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Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE I:\Obj-Ionic Equilibrium.doc

IONIC EQUILIBRIUM

READ THE INSTRUCTIONS CAREFULLY

1. The test is of 1 hour 30 minutes duration. 2. The maximum marks are 148. 3. This test consist 45 questions. 4. For each question in Section A & C you will be awarded 3 marks if you have darkened only

the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. Minus one (-1) mark will be awarded for wrong answer

5. For each question in Section E (Matrix Match), you will be awarded 2 marks for each row in which you have darkened the bubble(s) corresponding to the correct answer. Thus, each question in this section carries a maximum of 8 marks. There is no negative marks awarded for incorrect answer(s) in this Section.

6. For each question in Section F (Integer Type), you will be awarded 3 marks if you darken the bubble corresponding to the correct answer and zero mark if no bubble is darkened. No negative marks will be awarded for in this Section.

7. Keep Your mobiles switched off during Test in the Halls.

Section – A (Single Correct Choice Type)

This Section contains 34 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. (Mark only one choice) (34 × 3 = 102 Marks)

1. The aqueous solutions of four sodium salts NaA, NaB, NaC and NaD had pH 7.0, 9.0, 10.0 and 11.0

respectively, when each solution was 0.1 M. The strongest acid is:

a. HA b. HB c. HC d. HD

A

Sol. Aqueous solution of NaA had a pH 7.0 (lowest value). Hence, HA is the strongest acid.

2. In the hydrolytic equilibrium,

A– + H2O HA + OH–

Ka = 1.0 × 10–5. The degree of hydrolysis of a 0.001 M solution of the salt is :

a. 10 – 2 b. 10 – 3 c. 10 – 4 d.10 – 5

B

Sol. Degree of hydrolysis, c .K

Kh

a

w

= 001.0101

1015

14

= 6

35

14

10101101

101

= 10 – 3

3. A weak acid HX has the dissociation constant 1 × 10–5 M. It forms a salt NaX on reaction with alkali. The

degree of hydrolysis of 0.1 M solution of NaX is

a. 0.0001% b. 0.01% c. 0.1% d. 0.15%

B

Sol. Hydrolysis reaction is X– + H2O HX + OH–. For a salt of weak acid with strong base,

Kh = 2

a

w chK

K

5

14

10

10

= 0.1 × h2

or h2 = 10–8 or h = 10–4

%hydrolysis = 10–4 × 100 = 10–2 = 0.01

Dr. Sangeeta Khanna Ph.D


4. Which of the following species is more soluble in water?

a. M(OH)3, (Ksp = 1 × 10–35) b. M(OH)2, (Ksp = 1 × 10–35) c. MOH, (Ksp = 1 × 10–35) d. All have same solublity

A

Sol. S of M (OH)3 = 435

4 sp

27

10

27

K

S of M (OH)2 = sp3 sp

KMOH of S ;4

K

5. Mark the incorrect solution:

a. KBr : Neutral b. NH4NO3 : Acidic c. NaNO2 : Acidic d. KF : Basic C

Sol. acid Weak

2base Strong

2 HNO NaOHNaNO

Since NaOH is a strong base so the solution is basic solution 6. On adding 0.1 M solution each of [Ag+], [Ba]+, [Ca]+ in a Na2SO4 solution, species first precipitated is:

[Ksp, BaSO4 = 10–11, Ksp CaSO4 = 10–6, Ksp Ag2SO4 = 10–5]

a. Ag2SO4 b. BaSO4 c. CaSO4 d. all of these B

Sol. For Ag2SO4 ; Ksp = [Ag+]2 ]SO[ 24

2

sp42

24

][Ag

K SO Agof ionprecipitat for needed SO

= M101.01.0

10 35

For CaSO4 ; Ksp = [Ca2+] ]SO[ 24

]SO[ 24 needed for precipitation of CaSO4 M10

1.0

10

]Ca[

K5

6

2

sp

1.0

10BaSO

112

4

SO M10 1024

.

Thus, minimum ]SO[ 24 is required for precipitation of BaSO4 and hence it is precipitated out first.

7. pH of the solution containing 50.0 mL of 0.3 M HCl and 50.0 ml of 0.4 M NH3 is :

[pKa (NH 4

) =9.26] [log 5 = 0.6990; log 3 = 0.4771]

a. 4.74 b. 9.26 c. 8.78 d. 4.63 C

Sol. OHClNHNHHCl 24mg20

3mg15

It is base buffer pH = 14 – pKb -log5

15

pKb = 14 – pKa pKb = 14 – pKa = 14 – 9.26 = 4.74 pH = 14 – 4.74 = 8.78 8. The precipitate of Ag2CrO4 (Ksp = 1.9 × 10-12) is obtained when equal volumes of the following are mixed

a. 10–4 M Ag+ + 10–4 M 24

CrO b. 10–2 M Ag+ + 10–3 M 24

CrO

c. 10–5 M Ag+ + 10–3 M 24

CrO d. 10–4 M Ag+ + 10–5 M 24

CrO

B 9. The solubility of solid silver chromate, Ag2CrO4 is determined in three solvents:

Substance Ksp Ag2CrO4 9 x 10– 12

I. Pure water II. 0.1 M AgNO3 III. 0.1 M Na2CrO4 IV. 0.1 M NH3 solution Predict the relative solubility of Ag2CrO4, in the three solvents: a. I = II = III = IV b. I < II < III < IV c. II = III < I < IV d. II < III < I < IV D



10. The pH of solution of both ammonium acetate and sodium chloride is 7 due to: a. Hydrolysis in both cases b. Cationic & anionic hydrolysis of the former and not the latter. c. No hydrolysis in both d. Hydrolysis of the latter but not the former B 11. Calculate the buffer capacity of a solution. If its pH changes from 4.745 to 4.832 on addition of 0.01 mole

of NaOH to 250 ml of a Buffer solution:

a. 0.087 b. 0.04 c. 0.46 d. 0.01 C

Sol. Buffer capacity = pH in Change

buffer of litre per added base or acid of Moles

= 46.0087.0

25.0/01.0

12. Ksp of Fe(OH)3 in aqueous solution is 3.8 × 10–38 at 298 K. the concentration of Fe3+ will increase when:

a. pH is increased b. pH is 7 c. pH is decreased d. it is exposed to air C Sol. Fe(OH)3 Fe+3 + 3OH– 3HCl 3Cl– + 3H+

H+ + OH– H2O On adding acid (decrease in pH), the equilibrium shifts in the backward direction.

[Fe+3] increases. 13. In aqueous solution, the ionisation constants for carbonic acid are

K1 = 4.2 × 10-7 and K2 = 4.8 × 10-11 Select the statement for a saturated 0.034 M solution of the carbonic acid:

a. the concentration of H+ ion is double of 23

CO

b. the concentration of CO 23

is 0.034 M

c. the concentration of 23

CO is greater than that of 3HCO

d. the concentration of H+ and 3HCO are approximately equal

D

Sol. Concentration of H+ ion is approximately same as that of 3HCO because K2 << K1.

14. Ag+ + 2NH3 Ag (NH3)2

; K1 = 1.8 × 107

Ag+ + Cl– AgCl; K2 = 5.6 × 109

AgCl + 2NH3 Ag(NH3)2

+ Cl– ; K = ?

a. (1.8 × 107) × (5.6 × 109) b. (1.8 × 10)7 + (5.6 × 109)

c. (1.8 × 107) (5.6 × 109) d. (1.8 × 10)9 (5.6 × 107) C

Sol. First eqn. is obtained by adding the 2nd and 3rd equations. So,

K1 = K2K 15. What is the percentage dissociation of a substance if its vapour densities before and after dissociation

are 30 and 15 respectively and 1 mole of it dissociates to 3 moles of products?

a. 25% b. 50% c. 65% d. 75% B

Sol. AB2 A + 2B 1 mole

1 – 2

(1 + 2) 15 = 30

= 0.5 16. The conjugate base of [Al(H2O)3(OH)3] is:

a. [Al(H2O)3(OH)2]+ b. [Al(H2O)3(OH)2O]– c. [Al(H2O)3(OH)3]

– d. [Al(H2O)2(OH)4]–

D

Sol. Loss of H+ from [Al(H2O)3(OH)3] gives [Al(H2O)2(OH)4]– as conjugate base.



17. In which of the following, the reactants are Lewis acid base?

a. NH3 + BH3 (H3N BH3) b. Cu2+ + 4NH3 [Cu(NH3)4]2+

c. Ni + 4CO [Ni(CO)4] d. All D

Sol. In the choice (a), the first compound NH3 is Lewis base and BH3 is Lewis acid as per the order shown 18. The correct increasing order of basic character of the following compounds is:

(I) CH3NH2; pKb = 3.4 (II) NH3; pKb = 4.74 (III) C6H5NH2; pKb = 9.37

a. III < II < I b. I < II < III c. III < I < II d. III II < I A

Sol. Lower the pKb, stronger is the base. 19. The pH of a 0.1 M solution of the acid HQ is 3. The value of the ionisation constant Ka of the acid is:

a. 3 × 10-1 b. 1 × 10-3 c. 1 × 10-5 d. 1 × 10-7

C Sol. HQ H+ + Q– At t = 0 0.1 - - At Eqm. (0.1 – x) x x

pH = 3 [H+] = x = 10–3

H+ = C.Ka

(10-3) = Ka ×0.1 Ka = 10–5 20. 100 mL of 0.2 N NaOH is mixed with 100 mL 0.1 N HCl and the solution is made 1L. The pH of the

solution is:

a. 4 b. 8 c. 10 d. 12 D

Sol. 2101000

1.01002.0100]OH[

pOH = 2

pH = 12 21. Dissociation constants of HCOOH and CH3COOH at certain temperature are 1.8 × 10-4 and 1.8 × 10-5

respectively. At what concentration would CH3COOH have the same [H+] concentration as 0.01 M HCOOH?

a. 1 × 10-4 M b. 1 × 10-5 M c. 0.1 M d. 0.01 M C

Sol. For [H+] concentration of CH3COOH and HCOOH to be equal,

HCOOHCOOHCH )KC(KC3

01.0108.1C108.1 45

M 1.0108.1

01.0108.1C

5

4

)COOHCH( 3

22. A solution having pH value 3 is diluted to twice its volume. pH of the resultant solution will be:

a. 2 b. 2.3010 c. 3.3010 d. 4.0 C

Sol. pH = 3 [H+] = 10-3 M On dilution,

M 105102

1]H[ 43

New pH = - log (5 × 10-4) = - (0.699 – 4) = 3.301 23. The percentage hydrolysis of decinormal solution of ammonium acetate is:

(Ka = 1.75 × 10-5, Kb = 1.80 ×10-5, Kw = 1.0 × 10-14) a. 0.365 b. 0.305 c. 0.563 d. 0.666 C

Sol. ba

wh

KK

KK (for salt of weak acid and weak base)



5

55

14

10175.31080.11075.1

101

h% = 10010175.3100K 5h = 5.63 × 10-1 = 0.563

24. What is the pH of 0.1 M H2S if the Ka values of H2S to HS– and HS– to S2– are 10-7 and 10-13 respectively?

a. 2.5 b. 3 c. 4 d. 4.5 C

Sol. The Ka of H2S >>> Ka of HS–. H+ ions are mainly produced by H2S.

47a 101.010C.K]H[

pH = - log 10–4 = 4

Alternatively, pH = 482

1)17(

2

1C] logpK[

2

1a

25. Addition of a small quantity of an acid or a base to a basic buffer will not change its pH at all, if:

a. pH = pKb b. 1]base[

]salt[ c. pOH = pKb d. both (b) and (c)

D

Sol. Best basic buffer has pOH = pKb, which is possible only when 1]base[

]salt[

i.e., 0]base[

]salt[log

26. Which of the following work(s) as a single salt buffer:

a. (NH4)2SO4 b. CH3COONa c. NH3 solution d. Borax

D

27. For preparing a buffer solution of pH 6 by mixing sodium acetate and acetic acid, the ratio of

concentrations of salt and acid should be (Ka = 10–5):

a. 1 : 10 b. 10 : 1 c. 100 : 1 d. 1 : 100

B

Sol. pH = ]acid[

]salt[logpKa

156)10 log(6]acid[

]salt[log 5-

10]acid[

]salt[

So, [salt] : [acid] = 10 : 1 28. The molar solublity (in mol L-1) of a sparingly soluble salt MX4 is ‘S’. The corresponding solubility product

is Ksp. S is given in terms of Ksp by the relation:

a.

4/1sp

128

KS

b.

5/1sp

256

KS

c. S = [265 Ksp]

1/5 d. S = [128 Ksp]1/4

B

Sol. 4SS

4 4X M MX

Ksp = (S) (4S)4 = 256 S5

S =

5/1sp

256

K

29. Which of the following is the most soluble in water?

a. Bi2S3(Ksp = 1 × 10–70) b. Ag2S (Ksp = 6 × 10-51) c. CuS (kSP = 8 × 10-37) d. MnS (Ksp = 7 × 10-16) D

Sol. Solubility is directly proportional to Ksp MnS has highest Ksp among the given substances and hence has highest solubility.



30. Solubility of a salt A2B3 is 1 × 10-2 M, its solubility product is:

a. 1.08 × 108 b. 1.08 × 10-8 c. 1.08 × 10-10 d. 1.0 × 10-10 B

Sol. For solubility S, Ksp of A2B3 = (2)2 × (3)2 × S2 × S3 = 108 × (1 × 10-2)5 = 108 × 10-10 = 1.08 × 10-8 31. What is the concentration of Cl– ions in saturated solution if Ksp of HgCl2 at 298 K is 4 × 10–15?

a. 1 × 10-5 M b. 2 × 10-5 M c. 4 × 10-5 M d. 4 × 10-15 M B

Sol. HgCl2 Hg2+ + 2Cl– Ksp = S × (2S)2 = 4S3 4S3 = 4 × 10-15 S = 10– 5 [Cl–] = 2S = 2 × 10-5 M

32. The solubility product of Al(OH)3 is 2 × 10-33. When the precipitate of Al(OH)3 is shaken with 2 dm3 of 0.1 M NaOH, concentration of Al3+ will be:

a. 2 × 10-34 M b. 2 × 10-32 M c. 2 × 10–31 M d. 2 × 10-30 M D

Sol. Al[OH]3 Al3+ + 3[OH]–, sparingly dissociated

NaOH Na+ + OH–, fully dissociation Ksp = [Al3+] [OH–]3 2 × 10-33 = [Al3+] [0.1]3

[Al3+] = 2 × 10-30 M

33. What is the conjugate acid of the Bronsted-Lowry base HAsO42– ?

a. H2AsO4

– b. AsO43– c. H2O d. H3O

+ A 34. Correct relation between solubility (S) & Ksp of Hg2Cl2 is

a. Ksp = 4s2 b. Ksp = 4s3 c. 16s4 d. none B

Sol. Hg2Cl2 Hg2+2 + 2Cl–

SECTION – C (Linked Comprehension Type)

This Section contains 2 paragraphs. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. (Mark only one) (6 × 3 = 18 Marks)

Passage -1 The following pictures represent solutions that contain a weak acid HA and/or its potassium salt KA.

Unshaded spheres represent H atoms and shaded represent A¯ ions. (K+, H3O+, OH¯, and solvent H2O

molecules have been omitted for clarity.) 1. Which solution has the highest pH? a. (1) b. (2) c. (3) d. (4) D 2. Which of the solutions are buffer solutions? a. (1) and (2) b. (1) and (3) c. (2) and (3) d. (2) and (4) B 3. For which solution(s) pH = pKa? a. only solution (1) b. only solution (2) c. only solution (3) d. solution (1) and (3) D

= HA = A¯



Passage-2

Consider an aqueous 0.01 M sodium acetate solution. Given : log 1.85 = 0.27 , Ka of acetic acid = 1.85 × 10–5 at 298 K.

4. pH of the solution is:

a. 7.0 b. 8.36 c. 9.2 d. 6.0 B

Sol. c log2

1pK

2

1pK

2

1pH aw

= )(10 log2

1)73.4(

2

1)14(

2

1 2-

= 7 + 2.365 – 1 = 8.365 5. The hydrolysis constant is:

a. 5.45 × 10–10 b. 5.45 × 1010 c. 54.5 × 108 d. 54.5 × 10 – 10 A

Sol. 10

5

14

a

w 1045.5~1085.1

10

K

K

6. Degree of hydrolysis is:

a. 23.4 × 104 b. 23.4 × 10– 4 c. 2.34 × 10– 4 d. 2.34 × 104 C

Sol. 4

2

10h 1034.2

10

1045.5

c

Kh

SECTION – E (Matrix Type) No Negative Marking

This Section contains 2 questions. Each question has four choices (A, B, C and D) given in Column I and five statements (p, q, r, and s) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. 8 × 2 = 16 Marks

1. Match Column – I with Column – II

Column – I Column – II

(A) Anionic part is hydrolysed (P) CH3COONa (pKa of CH3COOH = 4.78)

(B) Cationic part is hydrolysed (Q) NH4Cl

(pKa OF 4

NH = 10)

(C) Does not show hydrolysis (R) 333 CHHN OOCH

(Kb of CH3NH2 = 5 × 10–5)

(D) pH 7 (S) NaCl

Sol. (A) P, R; (B) Q, R; (C) S; (D) P, Q, R Only a component of weak acid or weak base is hydrolyzed e.g. (P) As CH3COO– is from CH3COOH which is a weak acid. Hence CH3COO– (Anion) is hydrolyzed. (Q)

As 4

NH is from NH4OH which is a weak base

Hence 4

NH (cation) is hydrolyzed

(R)

CH3COONa CH3COOH (weak acid)

NaOH (Strong Base) pH > 7

NH4Cl NH4OH (Weak Base)

HCl (Strong acid) pH < 7

333 CHHN OCOCH

CH3COOH (Weak acid)

CH3NH2 (Weak base)



As Θ

3 OCOCH and 33 HNCH

are from CH3COOH and CH3NH2 which are weak acid and weak base

respectively.

Hence both OCOCH3 and 33 HNCH

are hydrolyzed.

For such salts if pKa = pKb then pH = 7 pKb of CH3COOH = 4.78 pKb of CH3NH2 = 5 – log 5 = 4.30 pH pKb < pka means pH > 7

(S) Hence neither Na+ nor Cl– is hydrolyzed pH = 7

2. Match Column – I with Column – II

Column – I Column – II

(A) 500 ml of 0.01 M soltuion of CH3COONa was treated with 500 ml of 0.01 M solution of HCl

(P) pH of resulting solution less than 7

(B) 500 ml of 0.01 M solution of CH3COOH was treated with 500 ml of 0.01 solution of NaOH

(Q) pH of resulting solution greater than 7

(C) 500 ml of 0.01 M solution of NH4OH was treated with 250 ml of 0.01 M HCl solution

(R) Resulting solution will work as buffer

(D) 500 ml of 0.01 M solution of CH3COOH was treated with 500 ml of 0.01 M solution of CH3COONa

(S) pH of resulting solution will increase on adding 50 ml water

Ka of CH3COOH = 2 × 10-5 Kb of NH4OH = 2 × 10–5

(T) The resulting solution will behave like weak acid

Sol. (A) P, S, T; (B) Q; (C) Q, R; (D) P, R (i) milliequivalent of CH3COONa = 500 × 0.01 = 5 milliequivalent of HCl = 500 × 0.01 = 5

NaCl COOHCH HCl COONaCH 35

3

final solution will carry 5 milliequivalent of CH3COOH and 5 milliequivalent of NaCl

5 milliequivalent of CH3COOH (weak acid) pH < 7, on adding water pH will increase. (ii) milliequivalent of CH3COOH = 500 × 0.01 = 5 milliequivalent of NaOH = 500 × 0.01 = 5

OHCOONaCHNaOH COOHCH 2355

3

final solution will have 5 milliequivalent of CH3COONa (salt of weak acid strong base)

pH > 7, pH will decrease on dilution. (iii) milliequivalent of NH4OH = 500 × 0.01 = 5 milliequivalent of HCl = 250 × 0.01 = 2.5

NH4OH + HCl NH4Cl + H2O 5 2.5 2.5 x 2.5

Final solution will have 2.5 milliequivalent of NH4OH (weak base) and 2.5 milliequivalent of NH4Cl (salt of weak base and strong acid) basic buffer pH > 7, In case of buffer no change of pH occur on dilution.

(iv) milliequivalent of CH3COOH = 500 × 0.01 = 5 milliequivalent of CH3COONa = 500 × 0.01 = 5

Final solution carry ‘5’ milliequivalent of CH3COOH (weak acid) and ‘5’ milliequivalent of CH3COONa (salt of weak acid & strong base) Buffer (Acid) pH < 7, No change in pH in case of buffer solution on dilution

NaCl NaOH (strong base)

HCl (Strong acid)



SECTION – F (Integer Type)

This Section contains 4 questions. The answer to each question is a single digit integer ranging from 0

to 10. The correct digit below the question number in the OMR is to be bubbled. No negative Marking. 4 × 3 = 12 Marks

1. The Ksp of Mg(OH)2 is 1 × 10–12. At what limiting pH, 0.01M Mg+2 will be about to precipitised Sol. Ksp for Mg(OH)2 = [Mg2+] [OH–]2

10–12 = (0.01) [OH–]2

or [OH–]2 = 10–10 M or [OH–] = 10–5 M

[H+] = 10–9 M or pH = 9 2. What will be pH of a solution prepared by mixing 0.01 N, 100 ml H2SO4 & 0.01 M 100 ml Base NaOH at

25°C Sol. 7 3. From the value of pKa given number of acids in which 0.01 M solutions have pH greater than 3.3 or H+ =

5 × 10-4. [log 5 = 0.6990, log 2 = 0.3010]

HCN 9.31

CH3COOH 4.75

CH3CH2COOH 4.87

CH3CH2CH2COOH 4.82

COOHCHH

Cl

CCH 2|

3 4.05

CH2 = CH – COOH 4.25

C6H5COOH 4.19

HF 3.45

C6H5OH 9.89 Sol. 5 pH = 3.3, [H+] = 5 × 10–4

[H+] = CKa

5 × 10–4 = 2a 10K

25 × 10-8 = Ka × 10–2 Ka = 2.5 × 10–5 pKa = 5 – log 2.5 = 4.6 pH > 3.3 means pKa > 4.6 5 acids in above table have pKa > 4.6 4. Total number of solution from the following which has pH > 7 is:

(KaCH3COOH = 1.8 × 10-5, Kb NH4OH = 1.8 × 10–5) (i) A solution having (H+) = 10–10 (ii) A solution having (OH–) = 10–10

(iii) 10–8 M HCl solution (iv) 10-6 M HCl solution (v) 0.01 M solution of NH4Cl (vi) 0.01 M solution of CH3COONa (vii) 0.01 M solution of CH3COONH4 (viii) 0.01 M solution of CH3COONH3CH3 (ix) Pure water at 0°C

Sol. 4 (i) (H+) = 10-10, pH = 10 ; OH– = 10-4

(ii) (OH–) = 10–10, pOH = 10, pH = 4 (iii) 10-8 M solution of HCl, pH < 7

(iv) 10–6 M solution of HCl, pH < 7 (v) 0.01 M solution of NH4Cl (salt of weak base, strong acid) pH < 7 (vi) 0.01 M solution of CH3COONa (salt of weak acid, strong base) pH > 7 (vii) 0.01 M CH3COONH4 (salt of weak acid and weak base) Ka of CH3COOH = Kb of NH4OH pH = 7

(viii) 0.01 MCH3COONH3CH3 (salt of weak acid and weak base) CH3NH2 is more basic than NH3 (NH4OH) as given in the problem

Ka of CH3COOH = Kb of NH4OH means Kb of CH3NH2 > Ka of CH3COOH pH > 7

(ix) At 25°C, pH of water = 7 At a temperature < 25°C, pH water > 7 At a temperature > 25°C, pH water < 7

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