Choose the correct answer

Chemistry : Ionic Equilibrium

Choose the correct answer:

1. Concentration of the Ag+ ions in a saturated solution of Ag2 C2O4 is 2.24 Ã—10-4mol L-1 solubility product of Ag2 C2O4 is

a) 2.42 Ã—10-8 mol3L-3

b) 2.66 Ã—10-12 mol3L-3

c) 4.5 Ã—10-11 mol3L-3

d) 5.619 × 10–12 mol3L-3

Solution:

Ag2C2O4 â†” 2Ag+ + C2O42-

[ Ag+ ] = 2.24 × 10-4 mol L-1

[ C2 O42- ] = { 2.24×10-4 }/2 = mol L-1

 = 1.12×10-4 mol L-1

Ksp = [Ag+]2[C2O42-]

 = (2.24×10-4 mol-4 L-1 )2 (1.12×10-4 mol L-1)

= 5.619×10-12mol3 L-3

[Option (d)]

2. Following solutions were prepared by mixing different volumes of NaOH of HCl different concentrations.

i. 60 mL (M/10) HCl + 40mL (M/10) NaOH

ii. 55 mL (M/10) HCl + 45 mL (M/10) NaOH

iii. 75 mL (M/5) HCl + 25mL (M/5) NaOH

iv. 100 mL (M/10) HCl + 100 mL (M/10) NaOH

pH of which one of them will be equal to 1?

a) iv

b) i

c) ii

d) iii

Solution:

iii) 75 ml M/5 HCl + 25ml M/5 NaOH

No of moles of HCl = 0.2×75×10-3 = 15 × 10-3

No of moles of NaOH = 0.2 × 25 × 10-3 = 5 × 10-3

No of moles of HCl after mixing = 15 × 10-3 - 5 × 10-3

= 10 × 10-3

concentration of HCl = No of moles of HCl / Vol in litre

= 10 Ã—10−3 / 100 Ã—10−3 = 0.1M

for (iii) solution, pH of 0.1M HCl = -log10(0.1)

= 1.

[Option (d)].

3. The solubility of BaSO4 in water is 2.42 Ã—10-3 gL-1 at 298K. The value of its solubility product ( Ksp ) will be. (Given molar mass of BaSO4 =233g mol-1 )

a) 1.08 Ã—10-14 mol2L-2

b) 1.08 Ã—10-12 mol 2 L-2

c) 1.08 Ã—10-10 mol2L-2

d) 1.08 Ã—10-8mol2 L-2

Solution:

BaSO4 â†” Ba2+ +SO42-

Ksp =(s) (s)

Ksp =(s)2

= ( 2.42×10-3 g L-1 )2

= ( 2.42×10-3 g L-1 / 233g mol-1 )

= ( 0.01038×10-3 ) 2

= (1.038×10-5 )2

= 1.077×10-10

= 1.08×10-10 mol2 L-2

[Option (c)]

4. pH of a saturated solution of Ca(OH)2 is 9. The Solubility product ( Ksp )of Ca(OH)2

a) 0.5 Ã—10-15

b) 0.25 Ã—10-10

c) 0.125 Ã—10-15

d) 0.5 Ã—10-10

Solution:

Ca(OH)2 â†” Ca2+ + 2OH-

Given that pH = 9

pOH = 14-9 = 5

[pOH = -log10 [OH- ]]

∴[OH- ] = 10- pOH

[OH- ]=10-5 M

Ksp =[Ca2+ ][OH- ]2

= (10-5 /2 )×(10-5 )2

=0.5 Ã—10-15

[Option (a)]

5. Conjugate base for Bronsted acids H2O and HF are

a) OH- and H2FH+ , respectively

b) H3O+ and F- , respectively

c) OH- and F- , respectively

d) H3 O+and H2F+ , respectively

Solution:

H2O + H2O â†” H3O+ + OH-

acid 1 + base 1 â†” acid 2 + base 2

HF + H2O â†” H3O+ + F-

acid 1 + base 1 â†” acid 2 + base 2

Conjugate bases are OH- and F- respectively

i.e.,[ Option (c)]

6. Which will make basic buffer?

a) 50 mL of 0.1M NaOH+25mL of 0.1M CH3COOH

b) 100 mL of 0.1M CH3COOH+100 mL of 0.1M NH4OH

c) 100 mL of 0.1M HCl+200 mL of 0.1M NH4OH

d) 100 mL of 0.1M HCl + 100 mL of 0.1M NaOH

Solution:

Basic buffer is the solution which has weak base and its salt

 NH4OH(200ml) + HCl(100ml) â†’NH4Cl(Salt)  + H2O + NH4OH (100ml weak base)

i.e.,[ Option (c)]

7. Which of the following fluro compounds is most likely to behave as a Lewis base?

a) BF3

b) PF3

c) CF4

d) SiF4

Solution:

BF3 â†’ elctron deficient â†’ Lewis acid

PF3 â†’ electron rich â†’ lewis base

CF4 â†’ neutral â†’ neither lewis acid nor base

SiF4- â†’ neutral â†’ neither lewis acid nor base

[option (b)]

8. Which of these is not likely to act as Lewis base?

a) BF3

b) PF3

c) CO

d) F–

Solution:

BF3 â†’ elctron deficient â†’ Lewis acid

PF3 â†’ electron rich â†’ lewis base

CO â†’ having lone pair of electron â†’ lewis base

F- â†’ unshared pair of electron â†’ lewis base

[option (a)]

9. The aqueous solutions of sodium formate, anilinium chloride and potassium cyanide are respectively

a) acidic, acidic, basic

b) basic, acidic, basic

c) basic, neutral, basic

d) none of these

Solution:

HCOONa Basic in nature. + H â‹…OH â†” NaOH strong base  + H-COOH weak acid

C6H5NH3Cl- + H â‹… OH â†” H3O+Acidic +C6H5 -NH2 + Cl-

KCN basic + H− OH â†” KOH strong base + HCN weak acid

basic, acidic, basic is correct.

[option (b)]

10. The percentage of pyridine (C5H5N) that forms pyridinium ion (C5 H5NH) in a 0.10M aqueous pyridine solution (Kb for C5 H5 N= 1.7 Ã—10-9 ) is

a) 0.006%

b) 0.013%

c) 0.77%

d) 1.6%

Solution:

C5H5N + H-OH â†” C5H5 NH + OH-

(α2C) / (1-α ) =Kb

α2C =∼ Kb

α = âˆš [ Kb / C ] = âˆš[ 1.7 ×10-9  / 0.1 ]

= âˆš1.7 × 10-4

= âˆš1.7 Ã— 10−4 Ã— 100

Percentage of dissociation  = 1.3 ×10-2 = 0.013 %

[Option (b)]

11. Equal volumes of three acid solutions of pH 1,2 and 3 are mixed in a vessel. What will be the H+ ion concentration in the mixture?

a) 3.7 Ã—10-2

b) 10-6

c) 0.111

d) none of these

Solution:

pH = -log10[H+]

∴[H+]=10-pH

Let the volume be x mL

V1M1 + V2M2 + V3M3 =VM

∴ x mL of 10-1M+ x mL of 10-2M + x mL of 10-3 M

 = 3x mL of [H+]

∴[H+ ] =  x[0.1+0.01+0.001]  / 3x

= ( 0.1+ 0.01+ 0.001 ) /  3

= 0..111 /  3

= 0.037

= 3.7 ×10−2

[Option (a)]

12. The solubility of AgCl (s) with solubility product 1.6 Ã—10-10 in 0.1M NaCl solution would be

a) 1.26 Ã— 10-5M

b) 1.6 Ã—10-9M

c) 1.6 Ã—10-11M

d) Zero

Solution:

AgCl(s)  â†” Ag+ (aq) + Cl-(aq)

NaCl (0.1 M) â†’ Na+(0.1 M) + Cl- (0.1 M)

Ksp = 1..6 ×10-10

Ksp = [Ag+][Cl-]

Ksp = (s)(s+0.1)

0.1>>>s

∴ s + 0.1  = 0.1

∴ S = [ 1..6 Ã—10-10 ] / 0.1 = 1.6 Ã—10-9

 [Option (b)]

13. If the solubility product of lead iodide is 3.2 Ã—10-8 , its solubility will be

a) 2×10-3M

b) 4 Ã—10-4M

c) 1.6 Ã—10-5M

d) 1.8 Ã—10-5M

Solution:

PbI2 (s) â†” Pb2+ (aq) + 2I- (aq)

Ksp = (s)(2s)2

3.2 Ã— 10-8 = 4s3

 s =  (3.2×10-8 / 4)1/3

 = (8 × 10-9 )1/3

 = 2 × 10-3M

= 2 Ã— 10-3M

 [Option (a)]

14. MY and NY3 , are insoluble salts and have the same Ksp values of 6.2 Ã—10-13 at room temperature. Which statement would be true with regard to MY and NY3?

a) The salts MY and NY3 are more soluble in 0.5M KY than in pure water

b) The addition of the salt of KY to the suspension of MY and NY3 will have no effect on their solubility’s

c) The molar solubilities of MY and NY3 in water are identical

d) The molar solubility of MY in water is less than that of NY3

Solution:

Addition of salt KY (having a common ion Y–) decreases the solubility of MY and NY due to common ion effect.

Option (a) and (b) are wrong.

 For salt MY , MY â†”  M+ + Y-

 Ksp = (s)(s)

 6.2 × 10-13 = s2

 âˆ´ s= âˆš(6.2 Ã— 10-13) = 10-7

for salt NY3 ,

NY3 â†” N3+ + 3Y-

Ksp = (s )(3s)3

 Ksp = 27s4

s = [ (6.2 ×10−13)/27 ]1/4

 s = 10-4

 The molar solubility of MY in water is less than of NY3

 [Option (d)]

15. What is the pH of the resulting solution when equal volumes of 0.1M NaOH and 0.01M HCl are mixed?

a) 2.0

b) 3

c) 7.0

d) 12.65

Solution:

x ml of 0.1 M NaOH + x ml of 0.01 M HCl

No of moles of NaOH = 0.1 × x Ã— 10–3 =  0.1x Ã— 10-3

No of moles of HCl = 0.01 × x Ã— 10-3 = 0.01x × 10-3

No of moles of NaOH after mixing = 0.1x Ã— 10-3 - 0.01x Ã— 10-3

 = 0.09x Ã— 10-3

Concentration of NaOH= [0.09x Ã— 10−3 ] /  [2x × 10−3] = 0.045

[OH-] = 0.045

POH = −log (4.5 × 10−2 )

 = 2 −log 4.5

 = 2- 0.65 = 1.35

pH = 14-1.35=12.65

Solution: Option (d)

16. The dissociation constant of a weak acid is 1 Ã—10-3. In order to prepare a buffer solution with a pH = 4, the [Acid]/[Salt] ratio should be

a) 4:3

b) 3:4

c) 10:1

d) 1:10

Solution:

K a =1 Ã—10-3

pH=4

[salt] / [Acid] = ?

pH = pKa +log {[Salt]/[Acid]}

 4 = - log10 (1 Ã—10-3 ) + log {[Salt]/ [Acid]}

 4 = 3 + log ([Salt]/ [Acid])

 1= log10 ([Salt]/[Acid])

 [Salt] / [Acid] =101

i.e., [Salt] / [Acid] = 1 / 10

 1:10

 [Option (d)]

17. The pH of 10-5M KOH solution will be

a) 9

b) 5

c) 19

d) none of these

Solution:

KOH    →    K+      +  OH-

10-5m           10-5m          10-5m

KOH (10-5m)   →    K+ (10-5m)     +  OH- (10-5m)

 [OH-] = 10-5M.

 pH= 14 - pOH

pH = 14 - ( - log [OH-] )

 = 14 + log [OH- ]

= 14 + log10−5

= 14−5

= 9.

[Option (a)]

18. H2PO4- the conjugate base of

a) PO43−

b) P2O5

c) H3PO4

d) HPO42-

Solution:

H3PO4 +H âˆ’ OH â†” H3O+ + H2PO4-

acid 1 + base 1 â†” acid 2 + base 2

∴H2PO4− is the conjugate base of H3PO4

[Option (c)]

19. Which of the following can act as Lowry – Bronsted acid as well as base?

a) HCl

b) SO42−

c) HPO42−

d) Br-

Solution:

HPO42− can have the ability to accept a proton to form H2PO−4.

It can also have the ability to donate a proton to form PO4-3

[Option (c)]

20. The pH of an aqueous solution is Zero. The solution is

a) slightly acidic  

b) strongly acidic

c) neutral

d) basic

Solution:

pH = -log10[H+ ]

∴[H+ ]=10-pH

=100 =1

[H+] = 1M

The solution is strongly acidic

[Option (b)]

21. The hydrogen ion concentration of a buffer solution consisting of a weak acid and its salts is given by

a) [H+] = Ka ([acid][salt])

b) [H+] = Ka[salt]

c) [H+] = Ka[acid]

d) [H+] = Ka ([salt][acid])

Solution:

According to Henderson equation

pH = pKa + log ([salt]/[acid])

ie. - log [H+ ] = - log Ka +log ([salt] / [acid])

 -log[H+] = log ([salt]/ [acid]) × (1/Ka)

  log (1/[H+]) = log ([salt]/ [acid]) × (1/Ka)

∴[H+] = Ka {[acid] / [salt] }

[option (a)]

22. Which of the following relation is correct for degree of hydrolysis of ammonium acetate?

[Option (c)]

Solution:

 h= âˆš{ (Kh) / (Ka .Kb) }

[Option (c)]

23. Dissociation constant of NH4OH is 1.8 Ã—10-5 the hydrolysis constant of NH4Cl would be

a) 1.8 Ã—10-19

b) 5.55 Ã—10-10

c) 5.55 Ã—10-5

d) 1.80 Ã—10-5

Solution:

 Kh = Kw/Kb = 1×10-14 / 1.8×10-5

 = 0.55×10-9

 = 5.5×10-10

[option (b)]

PTA Question Oneword:

1. In which of the following cases, the sparingly soluble salt solution is unsaturated?

a) Ionic product > solubility product (K_sp)

b) Ionic product < solubility product (K_sp)

c) Ionic product = solubility product (K_sp)

d) Both (a) and (b)

Answer: b)

2. Which of the following salts do not undergo salt hydrolysis?

a) Sodium acetate

b) Ammonium acetate

c) Ammonium chloride

d) Sodium nitrate

Answer: d)

3. The relationship between the solubility product (K_sp) and molar solubility (S) for Ag₂(CrO₄) is

a) K_sp = s³

b) K_sp = s²

c) K_sp = 4s³

d) K_sp = 3s²

Answer: c)

4. The aqueous solution sodium acetate, ammonium chloride, sodium nitrate are respectively

a) Neutral, acidic, basic

b) acidic, basic, neutral

c) basic, acidic, neutral

d) basic, acidic, basic

Answer: d)