reaction, while the low temperatures are favourable for the exothermic reactions.
• Since the formation of SO₃ is exothermic reaction, low temperatures are favourable. So, SO₃ is prepared at a temperature of about 673 K.
  Effect of catalyst: To increase the rate of formation of SO₃ catalyst like V₂O₅ or platinised asbestos is used.

Effect of concentration :

• Increase in the concentration of the reactants, increase the rate of forward reaction.
• Removal of products from reaction mixture decreases the rate of backward reaction.
• So more SO₃ can be obtained by increasing the concentrations of SO₂ and O₂ and by removing, the SO₃ is formed at regular intervals.
  Optimum conditions for the synthesis of ammonia are
  Pressure : 1.5 – 1.7 atmospheres Temperature : 673 K
  Catalyst: V₂O₅ (Vanadium pentoxide) or Platinised asbestos.

Question 4.
The species H₂O, HCO₃^–, HSO₄^– and NH₃ can act as both bronsted acids and bases. Give the corresponding conjugate acid and base for each of them.
Answer:

Question 5.
Write notes on
(i) Common ion effect.
(ii) Solubility product
Answer:
i) Common ion Effect : The decrease in the dissociation of the aq. solution of weak electrolyte by adding another strong electrolyte having one common ion is called common ion effect. Eg : To the saturated solution of AgCl, electrolyte NaCl is added to decrease the dissociation.
ii) Solubility Product (K_sp) : The solubility product is the product of molar equilibrium concentrations of cation and anion at a particular temperature in a saturated solution of sparingly soluble salt.
Applying the concept at equilibrium, A_xB_y(s) ⇌ xA^y+ + yB^x-
Applying the concept at equilibrium, K_sp = [A^y+]^x [B^x-]^y
AgCl ⇌ Ag⁺ + Cl^–
K_sp = [Ag⁺] [Cl^–]
⇒ K_sp = s × s [where s is the solubility]
⇒ K_sp = s²
Solubility s = \(\sqrt{\mathrm{K}_{\mathrm{sp}}}\)

Question 6.
Explain the concept of Bronsted-Lowry acids and bases. Illustrate with suitable examples.
Answer:
Bronsted – Lowry theory : Any substance that can lose a proton or protons is an acid.
Eg : HCl, H₂SO₄, H₃PO₄, CH₃COOH.

• Any substance that can gain a proton or protons is a base. Eg : NH₃, H₂O, OH^– etc.
• When an acid react with base neutralization takes place. As per Bronsted – Lowry theory proton transfer from an acid to a base is called neutralization.
  Eg : HCl + H₂O ⇌ H₃O⁺ + Cl^–
• The above reaction is reversible. In this reaction,
  1. HCl donates proton to water. So it is a bronsted acid.
  2. H₂O gains the proton. So it is a bronsted base.
  3. H₂O⁺ donates a proton to Cl^–. So it is bronsted acid.
  4. Cl^– can gain a proton. So it is a bronsted base.
• The acid base pair which differ by a single proton is said to be conjugate acid-base pair.
• In the above reaction HCl & Cl^– and H₃O⁺ and H₂O are conjugate acid-base pairs.
• Every bronsted acid has its conjugate base and every bronsted base has a conjugate protonic acid.
• According to bronsted – Lowry theory greater the tendency to donate proton stronger is the acid. Higher the tendency of the base to accept protons, stronger is the base.
• In the above reaction, HCl gives proton easily than H₃O⁺. So HCl is stronger acid than H20. H20 accepts the proton more easily than Cl^–. So H₂O is a stronger base than NH₃
• In all acid – base reactions, the reaction takes place in the direction of formation of weaker acid and a weaker base.
• A stronger acid has a weaker conjugate base and a stronger base has a weaker conjugate acid.

Advantages of Bronsted – Lowry theory :

• This theory explains the behaviour of acids and bases in both aqueous and non-aqueous solvents.
• It explains the behaviour of NH₃, CaO, etc. as bases and CO₂, SO₂, etc. as acids.

Drawbacks of Bronsted – Lowry theory :

• Proton donation or acceptance happens only in the presence of other substances.
• It could not explain the behaviour of electron deficient molecules like AlCl₃, BCl₃, etc. as acids.

Question 7.
Explain Lewis acids and bases with suitable examples. Classify the following species into Lewis acids and Lewis bases.
a) OH^–
b) F^–
c) H⁺
d) BCl₃.
Answer:
Lewis acid : The substance which can accept an electron pair to form a coordinate covalent bond is called Lewis acid.
Types of Lewis acids :
1) All +vely charged metal ions or cations act as Lewis acids.
Eg : Ag⁺, CO⁺³, Cu⁺², Fe⁺³, etc.
2) Electron deficient compounds whose central atom has an incomplete octet, possessing an empty orbital acts as Lewis acid. Eg : BF₃, BCl₃, AlCl₃, FeCl₃.
3) Compounds in which central atom has vacant’d’ orbitals and which can expand its octet. Eg : SiF₄, SnCl₄, SF₄, TeF₄, FeCl₃.
4) Molecules having multiple bonds between atoms of different electronegativities.
Eg : CO₂, SO₂, SO₃, NO₂, Cl₂O₇, P₄O₁₀.
5) Elements with an electron sextet. Eg : S, O.

Lewis base : The substance which donates electron pair to form a coordinate covalent bond is called Lewis base.
Types of Lewis bases :
1) All anions will act as Lewis bases. Eg : Cl^–, OH^–, CN^–, NH₂^–, F^–, SCN^–.
2) Molecules with one or two lone pairs on the central atom.
Eg : H₂O, NH₃, R – OH, R – NH₂,
R-O-R, C₅H₅N.
3) Molecules with multiple bonds. Eg : CO, NO, HC ☰ CH, H₂C ☰ CH₂
a) OH^– is Lewis base, b) F^– is Lewis base, c) H⁺ is Lewis acid, d) BCl₃ is Lewis acid.

Question 8.
Calculate the pH of
a. 10^-3 M of HCl
b. 10^-3 M of H₂SO₄ [log 2 = 0.3010]
Solution:
a) 10^-3M of HCl [H⁺] = 10^-3M
pH = – log [H⁺] = – Log 10^-3 = 3 log 10 = 3

b) 10^-3 M of H₂SO₄
[H⁺] = 2 × 10^-3
pH = -log [H⁺] = -log 2 × 10^-3 = – log 2 – log 10^-3
= – log 2 + 3 log 10 = -log 2 + 3
= 3 – log 2
= 3 – 0.3010 = 2.6990

Question 9.
Calculate the pH for
a. 0.001 M of NaOH
b. 0.01 M of Ca(OH)₂
Solution:
a) 0.001 M of NaOH
[OH⁺] = 0.001M
P^OH = -log [OH^–] = – log 10^-3 = 3 log 10 = 3
pH + POH =14
pH = 14 – 3 = 11.
∴ pH = 11

b) 0.01 M of Ca(OH)₂
[OH^–] = 0.01 × 2 = 2 × 10^-2
P^OH = – log [OH^–]
= – log 2 × 10^-2
= – log 2 – log 10^-2 = – log2 + 2 log 10
= 2 – log2 = 2 – 0.3010 = 1.6990
pH + P^OH = 14
pH = 14 – 1.699 = 12.3010

Question 10.
The pH of 0.1M solution of weak mono protic acid is 4.0. Calculate its [H⁺] and K_a.
Solution:
HA ⇌ H⁺ + A^–
pH = 4 ; [H⁺] = 10^-4
K_a = \(\frac{c \alpha^2}{1-\alpha}=\frac{(0.1)\left(10^{-4}\right)^2}{1}\) [Neglecting α in the denominator]
K_a = 10^-7

Question 11.
One litre of buffer solution contains 0.1 mole of acetic acid and 1 mole of sodium acetate. Find its pH if pK_a of CH₃COOH is 4.8.
Solution:
Henderson’s Equation, pH = PK_a + log \(\frac{[\text { Salt }]}{[\text { Acid }]}\)
Given [Salt] = 1M; [Acid] = 0.1M; PK_a = 4.8
= 4.8 + log \(\frac{1}{0.1}\)
= 4.8 + log 10
= 4.8 +1 = 5.8

VI. Long Answer Questions

Question 1.
What are equilibrium processes? Explain equilibrium in Physical and Chemical processes with examples.
Answer:

• In a reversible process, a state will be attained such that the rate of forward reaction is equal to the rate of backward reaction.
• Such a state is known as equilibrium state.
• Equilibrium processes are of two types. They are

1. Physical equilibrium process : In such a process, equilibrium occurs in a phase transformation process.
Ex: Solid ⇌ Liquid
Solid ⇌ Vapour
Vapour ⇌ Liquid

2. Chemical equilibrium process: If there exists an equilibrium in between reactants and products, such reactions are said to be chemical equilibrium process.
1) N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)
2) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

Question 2.
What is meant by dynamic equilibrium? Explain with suitable examples.
Answer:

• In a reversible process, a state will be attained such that the rate of forward reaction is equal to the rate of backward reaction. Such a state is known as equilibrium state.
• At equilibrium we find no change in the concentration of either reactants (or) products then we may feel that the reaction has stopped. But actually the reaction will be going on. Such a state of reaction is called dynamic equilibrium.

Equilibrium processes are of two types. They are :
1. Physical equilibrium process : In such a process, equilibrium occurs in a phase transformation process.
Ex : Solid ⇌ liquid
Solid ⇌ vapour
Vapour ⇌ liquid

2. Chemical equilibrium process: If there exists an equilibrium in between reactants and products, such reactions are said to be chemical equilibrium process.
1) N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)
2) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

The forward and backward reactions of a reversible reaction continuously takes place with equal rates simultaneously at the equilibrium stage also. Hence the equilibrium is called dynamic equilibrium.

At this stage, the reaction appears to be stopped.

All chemical equilibrium processes are dynamic equilibriums in nature.

Question 3.
What are the important features of equilibrium constant? Discuss any two applications of equilibrium constant.
Answer:

• The equilibrium constant has a definite value for every chemical reaction at a particular temperature. The value of equilibrium constant is independent of initial concentrations of reacting species.
• The value of equilibrium constant K changes with change in temperature.
• For a reversible reaction, the equilibrium constant for the backward reaction is inverse of the equilibrium constant for the forward reaction.
• The equilibrium constant is independent of the presence of catalyst.
• The equilibrium constant is expressed in terms of concentration, it has different units for different reactions.
• When addition of two equilibria leads to another equilibrium then the product of their equilibrium constants gives the equilibrium constant of the resultant equilibrium.
  Eg : N₂ + O₂ ⇌ 2NO ; K₁
  2NO + O₂ + O₂ ⇌ 2NO₂; K₂
  N₂ + 2O₂ ⇌ 2NO₂; K₃
  K = K₁ × K₂ × K₃

Applications of equilibrium constant:
i) Predicting of extent of reaction: The magnitude of equilibrium constant tells us about the extent to which the reactants are converted into the products before the equilibrium is attained.

ii) Predicting the direction of the reaction : From the knowledge of equilibrium con¬stant it is possible to predict the direction in which the net reaction is taking place at a given concentrations or partial pressures of reactants and products.

• If Q_c > K_c the reaction taking place in backward direction is more.
• If Q_c < K_c the reaction taking place in forward direction is more.
• If Q_c = K_c the reaction is at equilibrium. No reaction taking place.

Question 4.
What is Le Chatelier’s principle? Discuss briefly the factors which can influence the equilibrium.
Answer:
Le Chatelier’s principle : If a chemical reaction at equilibrium is subjected to a change in temperature, pressure or concentration that govern the chemical equilibrium, the equilibrium position shifts in the direction in which the change is reduced or nullified.
1) Effect of pressure : ,

• Pressure is inversely proportional to volume.
• When pressure is increased the equilibrium shifts in such a direction where the decrease in volume is observed.
• Similarly when pressure is decreased, the equilibrium shifts in such a direction where increase in volume is observed. This is applicable exclusively for gaseous reactions.

2) Effect of temperature :

• When temperature is increased, equilibrium shift in such a direction where the temperature change is nullified or equilibrium shifts towards the direction in which temperature is decreased or heat is absorbed. That is endothermic reaction is favoured.
• When temperature is decreased equilibrium shifts in such a direction where temperature is increased or heat is liberated. That is exothermic reaction is favoured.

3) Effect of concentration :

• Increase in concentration of the reactants favours forward reaction and decrease in concentration of the reactants favours the backward reaction.
• When products are added to the equilibrium reaction the concentration of the products increases and reaction proceeds in backward direction.
• Removal of products leads to decrease in concentration of the products and leads to forward reaction.

Question 5.
Describe the effect of:
a) addition of H₂
b) addition of CH₃OH
c) removal of CO
d) removal of CH₃OH on the equilibrium of the reaction 2H₂ (g) + CO (g) ⇌ CH₃OH (g)
Answer:
a) According to Le Chatelier’s principle increase in the concentration of reactant increase the forward reaction.
So the addition of H₂ results in the formation of more amount of CH₃OH.

b) According to Le Chatelier’s principle increase in the concentration of the product, shift the equilibrium in the backward direction. So the addition of CH₃OH (product) results in the decrease in concentration and CH₃OH and increase in the concentration of H₂ and CO.

c) According to Le Chatelier’s principle the decrease in the concentration of reactant equilibrium shift in the backward direction. So removal of CO (reactant) the equilibrium shift in the backward direction. So formation of CH₃OH decreases.

d) According to Le Chatelier’s principle the decrease in concentration of product shifts the equilibrium in the forward direction. So the removal of CH₃OH (product) shift the equilibrium in the forward direction. Thus more amount of CH₃OH is formed.

Question 6.
What is degree of ionization in respect of weak acids and weak bases? Derive the relationship between degree of ionization (α) and ionization constant (K_a) for the weak acid HX.
Answer:
Degree of ionization : It can be defined as the fraction of acid (a) base that undergoes ionization per one mole of acid or base. For strong acids or bases degree of ionization (a) is equal to ‘1’.
Let’s consider a weak acid HX with concentration ‘c’. It undergoes partial ionization.

where α is degree of ionization
Ionization constant K_a = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{X}^{-}\right]}{[\mathrm{HX}]}\)
⇒ K_a = \(\frac{(\mathrm{c} \alpha)(\mathrm{c} \alpha)}{\mathrm{c}(1-\alpha)}\) for weak acids α < < 1, so α is negligible.
⇒ K_a = \(\frac{c^2 \alpha^2}{c}\) ⇒ α = \(\sqrt{\frac{K_a}{c}}\)

Question 7.
Define pH. What is buffer solution? Derive Henderson – Hasselbalch equation for calculating the pH of an acid buffer solution.
Answer:
The pH of a solution may be defined as negative logarithm of the activity of hydrogen ion [H⁺]
pH = – log [H⁺]
Buffer solution : A buffer solution is the solution which can resist the change in pH on
addition of small amount of acid or base. The ability of a solution to resist change in pH on
addition of small amount of acid or base is called buffer solution.
Henderson – Hasselbalch equation :
HA + H₂ ⇌ H₃O⁺ + A^– Partial ionisation.
NaA → Na⁺ + A^– Complete ionisation
K_a = \(\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)
Since HA is weakly dissociated, the conc. HA ion solution can be taken equal to the initial conc. of the acid.
Since NaA is completely dissociated, conc. of A- can be taken equal to the conc. of NaA (salt)
K_a = \(\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right][\text {Salt }]}{[\text { Acid }]}\)
H₃O⁺ = \(\frac{\mathrm{K}_{\mathrm{a}} \text { [Acid] }}{[\text { Salt }]}\)
-log [H₃O⁺] = – logK_a – log\(\frac{\text { [Acid] }}{[\text { Salt }]}\)
pH = pKa + log \(\frac{\text { [Salt] }}{[\text { Acid }]}\)

Question 8.
Write notes on
(i) Common ion effect.
(ii) The relation between K_sp and solubility (S) of a sparingly soluble salt BaSO₄.
Answer:
i) Common ion effect: The solubility of an electrolyte in water decreases by the addition of another electrolyte which has one ion common with the electrolyte.

• For example the solubility of NaCl decreases on addition of HCl to the solution of NaCl in water which is known as salting out of water.
• Acetic acid is weak acid and it undergoes partial ionization as follows.
  CH₃COOH ⇌ CH₃COO^– + H⁺
• In case of such electrolytes there exists an equilibrium in between ions and molecular form. In such a condition, the addition of electrolyte with common ions shifts the equilibrium in backward direction which causes the decrease in solubility.

Significance :
1) Common ion effect is an important phenomenon in qualitative analysis.
2) The common ion effect principle is also used in controlling the H⁺ ion concentration in buffer solution.

ii) The relation between K and solubility (S) of a sparingly soluble salt BaSO₄ : The sparingly soluble salt like BaSO₄, when added to water form saturated solution. There exists an equilibrium between undissolved solid and the ions in saturated solution.
BaSO₄ ⇌ Ba²⁺ + SO₄^2-
The equilibrium constant, K = \(\frac{\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]}{\left[\mathrm{BaSO}_4\right]}\)
K = \(\frac{\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]}{\mathrm{K}_1}\) (∴ Activity of solid BaSO₄ is constant)

K × K₁ = [Ba²⁺][SO^2-] or K_sp = [Ba²⁺][SO₄^2-]
The new constant K_sp is called solubility product.

If S moles lit^-1 is the solubility of BaSO₄.

K_sp = s × s = s²
or solubility of BaSO₄ is s = \(\sqrt{\mathrm{K}_{\mathrm{sp}}}\) .

Equilibrium Class 11 NCERT Solutions

Question 1.
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature.
The volume of the container is suddenly increased.
i) What is the initial effect of the change on vapour pressure ?
ii) How do rates of evaporation and condensation change initially ?
iii) What happens when equilibrium is restored finally and what will be the final vapour pressure ?
Answer:
i) The vapour pressure will decrease due to increase in volume.
ii) Since, the temperature is fixed, the rate of evaporation will remain constant. However, the rate of condensation will be low initially due to lesser number of molecules per unit volume in the vapour phase.
iii) When equilibrium is restored, the final vapour pressure will be the same as original vapour pressure because at equilibrium the rates of evaporation and condensation are equal.

Question 2.
W hat is K_c for the following equilibrium when the equilibrium concentration of each substance is : [SO₂] = 0.60 M, [O₂] = 0.82 M and [SO₃] = 1.90 M ?
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Answer:
K_c = \(\frac{\left[\mathrm{SO}_3(\mathrm{~g})\right]^2}{\left[\mathrm{SO}_2(\mathrm{~g})\right]^2\left[\mathrm{O}_2(\mathrm{~g})\right]}=\frac{(1.90)^2}{(0.60)^2 \times(0.82)}\) = 12.229 L mol^-1

Question 3.
At a certain temperature and a total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms.
I_2(g) ⇌ 2I_(g)
Calculate K_p, for the equilibrium.
Answer:
Partial pressure of I atom (pI₁) = \(\frac{40}{100}\) × 10⁵
Partial pressure of I₂ (pI₂) = \(\frac{(100-40)}{100}\) × 10⁵
∴ K_p = \(\frac{\mathrm{p}_{\mathrm{I}}^2}{\mathrm{p}_{\mathrm{I}_2}}=\frac{\left(4.0 \times 10^4\right)^2}{6.0 \times 10^4}\)
= 2.67 × 10⁴ Pa

Question 4.
Write the expression for the equilibrium constant, K_c for each of the following reactions.
i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g)
ii) 2Cu(NO₃)₂(s) ⇌ 2Cu0(s) + 4NO₂(g) + O₂(g)
iii) CH₃COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(aq)
iv) Fe³⁺(aq) + 3OH^–(aq) ⇌ Fe(OH)₃(s)
v) I₂(s) + 5F₂(g) ⇌ 2IF₅(g)
Answer:
i) K_c = \(\frac{[\mathrm{NO}(\mathrm{~g})]^2\left[\mathrm{Cl}_2(\mathrm{~g})\right]}{[\mathrm{NOCl}(\mathrm{~g})]^2}\)
ii) K_c = [NO₂(g)]⁴[O₂(g)] {∵ [CuO(s) = [Cu(NO₃)₂(s)] = 1}
iii) K_c = \(\frac{\left[\mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq})\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{aq})\right]}{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(\mathrm{aq})\right]}\) (∵ [H₂O(l)] = 1)
iv) K_c = \(\frac{1}{\left[\mathrm{Fe}^{3+}(\mathrm{aq})\right]\left[\mathrm{OH}^{-}(\mathrm{aq})\right]^3}\) (∵ [Fe(OH)₃(s)] = 1)
v) K_c = \(\frac{\left[\mathrm{IF}_5\right]^2}{\left[\mathrm{~F}_2\right]^5}\) (∵ [I₂(s) = 1)

Question 5.
Find out the value of Kc for each of the following equilibria from the value of K_p.
i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) ;K_p = 1.8 × 10^-2 at 500 K
ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); K_p = 167 at 1073 K
Answer:
i) For the reaction, 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) ; ∆n = (2 + 1) – 2 = 1
K_p = K_c (RT)^∆n
∴ K_c = \(\frac{\mathrm{K}_{\mathrm{p}}}{(\mathrm{RT})^{\Delta \mathrm{n}}}=\frac{1.8 \times 10^{-2}}{(0.0831 \times 500)}\) = 4.33 × 10^-4 (∵ R = 0.0831 L bar mol^-1K^-1)

ii) For the reaction, CaCO₃(s) ⇌ CaO(s) + CO₂(g) ; ∆n = 1 – 0 = 1
∵ K_c = \(\frac{\mathrm{K}_{\mathrm{p}}}{(\mathrm{RT})^{\mathrm{An}}}=\frac{167}{(0.0831 \times 1073)}\) = 1.873

Question 6.
For the following equilibrium, K_c = 6.3 × 10¹⁴ at 1000 K.
NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reactions ?
Answer:
K_c(reverse) = \(\frac{1}{\mathrm{~K}_{\mathrm{c}(\text { forward })}}\) = \(\frac{1}{3.3 \times 10^{14}}\) = 1.59 × 10^-15

Question 7.
Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression ?
Answer:
[Pure solid] or [Pure liquid] = \(\frac{\text { No.of moles }}{\text { Volume (in } \mathrm{L})}\) = \(\frac{\text { Mass } / \text { molecular mass }}{\text { Volume }}\)
= \(\frac{\text { Density }}{\text { Molecular mass }}\) = Constant
(The density of a pure solid or a pure liquid is constant at constant temperature)

Question 8.
Reaction between N₂ and O₂ takes place as follows.
2N₂(g) + O₂(g) ⇌ 2N₂O(g)
If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a 10L reaction vessel and allowed to form N₂O at a temperature for which K_c = 2.0 × 10^-37, determine the composition of equilibrium mixture.
Answer:
Suppose x moles of N₂ react till equilibrium is attained.

Since, the value of Kc for the reaction (K_c = 2.0 × 10^-37) is very small, the value of x must be very small. Therefore,
[N₂(g)] = \(=\frac{0.482-x}{10}=\frac{0.482}{10}\) = 0.0482 mol L^-1
[O₂(g)] = \(\frac{0.933-x / 2}{10}=\frac{0.933}{10}\frac{0.933-x / 2}{10}=\frac{0.933}{10}\) = 0.0933 mol L^-1 and [N₂O(g)] = \(\frac{\mathrm{x}}{10}\)
From K_c = \(\frac{\left[\mathrm{N}_2 \mathrm{O}(\mathrm{~g})\right]^2}{\left[\mathrm{~N}_2(\mathrm{~g})\right]^2\left[\mathrm{O}_2(\mathrm{~g})\right]}\)
We have, 2.0 × 10^-37 = \(\frac{(x / 10)^2}{(0.0482)^2 \times 0.0933}\)
Which gives, x = 6.6 × 10^-20
∴ [N₂O(g)] = \(\frac{x}{10}=\frac{6.6 \times 10^{-20}}{10}\) = 6.6 × 10^-21

Question 9.
Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below.
2NO(g) + Br₂(g) ⇌ 2NOBr(g)
When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br₂.
Answer:

Given that x = 0.0518
No. of moles of NO(g) at equilibrium = 0.087 – x = 0.087 – 0.518 = 0.0352
and number of moles of Br₂(g) at equilibrium = 0.0437 – \(\frac{\mathrm{x}}{2}\) = 0.0437 – \(\frac{0.0518}{2}\) = 0.0178

Question 10.
At 450 K, K_p = 2.0 × 10¹⁰ bar for the given reaction at equilibrium.
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
What is K_c at this temperature ?
Answer:
For the given reaction, ∆n = 2 – (2 + 1) = -1
K_p = K_c(RT)^∆n
⇒ K_c = \(\frac{\mathrm{K}_{\mathrm{p}}}{(\mathrm{RT})^{\Delta \mathrm{n}}}=\frac{2.0 \times 10^{10}}{(0.0831 \times 450)^{-1}}\) L mol^-1, (∵ R = 0.0831 L bar K^-1 mol^-1)
= 2.0 × 10¹⁰ × 0.831 × 450 L mol^-1
= 7.48 × 10¹¹ L mol^-1

Question 11.
A sample of HI(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is K_p for the given equilibrium ?
2HI(g) ⇌ H₂(g) + I₂(g)
Answer:

[∵ (0.2 – 0.04) is the decrease in the pressure of HI]
K_p = \(\frac{\mathrm{p}_{\mathrm{H}_2} \times \mathrm{p}_{\mathrm{I}_2}}{\mathrm{p}_{\mathrm{HI}}^2}=\frac{0.08 \times 0.08}{(0.04)^2}\) = 4.0

Question 12.
A mixture of 1.57 mol of N₂, 1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 1.7 × 10². Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction ?
Answer:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Q_c = \(\frac{\left[\mathrm{NH}_3(\mathrm{~g})\right]^2}{\left[\mathrm{~N}_2(\mathrm{~g})\right]\left[\mathrm{H}_2(\mathrm{~g})\right]^3}=\frac{(8.13 / 20)^2}{\left(\frac{1.57}{20}\right) \times\left(\frac{1.92}{20}\right)^3}\) = 2.38 × 10³ (V = 20 L)
Given that K_c = 1.7 × 10²
Obviously, Q_c ≠ K_c
Therefore, the reaction is not in equilibrium because Q_c ≠ K_c. It will proceed in the reverse direction.

Question 13.
The equilibrium constant expression for a gas reaction is, K_c = \(\frac{\left[\mathrm{NH}_3\right]^4\left[\mathrm{O}_2\right]^5}{[\mathrm{NO}]^4\left[\mathrm{H}_2 \mathrm{O}\right]^6}\)
Write the balanced chemical equation corresponding to this expression.
Answer:
4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g)

Question 14.
One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with CO according to the equation,
H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)
Calculate the equilibrium constant for the reaction.
Answer:
The given reaction is H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)
Initial number of moles of H₂O(g) = 1
Number of moles of H₂O(g) (reacted) = \(\frac{1 \times 40}{100}\) = 0.4
∴ Number of moles of H₂O(g) left (unreacted) = 1 – 0.4 = 0.6
From the given equation it is clear that 0.4 moles of H₂O(g) react with 0.4 moles of CO(g) to form 0.4 moles of H₂(g) and 0.4 moles of CO₂(g). The total volume is 10 litres.
∴ At equilibrium,
[H₂O(g)] = \(\frac{1-0.4}{10}\) = \(\frac{0.6}{10}\) = 0.06 mol L^-1 ; [CO(g)] = \(\frac{1-0.4}{10}\) = \(\frac{0.6}{10}\) = 0.06 mol L^-1
[H₂(g)] = \(\frac{0.4}{10}\) = 0.04 mol L^-1 ; [CO₂(g)] = \(\frac{0.4}{10}\) = 0.04 mol L^-1
Hence, K = \(\frac{\left[\mathrm{H}_2(\mathrm{~g})\right]\left[\mathrm{CO}_2(\mathrm{~g})\right]}{\left[\mathrm{H}_2 \mathrm{O}(\mathrm{~g})\right][\mathrm{CO}(\mathrm{~g})]}=\frac{0.04 \times 0.04}{0.06 \times 0.06}\) = 0 44

Question 15.
At 700 K, equilibrium constant for the reaction :
H₂(g) + I₂(g) ⇌ 2HI(g)
is 54.8. If 0.5 mol L_1 of HI(g) js present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K ?
Answer:
The given reaction can be written as 2HI(g) ⇌ H₂(g) + I₂(g)
For this reaction, K = \(\frac{1}{54.8}\) = 1.82 × 10^-2 (∵ because for the reverse reaction, K = 54.8)
As the number of moles of H₂ and N₂ are equal, their concentrations at equilibrium will also be equal.
Suppose [H₂(g)] = [I₂(g)] = x mol L^-1
Given that [HI(g)] = 0.5 mol L^-1
K = \(\frac{\left[\mathrm{H}_2(\mathrm{~g})\right]\left[\mathrm{I}_2(\mathrm{~g})\right]}{[\mathrm{HI}(\mathrm{~g})]^2}\) ⇒ 1.82 × 10^-2 = \(\frac{x \times x}{(0.5)^2}\) ⇒ x = [1.82 × 10^-2 × (0.5)²]^1/2 = 0.068 mol L^-1
Hence, at equilibrium, [H₂(g)] = [I₂(g)] = 0.068 mol L^-1

Question 16.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial, concentration of ICl was 0.78 M ?
2ICl(g) ⇌ I₂(g) + Cl₂(g); K_c = 0.14
Answer:
Suppose at equilibrium, [I₂(g)] = [Cl₂(g>] = x mol L^-1

Given that, K_c = 0.14
K_c = \(\frac{\left[\mathrm{I}_2(\mathrm{~g})\right]\left[\mathrm{Cl}_2(\mathrm{~g})\right]}{[\mathrm{ICl}(\mathrm{~g})]^2}\)
⇒ 0.14 = \(\frac{x \times x}{(0.78-2 x)^2}\) ⇒ x² = 0.14(0.78 -2x)²
⇒ x = \(\sqrt{0.14}\) (0.78 – 2x) = 0.374(0.78 – 2x)
⇒ x = 0.292 – 0.748 x
⇒ 1.748 x = 0.292
x = \(\frac{0.292}{1.748}\) = 0.167
Hence, equilibrium concentrations will be as follows.
[I₂(g)] – [Cl₂(g)] = x = 0.167 M
[ICl(g)] = 0.78 – 2x = 0.78 – (2 × 0.167) = 0.446 M

Question 17.
K_p = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
C₂H₆(g) ⇌ C₂H₄(g) + H₂(g)
Answer:
Suppose the partial pressure of C₂H₆ decreases by p atm till equilibrium is attained.

K_p = \(\frac{\mathrm{p}_{\mathrm{C}_2 \mathrm{H}_4} \cdot \mathrm{p}_{\mathrm{H}_2}}{\mathrm{p}_{\mathrm{C}_2 \mathrm{H}_6}}\) = \(\frac{\mathrm{p} \times \mathrm{p}}{4.0-\mathrm{p}}\) ⇒ 0.04 = \(\frac{\mathrm{p}^2}{4.0-\mathrm{p}}\) = 0.16 – 0.04 p ⇒ p² + 0.04 p – 0.16 = 0
p = \(\frac{-0.04 \pm \sqrt{(0.04)^2-4 \times 1 \times(-0.16)}}{2 \times 1}=\frac{-0.04 \pm 0.801}{2}\)
Taking positive value, p = 0.38
∴ [C₂H₆(g)]_eq = 4.0 – p = 4.0 – 0.38 = 3.62 atm

Question 18.
Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as,
CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)
i) Write the concentration ratio (reaction quotient), Q_c, for this reaction (note : water is not in excess and is not a solvent in this reaction)
ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
Answer:
i) Q_c = \(\frac{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(l)\right]\left[\mathrm{H}_2 \mathrm{O}(l)\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}(l)\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)\right]}\)

ii)

If the volume of the reaction mixture is V L, we have at equilibrium,
[CH₃COOH(l)] = \(\frac{0.829}{\mathrm{V}}\) mol L^-1 ; [C₂H₅OH(l)] = \(\frac{0.009}{\mathrm{V}}\) mol L^-1
[CH₃COOC₂H₅(l)] = \(\frac{0.171}{\mathrm{V}}\) mol L^-1 ; [H₂O(l)] = \(\frac{0.171}{\mathrm{V}}\) mol L^-1
K_c = \(\frac{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(l)\right]\left[\mathrm{H}_2 \mathrm{O}(l)\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}(l)\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)\right]}=\frac{\frac{0.171}{\mathrm{~V}} \times \frac{0.171}{\mathrm{~V}}}{\frac{0.829}{\mathrm{~V}} \times \frac{0.009}{\mathrm{~V}}}=\frac{0.171 \times 0.171}{0.829 \times 0.009}\) = 3.92

iii)

∴ Q_c = \(\frac{\frac{0.214}{V} \times \frac{0.214}{V}}{\frac{0.786}{V} \times \frac{0.286}{V}}\) = 0.204
Since, Q_c < K_c, equilibrium has not been reached.

Question 19.
A sample of pure PCl₅, was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 × 10^-1 mol L^-1. If value of K_c is 8.3 × 10^-5 what are the concentrations of PCl₃ and Cl₂ at equilibrium ?
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
Answer:
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
Since, the number of moles of PCl₃(g) and Cl₂(g) are equal, their concentrations at equilibrium will also be equal.
Suppose, at equilibrium, [PCl₃(g)] = [Cl₂(g)] = x mol L^-1
Given that at equilibrium, [PCl₅] = 0.5 × 10^-1 mol L^-1
∴ K_c = \(\frac{\left[\mathrm{PCl}_3(\mathrm{~g})\right]\left[\mathrm{Cl}_2(\mathrm{~g})\right]}{\left[\mathrm{PCl}_5(\mathrm{~g})\right]}\) ⇒ 8.3 × 10^-3 = \(\frac{\mathrm{x} \times \mathrm{x}}{0.5 \times 10^{-1}}\)
⇒ x = (8.3 × 10^-3 × 0.5 × 10^-1)^1/2 = 0.02
Hence, [PCl₃(g)] = [Cl₂(g)] = 0.02 mol L^-1.

Question 20.
One of the reaction that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO₂.
FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g); K_p = 0.265 atm at 1050 K
What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial partial pressures are : P _CO = 1.4 atm and P_CO2 = 0.80 atm?
Answer:

Q_p = \(\frac{\mathrm{P}_{\mathrm{CO}_2}}{\mathrm{P}_{\mathrm{CO}}}=\frac{0.80}{1.4}\) = 0.571
Since, Q_p > K_p, the reaction will proceed in the reverse direction. On account of this partial pressure of CO₂ will decrease while that of CO will increase.
Suppose, the decrease in the partial pressure of CO₂ is p atm.
∴ At equilibrium, P_CO2 = (0.80 – p) atm and P_CO = (1.4 + p) atm
∵ K_p = \(\frac{\mathrm{P}_{\mathrm{CO}_2}}{\mathrm{P}_{\mathrm{CO}}}\), we have
0.265 = \(\frac{0.80-\mathrm{p}}{1.4+\mathrm{p}}\) ⇒ 0.265 (1.4 + p) = 0.80 – p
⇒ 0.371 + 0.265 p = 0.80 – p
⇒ 1.265 p = 0.429 ⇒ p = \(\frac{0.429}{1.265}\) = 0.339 atm
∴ P _CO at equilibrium = 1.4 + 0.339 = 1.739 atm.
and P_CO2 at equilibrium = 0.80 – 0.339 = 0.461 atm.

Question 21.
Equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mole L^-1 N₂(g) 2.0 mol L^-1 H₂ and 0.5 mol L^-1 NH₃. Is the reaction at equilibrium ? If not in which direction does the reaction tend to proceed to reach equilibrium ?
Answer:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ; K_c = 0.061
For the given set of conditions, Q_c = \(\frac{\left[\mathrm{NH}_3(\mathrm{~g})\right]^2}{\left[\mathrm{~N}_2(\mathrm{~g})\right]\left[\mathrm{H}_2(\mathrm{~g})\right]^3}=\frac{(0.5)^2}{(3.00) \times(2.00)^3}\) = 0.0104
Since, Q_c < K_c, the reaction is not at equilibrium. It will proceed in the forward direction.

Question 22.
Bromine monochioride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
2BrCl(g) ⇌ Br₂(g) + Cl₂(g)
for which K_c = 32 at 500 K. If initially pure BrCl is present at a concentration 3.3 × 10^-3 mol L^-1, what is its molar concentration in the mixture at equilibrium?
Answer:

K_c = \(\frac{\left[\mathrm{Br}_2(\mathrm{~g})\right]\left[\mathrm{C} l_2(\mathrm{~g})\right]}{[\mathrm{BrCl}(\mathrm{~g})]^2}\) ⇒ 32 = \(\frac{\frac{x}{2} \times \frac{x}{2}}{\left(3.30 \times 10^{-3}-x\right)^2}\)
Taking square root on both sides, we get
5.66 = \(\frac{\frac{x}{2}}{3.30 \times 10^{-3}-x}=\frac{x}{2\left(3.30 \times 10^{-3}-x\right)}\)
⇒ 0.037 – 11.32 x = x ⇒ (1 + 11.32)x = 0.037 ⇒ x = \(\frac{0.037}{1+11.32}\) = 3.0 × 10^-3
∴ Equilibrium centration of BrCl = 3.30 × 10^-3 – x
= 3.30 × 10^-3 – 3.0 × 10^-3
= 3.0 × 10^-4 mol L^-1

Question 23.
At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass, C(s) + CO₂(g) ⇌ 2CO(g)
Calculate K_c for this reaction at the above temperature.
Answer:
If the total mass of the gaseous mixture at equilibrium is 100 g then
mass of CO(g) = 90.55 g
and mass of CO₂(g) = 100 – 90.55 = 9.45 g
Number of moles of COg) = \(\frac{90.55}{28}\) = 3.234,
(∵ molar mass of CO = 28 g/mol, molar mass of CO₂ = 44 g/mol)
and number of moles of CO₂(g) = \(\frac{9.45}{44}\) = 0.215
Total number of gaseous moles = 3.234 + 0.2 15 = 3.449
∴ P_CO = \(\frac{3.234}{3.449}\) × 1 = 0.938 atm (∵ Partial pressure = \(\frac{\text { No.of moles }}{\text { Total no.of moles }}\) × Total pressure)
and P_CO2 = \(\frac{0.215}{3.449}\) × 1 = 0.062 atm
Hence, for the reaction C(s) + CO₂(g) ⇌ 2CO(g)
K_p = \(\frac{\mathrm{p}{\mathrm{CO}}^2}{\mathrm{p}{\mathrm{CO}_2}}=\frac{(0.938)^2}{0.062}\) = 14.19
For the above reaction, ∆n = 2 – 1 = 1
K_p = K_p(RT)^∆n ⇒ K_c = \(\frac{\mathrm{K}_{\mathrm{p}}}{(\mathrm{RT})^{\mathrm{An}}}=\frac{14.19}{(0.0821 \times 1127)^1}\) = 0.153

Question 24.
Calculate (i) ∆G° and (ii) the equilibrium constant for the formation of NO₂ from NO and O₂ at 298
NO(g) + 1/2O₂(g) ⇌ NO₂(g)
where ∆_fG°(NO₂) = 52.0 kJ/mol ; ∆_fG°(NO) = 87,0 kJ/mol ; ∆_fG°(O₂) = 0 kJ/mol
Answer:
i) ∆_rG° = ∆_fG° (products) – ∆_fG° (reactants)
or = ∆_rG° [NO₂] – {∆_fG° [NO] + ½∆_fG° [O₂])
or ∆_r = 52.0 – (87.0 + 0) = -35.0 kJ mol ^-1

ii) ∆_rG° = -2.303 RT log10K
⇒ log₁₀K = –\(\frac{\Delta_{\mathrm{r}} G^{\circ}}{2.303 \mathrm{RT}}\) = \(\frac{(-35.0)}{-2.303 \times 8.314 \times 10^{-3} \times 298}\) = 6.1341
(∵ R = 8.314 × 10^-3 kJ K^-1 mol^-1)
⇒ K = antilog₁₀6.134l = 1.362 × 10⁶

Question 25.
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease In pressure by increasing the volume?
i) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
ii) CaO(s) + CO₂(g) ⇌ CaCO₃(s)
iii) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)
Answer:
According to Le Chateliers principle, if the pressure on the system is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more.
i) No.of moles of products will increase on decreasing pressure as the forward reaction will be favoured.
ii) No.of moles of products will decrease as the backward reaction will be favoured.
iii) No.of moles of products will remain the same as the number of moles of gaseous substances is the same on both the sides.

Question 26.
Which of the following reactions will get affected by increasing the pressure ? Also, mention whether change will cause the reaction to go into forward or backward direction.
i) COCl₂(g) ⇌ CO(g) + Cl₂(g)
ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g)
iii) CO₂(g) + C(s) ⇌ 2C0(g)
iv) 2H₂(g) + CO(g) ⇌ CH₃OH(g)
v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)
Answer:
i) Reaction will go in the backward direction as n_P > n_R
ii) Reaction will remain unaffected as n_P = n_R.
iii) Reaction will go in the backward direction as n_P > n_R
iv) Reaction will go in the forward direction as n_P < n_R
v) Reaction will go in the backward direction as n_P > n_R
vi) Reaction will go in the backward direction as n_P > n_R.

Question 27
The equilibrium constant for the following reaction is 1.6 × 10⁵ at 1024 K
H₂(g) + Br₂(g) ⇌ 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.
Answer:
For the reaction

For the above reaction, K = \(\frac{1}{1.6 \times 10^5}\) = 6.25 × 10^-6
Further, K = K = 6.25 × 10^-6 ⇒ K_p = \(\frac{\mathrm{p}{\mathrm{H}_2} \cdot \mathrm{p}{\mathrm{Br}2}}{\mathbf{p}{\mathrm{HBr}}^2}\)
Hence, 6.25 × 10^-6 = \(\frac{\frac{\mathrm{p}}{2} \cdot \frac{\mathrm{p}}{2}}{(10.0-\mathrm{p})^2}\)
Taking square mot on both sides, we have, 2.5 × 10^-3 = \(\frac{\frac{p}{2}}{10.0-p}=\frac{p}{2(10.0-p)}\)
⇒ 0.05 – 5.0 × 10^-3 p ⇒ p = p(1 +5.0 × 10^-3)= 0.05 ⇒ p = \(\frac{0.05}{1+5.0 \times 10^{-3}}\) = 4.98 × 10^-2 bar
Hence, at equilibrium, P_H2 = P_Br2 = \(\frac{p}{2}=\frac{4.98 \times 10^{-2}}{2}\) = 2.49 × 10^-2 bar
and P_HBr = (10.0 – p) = 10.0 – 4.98 × 10^-2 = 9.95 bar

Question 28.
Dihydrogen gas ¡s obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)
i) Write an expression for K_p for the above reaction.
ii) How will the values of K_p and composition of equilibrium mixture be affected by
a) increasing the pressure b) increasing the temperature c) using a catalyst?
Answer:
i) K_p = \(\frac{\mathrm{p}{\mathrm{CO}} \times \mathrm{p}{\mathrm{H}2}^3}{\mathrm{p}{\mathrm{CH}4} \times \mathrm{p}{\mathrm{H}_2 \mathrm{O}}}\)
ii) a) Equilibrium will shift in the backward direction as n_p > n_R.

b) Equilibrium will shift in the forward direction because the reaction is endothermic.

c) Composition of equilibrium mixture will remain unchanged. However, the equilibrium will be attained earlier,
The value of K_p will remain unchanged in (a) and (C). K_p will increase in (b).

Question 29.
Describe the effect of:
i) addition of H₂
ii) addition of CH₃OH
iii) removal of CO
iv) removal of CH₃OH
on the equilibrium of the reaction:
2H₂(g) + CO(g) ⇌ CH₃OH(g)
Answer:
i) Equilibrium will shift in the forward direction.
ii) Equilibrium will shift in the backward direction.
iii) Equilibrium will shift in the backward direction.
iv) Equilibrium will shift in the forward direction.

Question 30.
At 473 K, equilibrium constant K_c for decomposition of phosphorus pentachioride, PCl₅ is 8.3 × 10^-3. If decomposition is depicted as,
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); ∆_rH° = 124.0 kJ mol^-1
i) Write an expression for K_c for the reaction.
ii) What is the value of K_c for the reverse reaction at the same temperature?
iii) What would be the effect on K_c if (a) more PCl₅ is added (b) pressure is increased (c) the temperature is increased?
Answer:
i) K_c = \(\frac{\left[\mathrm{PCl}_3(\mathrm{~g})\right]\left[\mathrm{Cl}_2(\mathrm{~g})\right]}{\left[\mathrm{PCl}_5(\mathrm{~g})\right]}\)
ii) K’ = \(\frac{1}{K_c}=\frac{1}{8.3 \times 10^{-3}}\) = 120.48
iii) a) M constan temperature, the value of K_c is not affected by addition of reactants.
Therefore, K_c will remain unchanged.
b) K_c will remain unchanged.
c) Since, K_c = \(\frac{K_1}{K_2}\), an increase in temperature will favour the forward reaction as the reaction is endothermic. Thus, K₁ will Increase. Hence, the value of K_c will increase or increasing temperature.

Question 31.
Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)
If a reaction vessel at 4000 C is charged with an equimolar mixture of CO and steam such that p_CO = p_H2O = 4.0 bar, what will be the partial pressure of H₂ at equilibrium? K_p = 10.1 at 400°C.
Answer:
Suppose the partial pressure of H₂ at equilibrium is p bar.

∴ K_p = \(\frac{\mathrm{p}_{\mathrm{CO}_2} \cdot \mathrm{p}_{\mathrm{H}_2}}{\mathrm{p}_{\mathrm{CO}} \cdot \mathrm{p}_{\mathrm{H}_2 \mathrm{O}}}\)
⇒ 10.1 = \(\frac{p \times p}{(4.0-p) \times(4.0-p)}=\left(\frac{p}{4.0-p}\right)^2\) ⇒ \(\frac{p}{4.0-p}\) = \(\sqrt{10.1}\) = 3.178
⇒ p = 3.178 (40 – p) ⇒ p = 12.712 – 3.178 p ⇒ 3.178 + p = 12.712 bar
∴ p = 12.712/4.178 bar
p = 3.04 bar

Question 32.
Predict which of the following reaction will have appreciable concentration of reactants and products.
i) Cl₂(g) ⇌ 2Cl(g) ; K_c = 5 × 10^-39
ii) Cl₂(g) + 2NO(g) ⇌ 2NOCl(g) ; K_c = 3.7 × 10⁸
iii) Cl₂(g) + 2NO₂(g) ⇌ 2NO₂Cl(g) ; K_c = 1.8
Answer:
i) For reaction (a), K_c is very low. Therefore, the concentration of products will be very low.
ii) For reaction (b), K_c is quite high. Therefore, the reaction will proceed almost to completion and there will be no appreciable concentrations of reactants.
iii) The reaction (c) has a moderate value of K_c and will have appreciable concentrations of both reactants and products.

Question 33.
The value of K_c for the reaction 3O₂(g) ⇌ 2O₃(g) is 2.0 × 10^-50 at 25°C. If the equilibrium concentration of O₂ in air at 25°C is 1.6 × 10^-2, what is the concentration of O₃ ?
Answer:
K_c = \(\frac{\left[\mathrm{O}_3(\mathrm{~g})\right]^2}{\left[\mathrm{O}_2(\mathrm{~g})\right]^3}\) ⇒ 2.0 × 10^-30 = \(\frac{\left[\mathrm{O}_3(\mathrm{~g})\right]^2}{\left(1.6 \times 10^{-2}\right)^3}\)
⇒ [O₃(g)]² = 2.0 × 10^-50 × (1.6 × 10^-2)³ = 8.192 × 10-56
∴ [O₃(g)] = (8.192 × 10-56)¹⁰ = 2.86 × 10-28 M

Question 34.
The reaction, CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g) is at equilibrium at 1300 K in a 1 L flask. It also contain 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, K_c for the reaction at the given temperature is 3.90.
Answer:
K_c = \(\frac{\left[\mathrm{CH}_4(\mathrm{~g})\right]\left[\mathrm{H}_2 \mathrm{O}(\mathrm{~g})\right]}{[\mathrm{CO}(\mathrm{~g})]\left[\mathrm{H}_2(\mathrm{~g})\right]^3}\)
∴ 3.90 = \(\frac{\left[\mathrm{CH}_4(\mathrm{~g})\right] \times 0.02}{0.30 \times(0.10)^3}\) (V = 1 L)
⇒ [CH₄(g)] = \(\frac{3.90 \times 0.30 \times(0.10)^3}{0.02}\) = 0.0585 = 5.85 × 10^-2 mol L^-1

Question 35.
What is meant by the conjugate acid-base pair ? Find the conjugate acid/base for the following species.
HNO₂, CN^–, HClO₄, F^–, OH^–, \(\mathrm{CO}_3^2\) and S^2-
Answer:
A conjugate acid-base pair that differs only by one proton. Conjugate pairs of the given species are NO₂^–, HCN, ClO₄^–, HF, H₂O, HCO₃^– and HS^– respectively.

Question 36.
Which of the followings are Lewis acids ? H₂O, BF₃, H⁺, and \(\mathrm{NH}_4^{+}\)?
Answer:
BF₃, H⁺ and \(\mathrm{NH}_4^{+}\).

Question 37.
What will be the conjugate bases for the Bronsted acids : HF, H₂SO₄ and \(\mathrm{HCO}_3^{-}\) ?
Answer:
F^–, HSO₄^– and CO₃^2-
(Conjugate base = Conjugate acid – H⁺)

Question 38.
Write the conjugate acids for the following Bronsted bases : NH₂^–, NH₃ and HCO₃^–.
Answer:
NH₃, NH₄⁺, HCOOH (Conjugate acid = Conjugate base + H⁺)

Question 39.
The species : H₂O, HCO₃^–, HS₄^– and NH₃ can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.
Answer:

Question 40.
Classify the following species into Lewis acids and Lewis bases and show these act as Lewis acid/base :
(a) OH^–
(b) F^–
(c) H⁺
(d) BCl₃.
Answer:
a) OH^– is a Lewis base because it can donate an electron pair.
b) F^– is a Lewis base because it can donate electron pair.
c) H⁺ is a Lewis acid because it can accept an electron pair.
d) BCl₃ is a Lewis acid because it is an electron deficient molecule and can accept an electron pair.

Question 41.
The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^-3 M. What is its pH?
Answer:
pH = -log₁₀(H⁺) = -log₁₀(3.8 × 10^-3) = 2.42

Question 42.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Answer:
log₁₀[H⁺] = -3.76
[H⁺] = antilog₁₀(-3.76) = antilog₁₀\(\overline{4} .24\) = 1.74 × 10^-4 M

Question 43.
The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 × 10^-4, 1.8 × 10^-4 and 4.8 × 10^-9 respectively. Calculate the ionization constants of the corresponding conjugate base.
Answer:
i) The conjugate base of HF is F^–. It will react with water as follows.
F^– + H₂O ⇌ HF + OH^–
Obviously, this is the equation of hydrolysis of a salt of HF with a strong base. Hence,
K_F = K_h = \(\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{~K}_{\mathrm{a}}}=\frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}}\) = 1.5 × 10^-11

ii) Similarly, for HCOOH, K_HCOO^– = K_h = \(\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{~K}_{\mathrm{a}}}=\frac{1.0 \times 10^{-14}}{1.8 \times 10^{-9}}\) = 5.6 × 10^-11

iii) For HCN, K_CN^– = K_h = \(\frac{K_w}{K_a}=\frac{1.0 \times 10^{-14}}{4.8 \times 10^{-9}}\) = 2.1 × 10^-6

Question 44.
The ionization constant of phenol is 1.0 × 10^-10. What is the concentration of phenolate ion is 0.05 M solution of phenol ? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate ?
Answer:

∴ K_a = \(\frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\right]}=\frac{\mathrm{x} . \mathrm{x}}{0.05-\mathrm{x}}\) = 1.0 × 10^-10(given)
⇒ \(\frac{x^2}{0.05-x}\) = 1.0 × 10^-10
Since, phenol is not much dissociated, 0.05 – x ≈ 0.05.
Hence, \(\frac{x^2}{0.05}\) = 1.0 × 10^-10 ⇒ x = (0.05 × 1.0 × 10^-10)^1/2 = 2.24 × 10^-6
Hence, [C₆H₅O^–] in solution = x = 2.24 × 10^-6 M

In the presence of 0.01 M of C₆H₅ONa :
Suppose y is the amount of phenol dissociated. Therefore, at equilibrium
[C₆H₅OH] – 0.05 – y, [C₆H₅O^–] = 0.01 + y and [H⁺] = y
∴ K_a = \(\frac{(0.01+\mathrm{y})(\mathrm{y})}{(0.05-\mathrm{y})}\) = 1.0 × 10^-10 (given)
Since, in the presence of sodium phenolate, the dissociation of phenol will get further suppressed we can take
0.01 + y ≈ 0.01 and 0.05 – y ≈ 0.05
Hence, \(\frac{0.01+\mathrm{y}}{0.05}\) = 1.0 × 10^-10 ⇒ y = \(\frac{1.0 \times 10^{-10} \times 0.05}{0.01}\) = 5.0 × 10^-10
Hence, degree of dissociation of phenol
α = \(\frac{\text { Number of moles dissociated }}{\text { Total no.of moles initially taken }}=\frac{5.0 \times 10^{-10}}{0.05}\) = 1.0 × 10^-10

Question 45.
The first ionization constant of H₂S is 9.1 × 10^-8. Calculate the concentration of HS^– ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H₂S is 1.2 × 10^-13, calculate the concentration of Sounder both conditions.
Answer:
Calculation of [HS^–]
H₂S ionises in two stages. The first stage of ionisation is as follows.

K_a1 = 9.1 × 10^-8 (given)
According to the law of equilibrium, K_a1 = \(\)
⇒ 9.1 × 10^-8 = \(\frac{x \cdot x}{0.1-x}\) = \(\frac{x^2}{0.1}\) (as x is very small so 0.1 – x = 0.1)
⇒ x = (9.1 × 10^-8 × 0.1)^1/2 = 9.54 × 10^-5
Hence, [HS^–] in 0.1 M solution = x = 9.54 × 10^-5.
In the presence of 0.1 M HCl, suppose H₂S dissociates to an extent of y M.
Therefore, [H₂S] = 0.1 – y ≈ 0.1M, [H⁺] = y + 0.1 ≈ 0.1M and [HS^–] = y M
(Since, in the presence of HCl, the dissociation of H₂S will further get suppressed)
∴ K_a1 = \(\frac{0.1 \times \mathrm{y}}{0.1}\) = 9.1 × 10^-8 ⇒ y = 9.1 × 10^-8M
Hence, in the presence of 0.1 M of HCl,
[HS^–] = y = 9.1 × 10^-8M
Calculation of [S^2-]

The overall dissociation can be expressed as follows.
H₂S ⇌ 2H⁺ + S^2-
K_a = \(\frac{\left[\mathrm{H}^{+}\right]^2\left[\mathrm{~S}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{~S}\right]}\)
K_a = K_a1 × K_a2 = 9.1 × 10^-8 × 1.2 × 10^-13 = 1.1 × 10^-20

In the absence of 0.1 M of HCl:

⇒ x’ = \(\left(\frac{1.1 \times 10^{-20} \times 0.1}{4}\right)^{1 / 3}\) = 6.5 × 10^-8 M
Hence, [S^2-] in the absence of 0.1 M of HCl = 6.5 × 10^-8 M

In the presence of 0.1 M of HCl:
Suppose H₂S dissociates to an extent of y’ M
∴ [H₂S] = 0.1M – y’ ≈ 0.1 M, [H⁺] = 2y’ + 0.1 ≈ 0.1 M and [S^2-] = y
∴ K_a = \(\frac{(0.1)^2 \times y^{\prime}}{0.1}\) = 1.1 × 10^-20 ⇒ y’ = \(\frac{1.1 \times 10^{-20}}{0.1}\) = 1.1 × 10^-19 M
Hence, [S^2-] in the presence of 0.1 M of HCl = 1.1 × 10^-19 M

Question 46.
The ionization constant of acetic acid is 1.74 × 10^-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.
Answer:

According to the law of equilibrium,
⇒ K = \(\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}\) ⇒ 1.74 × 10^-5 = \(\frac{x . x}{0.05-x}=\frac{x^2}{0.05}\) (∵ x is very small)
⇒ x = (1.74 × 10^-5 × 0.05)^1/2 = 9.33 × 10^-4 M
[CH3COO^–] = x = 9.33 × 10^-4 mol L^-1
[H⁺] = x = 9.33 × 10^-4 mol L^-1
∴ pH = -log₁₀[H⁺] = -log₁₀(9.33 × 10^-4) = 3.03

Question 47.
It has been found that the pH of a 0.01 M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK_a.
Answer:
Suppose the organic acid is HA. HA ⇌ H⁺ + A^–
pH = -log₁₀[H⁺]
4.15 = -log₁₀[H⁺]
∴ [H⁺] = antilog10(-4.15) = 7.08 × 10^-5 M
[A] = [H⁺] = 7.08 × 10^-5 M
∴ K_a = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)
K_a = \(\frac{7.08 \times 10^{-5} \times 7.08 \times 10^{-5}}{0.01}\) = 5.01 × 10^-7 (∵ Dissociation of the acid is very low)
pK_a = -log₁₀ K_a = -log₁₀(5.01 × 10^-7) = 6.3002

Question 48.
Assuming complete dissociation, calculate the pH of the following solutions :
i) 0.003 M HCl
ii) 0.005 M NaOH
iii) 0.002 M HBr
iv) 0.002 M KOH
Answer:
i) [H₃O⁺] = [HCl] – 0.003 mol L^-1
∴ pH = -log₁₀[H₃O⁺] = -log₁₀ (0.003) = 2.52

ii) [OH^–] = [NaOH] = 0.005
∴ [H₃O⁺] = \(\frac{\mathrm{K}_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{0.005}\) = 20 × 10^-12
pH = -log₁₀[H₃O⁺] = -log₁₀(2.0 × 10^-12) = 11.7

iii) [H₃O⁺] = [HBr] = 0.002 mol L^-1
∴ pH = -log₁₀[H₃O⁺] = -log₁₀(0.002) = 2.70

iv) [OH^–] = [KOH] = 0.002 mol L^-1
∴ [H₃O⁺] = \(\frac{\mathrm{K}_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{0.002}\) = 5.0 × 10^-12
pH = -log₁₀[H₃O⁺] = -log₁₀(5.0 × 10^-12) = 11.3

Question 49.
Calculate the pH of the following solutions.
i) 2 g of TlOH dissolved in water to give 2 litres of solution.
ii) 0.3 g of Ca(OH)₂ dissolved in water to give 500 mL of solution.
iii) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
iv) 1 mL of 13.6 M HCl is diluted with water to give 1 litre of solution,
Answer:
i)

iv) Suppose the molarity of the diluted solution is M’.
M₁V₁ = M’V’ ⇒ 13.6 × 1 = M’ × 1000
⇒ M’ = \(\frac{13.6 \times 1}{1000}\) = 1.36 × 10^-2
∴ [H₃O⁺] = 1.36 × 10^-2 M
pH = -log₁₀[H₃O⁺] = -log₁₀(1.36 × 10^-2) = 1.87

Question 50.
The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the PK. of bromoacetic acid.
Answer:

∴ K_a = \(\frac{\left[\mathrm{BrCH}_2 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{BrCH}_2 \mathrm{COOH}\right]}=\frac{(0.1 \times 0.132) \times(0.1 \times 0.132)}{0.1(1-0.132)}\) = 2.01 × 10^-3
pK_a = -log₁₀K_a = -log₁₀(2.01 × 10^-3) = 2.70
[H⁺] = 0.1 × 0.132 = 0.0132 = 1.32 × 10^-2
pH = -log₁₀[H⁺] = – log₁₀(1.32 × 10^-2) = 1.88

Question 51.
The pH Of 0.005 M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionization constant as pK_b.
Answer:
Codeine is a base, therefore, it is expected to ionise in solution as follows.
Codeine + H₂ ⇌ Codeine H⁺ + OH^–
Given, pH = 9.95
∴ POH = 14 – 9.95 = 4.05
i.e., -log₁₀[OH^–] = 4.05
[OH^–] = antilog₁₀(-4.05) = 8.91 × 10^-5
[Codeine H⁺] = [OH^–] = 8.91 × 10^-5
[Codeine] = 0.005 M
K_b = \(\frac{\left[\text { Codeine } \mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\text {Codeine }]}=\frac{\left(8.91 \times 10^{-5}\right) \times\left(8.91 \times 10^{-5}\right.}{0.005}\) = 1.587 × 10^-5
[Codeine W] [OH ]
and pK_b = -log₁₀K_b = -log₁₀(1.587 × 10^-6) = 5.8

Question 52.
What is the pH of 0.001 M aniline solution ? The ionization constant of aniline can be taken as 4.27 × 10^-10. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Answer:
K_b for aniline = 4.27 × 10^-10
C₆H₅NH₂ + H₃O⁺ ⇌ C₆H₅NH₃⁺ + OH^–
K_b = \(\frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_3^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\right]}\)
Since, [C₆H₅NH₃⁺] = [OH^–] and dissociation of aniline is not much, we have
and
[OH^–] = {K_b[C₆H₅NH₂]}^1/2 = (4.27 × 10^-10 × 0.00)^1/2 = 6.53 × 10^-7 M
∴ POH = -log₁₀(6.53 × 10^-7) = 6.185
and pH = 14 – 6.185 = 7.815

∴ K_b = \(\frac{0.001 \alpha \times 0.001 \alpha}{0.001(1-\alpha)}=\frac{0.001 \alpha^2}{1-\alpha}\) = 0.001 α² (Since, α is very small, 1 – α = 1)
Thus, 0.001 α² = 4.27 × 10^-10 ⇒ α = \(\left(\frac{4.27 \times 10^{-10}}{0.001}\right)^{1 / 2}\) = 6.53 × 10^-4
For a pair of conjugate acid and base,
pK_b + pK_a = 14 ⇒ pK_a = 14 – pK_b = 14 – (-log₁₀ 4.27 × 10^-10) = 14 – 9.37 = 4.63
Hence, the ionisation constant of the conjugate acid.
K_a = antilog₁₀(-pK_a) (∵ pK_a = -log10K_a)
= antilog₁₀ (-4.63) = 2.34 × 10^-5

Question 53.
Calculate the degree of ionization of 0.05 M acetic acid if its pK_a value is 4.74. How is the degree of dissociation affected when its solution also contains
i) 0.01 M
ii) 0.1 M of HCl?
Answer:
pK_a = -log₁₀K_a = 4.74
∴ K_a = antilog₁₀(-4.74) = 1.82 × 10^-5
For a weak electrolyte, α = \(\sqrt{\frac{K_a}{c}}\); For acetic acid, α = \(\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{c}}}=\sqrt{\frac{1.82 \times 10^{-5}}{0.05}}\) = 1.908 × 10^-2
In the presence of HCl, the dissociation of acetic acid will be suppressed due to common ion effect.

i) When the solution contains 0.01 M of HCl also : Suppose the concentration of CH₃COOH decreases by x.

K_a = \(\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}\) ⇒ 1.80 × 10^-5 = \(\frac{x(x+0.01)}{(0.05-x)}=\frac{0.01 x}{0.05}\)
(∵ Since, x is very small, x + 0.01 = 0.01 and 0.05 – x = 0.05)
⇒ x = \(\frac{1.82 \times 10^{-5} \times 0.05}{0.01}\) = 9.1 × 10^-5
Degree of dissociation (α) = \(\frac{\text { Amount dissociated }}{\text { Initial amount }}=\frac{9.1 \times 10^{-5}}{0.05}\) = 1.85 × 10^-5

ii) When the solution contains 0.1 M of HCl:
Proceeding as above, we have, α’ = \(\frac{1.82 \times 10^{-5} \times 0.05}{0.1}\) = 9.1 × 10^-4 and
α’ = \(\frac{9.1 \times 10^{-6}}{0.05}\) = 1.82 × 10^-4

Question 54.
The ionization constant of dimethylamine is 5.4 × 10^-4. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.11 in NaOH?
Answer:
Dimethylamine is weak base. For a weak base, α = \(\sqrt{\frac{K_b}{c}}\)
Therefore, for dimethylamine, α = \(\sqrt{\frac{5.4 \times 10^{-4}}{0.02}}\) = 0.164
In presence of 0.1 M of NaOH, its dissociation will get suppressed.

∴ K_a = \(\frac{x(x+0.1)}{0.02-x}=\frac{0.1 x}{0.02}\) (∵ x is very small, x + 0.1 ≈ 0.1 and 0.02 – x ≈ 0.02)
⇒ 5.4 × 10^-4 = \(\frac{0.1 x}{0.02}\) ⇒ x = \(\frac{5.4 \times 10^{-4} \times 0.02}{0.1}\) = 1.08 × 10^-4
∴ Percent ionisation of dimethylamine = \(\frac{\text { Amount dissociated }}{\text { Total amount }}\) × 100
= \(\frac{x \times 100}{0.02}=\frac{1.08 \times 10^{-4}}{0.02}\) × 100 = 0.57%

Question 55.
Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below.
i) Human muscle-fluid, 6.83
iii) Human blood, 7.38
ii) Human stomach fluid, 1.2
iv) Human saliva, 6.4
Answer:
Given, pH = -log₁₀[H₃O⁺]
[H₃O⁺] = antilog₁₀(-pH)
i) [H₃O⁺] in human muscle fluid = antilog₁₀(-6.83) = 1.47 × 10^-7 mol L^-1
ii) [H₃O⁺] in stomach fluid = antilog₁₀(-1.2) = 6.3 × 10^-2 mol L^-1
iii) [H₃O⁺] in human blood = antilog₁₀(-7.38) = 4.1 × 10^-8 mol L^-1
iv) [H₃O⁺] in human saliva = antilog₁₀(-6.4) = 3.98 × 10^-7 mol L^-1

Question 56.
The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate the corresponding hydrogen ion concentration in each.
Answer:
[H⁺] = [H₃O⁺]
= = antilog₁₀(-pH) = antilog₁₀(-6.8) = 1.58 × 10^-7 mol L^-1
[H₃O⁺] in milk = antilog₁₀(-5.0) = 1.0 × 10^-5 mol L^-1
[H₃O⁺] in black coffee = antilog₁₀(-4.2) =1.0 × 10^-5 mol L^-1
[H₃O⁺] in tomato juice = antilog₁₀(-2.2) = 6.31 × 10^-5 mol L^-1
[H₃O⁺] in lemon juice = antilog₁₀(-7.8) = 6.31 × 10^-3 mol L^-1
[H₃O⁺] in egg white = antilog₁₀(-7.8) = 1.58 × 10^-8 mol L^-1

Question 57.
If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH ?
Answer:
Molarity of a solution is given by w = \(\frac{\mathrm{M} \times \mathrm{M}^{\prime} \times \mathrm{V}}{1000}\)
∴ 0.561 = \(\frac{\mathrm{M} \times 56 \times 200}{1000}\) (mol. mass of KOH = 56)
⇒ M = \(\frac{0.561 \times 1000}{56 \times 200}\) = 0.05 M
Since, KOH dissociates completely in the solution, KOH → K⁺ + OH^–
∴ [K⁺] = [OH^–] = Molarity of KOH = 0.05 M
and [H⁺] = \(\frac{K_w}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{0.05}\) = 2.0 × 10^-13 M
Now, pH = -log₁₀[H⁺] = -log₁₀(2 × 10^-13) = -log₁₀2 + 13 log 10 = -0.3010 + 13 = 12.70

Question 58.
The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Answer:
Molecular mass of Sr(OH)₂ = 87.6 + 2 × (16 + 1) = 121.6
Solubility of Sr(OH)₂ = 19.23 g/L = \(\frac{19.23}{121.6}\) mol/L^-1 = 0.1581 mol L^-1
Sr(OH)₂ ⇌ Sr²⁺ + 2OH^–
If Sr(OH)₂ is assumed to be completely ionised, we have
[Sr²⁺] = 0.1581 mol L^-1
and [OH^–] = 2 × 0.1581 = 0.3162 mol L^-1
∴ [H₃O⁺] = \(\) = 3.16 × 10^-14
and pH = -log₁₀[H₃O⁺] = -log₁₀(3.16 × 10^-14) = 13.50

Question 59.
The ionization constant of propanoic acid is 1.32 × 10^-5. Calculate the degree of ionization of the acid in its 0.05 M solution and also its pH. What will be its degree of ionization if the solution is 0.01 M in HCl also ?
Answer:

∴ K_a = \(\frac{\left[\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}\right]}\)
or 1.32 × 10^-5 = \(\frac{0.05 \alpha \times 0.05 \alpha}{0.05(1-\alpha)} \approx \frac{(0.05 \alpha)^2}{0.05}=\frac{0.05 \alpha^2}{1}\)
α = 1.63 × 10^-2
Then, [H₃O⁺] = 0.05α = 0.05 × 1.63 × 10^-2 = 8.15 × 10^-4 M
∴ pH = 3.09
In the presence of 0.1 M of HCl, let ‘α’ be the degree of ionization.
Then, [H₃O⁺] = 0.01; [A^–] = 0.05 α’ ; [H_A] = 0.05
K_a = \(\frac{0.01 \times 0.05 \alpha^{\prime}}{0.05}\) ⇒ 1.32 × 10^-5 = 0.01 α’
⇒ α’ = 1.32 × 10^-3

Question 60.
The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Answer:
Suppose the degree of ionisation of HCNO is α.

Given that pH of the solution is 2.34.
∴ -log₁₀(0.1 × α) = 2.34
⇒ log₁₀(0.1 × α) = -2.34 ⇒ 0.1 × α = antilog₁₀(-2.34) = 0.00457 = \(\frac{0.00457}{0.1}\) = 0.0457
K_a = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{CNO}^{-}\right]}{[\mathrm{HCNO}]}=\frac{(0.1 \times \alpha) \times(0.1 \times \alpha)}{0.1(1-\alpha)}\)
= \(\frac{(0.1 \times \alpha)^2}{0.1}=\frac{(0.1 \times 0.0457)^2}{0.1}\) = 2.09 × 10^-4

Question 61.
The ionization constant of nitrous acid is 4.5 × 10^-4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Answer:
Sodium nitrite is a salt of a strong base and a weak acid. For such a salt,
pH = \(\frac{1}{2}\)pK_w + \(\frac{1}{2}\)pK_a + \(\frac{1}{2}\)log₁₀c
= \(\frac{1}{2}\) × (-log₁₀1.0 × 10^-14) + \(\frac{1}{2}\) × (-log₁₀4.5 × 10^-4) + \(\frac{1}{2}\) × log₁₀0.04
= 7.0 + 1.673 – 0.698 = 7.975
The degree of hydrolysis of such a salt is given by
h = \(\sqrt{\frac{K_w}{K_a \cdot c}}=\sqrt{\frac{1.0 \times 10^{-14}}{4.5 \times 10^{-4} \times 0.04}}\) = 2,36 × 10^-5

Question 62.
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Answer:
Pyridinium hydrochloride is the salt of a weak base (pyridine) and a strong acid (HCl). For the salt of a weak base and strong acid.
pH = \(\frac{1}{2}\)pK_w + \(\frac{1}{2}\)pK_a + \(\frac{1}{2}\)log₁₀c
Substituting the values, we have
3.44 = [-\(\frac{1}{2}\)log₁₀1.0 × 10^-14 – \(\frac{1}{2}\) × (- log₁₀ K_b) – \(\frac{1}{2}\) × log₁₀ 0.02
⇒ 3.44 = – \(\frac{1}{2}\) × (-14) + \(\frac{1}{2}\)log₁₀ K_b – \(\frac{1}{2}\) × (-1.699)
⇒ 3.44 = 7 + \(\frac{1}{2}\)log₁₀K_b + 0.849
log₁₀K_b = (3.44 – 7 – 0.849) × 2 = -8.82
⇒ K_b = anti log₁₀(-8.82) = 1.5 × 10^-9

Question 63.
Predict if the solutions of the following salts are neutral, acidic or basic.
NaCl, KBr, NaCN, NH₄NO₃, NaNO₂ and KF
Answer:
Neutral solutions : NaCl, KBr.
Acidic solutions : NH₄NO₃.
Basic solutions : NaCN, NaNO₂, KF.

Question 64.
The ionization constant of chloroacetic acid is 1.35 × 10^-3. What will be the pH of 0.1 M acid and its 0.1 M of sodium salt solution ?
Answer:
pH of 0.1 M acid :
Suppose the degree of ionisation of chloroacetic acid is α.

∴ K_a = \(\frac{\left[\mathrm{Cl} \mathrm{CH}_2 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{ClCH}{ }_2 \mathrm{COOH}\right]}\)
⇒ 1.35 × 10^-3 = \(\frac{0.1 \alpha \times 0.1 \alpha}{0.1(1-\alpha)}\) ≈ \(\frac{(0.1 \alpha)^2}{0.1}\) ⇒ α = \(\left(\frac{1.35 \times 10^{-3} \times 0.1}{(0.1)^2}\right)^{1 / 2}\) = 0.116
∴ [H⁺] = 0.1 × α = 0.1 × 0.116 = 0.0116
and pH = -log₁₀[H⁺] = -log₁₀(0.0116) = 1.94
pH of 0.1 M of sodium salt:
The sodium salt of chloroacetic acid is the salt of a weak acid and a strong base.
For such as salt
pH = \(\frac{1}{2}\)pK_w + \(\frac{1}{2}\)pK_a + \(\frac{1}{2}\)log₁₀c
= \(\frac{1}{2}\)(-log₁₀1.0 × 10^-14) + \(\frac{1}{2}\)(-log₁₀1.35 × 10^-3) + \(\frac{1}{2}\) × log₁₀0.1
= 7.0 + 1.435 + (-0.5) = 7.94

Question 65.
Ionic product of water at 310 K is 2.7 × 10^-14. What is the pH of neutral water and temperature?
Answer:
In neutral water, [H₃O⁺] = [OH^–]
K_w = [H₃O⁺] [OH^–] = [H₃O⁺]²
⇒ [H₃O⁺]² = K_w = 2.7 × 10^-14
[H₃O⁺] = (2.7 × 10^-14)12 = 1.64 × 10^-7
Hence, pH = -log₁₀[H₃O⁺] = -log₁₀(1.64 × 10^-7) = 6.78

Question 66.
Calculate the pH of the resultant mixtures.
i) 10 mL Of 0.2 M of Ca(OH)₂ + 25 mL of 0.1 M of HCl
ii) 10 mL Of 0.01 M of H₂SO₄ + 10 mL of 0.01 M of Ca(OH)₂
iii) 10 mL of 0.1 M of H₂SO₄ + 10 mL of 0.1 M of KOH
Answer:
i) Normality of Ca(OH)₂ solution = 0.2 × 2 = 0.4 N
and normality of HCl solution = 0.1 × 1 = 0.1 N
Gram equivalents of is taken as = \(\frac{\mathrm{N} \times \mathrm{V}}{1000}=\frac{0.4 \times 10}{1000}\) = 0.004
Gram equivalents of HCl is taken as = \(\frac{N \times V}{1000}=\frac{0.1 \times 25}{1000}\) = 0.0025
0.0025 gram eq. of HCl would neutralise 0.0025 gram eq. of Ca(OH)₂.
∴ Amount of left unreacted = 0.004 – 0.0025 = 0.0015 gram equivalents
Total volume of the mixture = 10 + 25 = 35 mL
Suppose the normality of the resultant solution is N’.
No.of gram eq. = \(\frac{\mathrm{N}^{\prime} \times \mathrm{V}}{1000}\)
⇒ 0.0015 = \(\frac{\mathrm{N}^{\prime} \times 35}{1000}\)
⇒ N’ = \(\frac{0.0015 \times 1000}{35}\) = 0.0429
∴ [OH^–] = N’ = 0.042
∴ [H₃O⁺] = \(\frac{\mathrm{K}_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{0.0429}\) = 2.331 × 10^-13
pH = -log₁₀[H₃O⁺] = -log₁₀2.331 × 10^-13 = 12.63