Ionic Equilibrium

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Questions:

Multiple Choice Question

25 x 1 = 25

Dissociation constant of NH₄OH is 1.8 x 10^-5 the hydrolysis constant of NH₄Cl would be _______.

(a)

1.8 × 10^-19

(b)

5.55 × 10^-10

(c)

5.55 × 10^-5

(d)

1.80 × 10^-5

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

(a)

\(h=\sqrt { \frac { { K }_{ h } }{ C } } \)

(b)

\(h=\sqrt { \frac { { K }_{ a } }{ K_b } } \)

(c)

\(h=\sqrt { \frac { { K }_{ w } }{ { K }_{ a }.{ K }_{ b } } } \)

(d)

\(h=\sqrt { \frac { { { K }_{ a }.{ K }_{ b } } }{ { K }_{ w } } } \)

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

(a)

\([{ H }^{ + }]=\frac { { K }_{ a }[acid] }{ [salt] } \)

(b)

\([{ H }^{ + }]={ K }_{ a }[salt]\)

(c)

\([{ H }^{ + }]={ K }_{ a }[acid]\)

(d)

\([{ H }^{ + }]=\frac { { K }_{ a }[salt] }{ [acid] } \)

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

(a)

slightly acidic

(b)

strongly acidic

(c)

neutral

(d)

basic

Which of the following can act as Lowery – Bronsted acid well as base?

(a)

HCl

(b)

SO₄²⁻

(c)

HPO₄²⁻

(d)

Br^-

H₂PO₄^- the conjugate base of _______.

(a)

PO₄³⁻

(b)

P₂O₅

(c)

H₃PO₄

(d)

HPO₄^2-

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

(a)

(b)

(c)

(d)

none of these

The dissociation constant of a weak acid is 1 × 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

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)

(c)

7.0

(d)

12.65

10.

MY and NY₃, are insoluble salts and have the same K_sp values of 6.2 × 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 solubility of MY and NY₃in water are identical

(d)

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

11.

Using Gibb’s free energy change, ∆G^o=57.34 kJ mol^-1, for the reaction, X₂Y_(s)⇌2X⁺+Y^2-(aq), calculate the solubility product of X₂Y in water at 300 K_______. (R = 8.3 J K^-1Mol^-1)

(a)

10^-10

(b)

10^-12

(c)

10^-14

(d)

can not be calculated from the given dat

12.

If the solubility product of lead iodide is 3.2 × 10^-8, its solubility will be _______.

(a)

2 × 10^-3M

(b)

4 × 10^-4M

(c)

1.6 × 10^-5M

(d)

1.8 × 10^-5M

13.

The solubility of AgCl (s) with solubility product 1.6 × 10^-10 in 0.1M NaCl solution would be _______.

(a)

1.26 × 10^-5M

(b)

1.6 × 10^-9M

(c)

1.6 × 10^-11M

(d)

Zero

14.

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 × 10^-2

(b)

10^-6

(c)

0.111

(d)

none of these

15.

The percentage of pyridine (C₅H₅N) that forms pyridinium ion (C₅H₅NH) in a 0.10M aqueous pyridine solution _______.(K_b for C₅H₅N = 1.7×10^-9) is

(a)

0.006%

(b)

0.013%

(c)

0.77%

(d)

1.6%

16.

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

17.

What is the decreasing order of strength of bases
OH, NH₂^- H - C ≡ C and CH₃ - CH₂^-

(a)

OH^->NH₂^- >H-C≡C >CH₃^-CH₂^-

(b)

NH₂^->OH^->CH₃^-CH₂^- >H-C≡C

(c)

CH₃^-CH₂^->NH₂^->H-C≡C^->OH^-

(d)

OH^->H-C ≡ C^->CH₃^-CH₂^- >NH₂^-

18.

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

(a)

BF₃

(b)

PF₃

(c)

(d)

F^–

19.

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

(a)

BF₃

(b)

PF₃

(c)

CF₄

(d)

SiF₄

20.

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

21.

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)

H₃O⁺ and H₂F⁺, respectively

22.

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

(a)

0.5 × 10^-15

(b)

0.25 × 10^-10

(c)

0.125 × 10^-15

(d)

0.5 × 10^-10

23.

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

(a)

1.08 × 10^-14mol²L^-2

(b)

1.08 × 10^-12mol²L^-2

(c)

1.08 × 10^-10mol²L^-2

(d)

1.08 × 10^-8mol²L^-2

24.

Following solutions were prepared by mixing different volumes of NAOH of HCL different concentrations
1) 60 mL\(\frac{M}{10}\)HCl + 40mL\(\frac{M}{10}\)NaOH
2) 55 mL\(\frac{M}{10}\)HCl + 45mL\(\frac{M}{10}\)NaOH
3) 75 mL\(\frac{M}{5}\)HCl + 25mL\(\frac{M}{5}\)NaOH
4) 100 mL\(\frac{M}{10}\)HCl + 100mL\(\frac{M}{10}\)NaOH
pH of which one of them will be equal to 1?

(a)

(iv)

(b)

(i)

(c)

(ii)

(d)

(iii)

25.

Concentration of the Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.24 ×10^-4mol L^-1 solubility product of Ag₂C₂O₄ is_______.

(a)

2.42 × 10^-8mol³L^-3

(b)

2.66 × 10^-12mol³L^-3

(c)

4.5 × 10^-11mol³L^-3

(d)

5.619 × 10^-12mol³L^-3

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Answers Key:

Multiple Choice Question

25 x 1 = 25

\(K_h= { \frac { { K }_{ w } }{ K_b } } = \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}}\)
= 0.55 x 10^-9 = 5.5 x 10^-10

According to Henderson equation
pH = pK_a+ log [Acid]/[Salt]
i.e., -log [H⁺] = - log K_a + log [Acid]/[Salt]
-log [H⁺] = log [Acid]/[Salt] x 1/ K_a
log 1/[H⁺] = log [Acid]/[Salt] x 1/ K_a
[H⁺] = K_a [Acid]/[Salt]

pH = -log₁₀ [H⁺]
\(\therefore\) [H⁺] = 10^-pH
10⁰ = 1
[H⁺] = 1 M
The solution is strongly acidic

HPO₄²⁻ can have the ability to accept a proton to form H₂PO₄^-
It can also have the ability to donate a proton to form PO₄^-3

H₃PO₄ + H - OH ⇌ H₃O⁺ + H₂PO₄^-
acid 1 base 1 acid 2 base 2
\(\therefore\) H₂PO₄^- the conjugate base of H₃PO₄

KOH → K⁺ + OH^-
10^-5M 10^-5M 10^-5M
[OH^-] = 10^-5M
pH = 14 - pOH
pH= 14-(-log [OH^-])
= 14 + log [OH^-]
= 14 + log 10^-5
= 14 - 5 = 9

K_a = 1 × 10^-3
pH = 4
[Acid]/[Salt] = ?
pH = pK_a+ log [Acid]/[Salt]
4 = -log₁₀ (1 x 10^-3) + log [Acid]/[Salt]
4 = 3 + log [Acid]/[Salt]
1 = log₁₀ [Acid]/[Salt]
[Acid]/[Salt] = 10¹
i.e., [Acid]/[Salt] = 1/10
1 : 10

x ml of 0.1 M NaOH + x mL of 0.01 M HCI
No. of moles of NaOH = 0.1 x X x 10^-3
= 0.1 X x 10^-3
No. of moles of HCI = 0.01 x X x 10^-3
= 0.01 X x 10^-3
No. of moles of NaOH after mixing
= 0.1 X x 10^-3 - 0.01 X x 10^-3
= 0.09 X x 10^-3
Concentration of NaOH\(= (\frac{0.09x \times 10^{-3}}{2x \times 10^{-3}}) = 0 .045\)
[OH^-] = 0.045
pOH =-log (4.5 x 10^-2)
= 2 -log 4.5
= 2 - 0.65 = 1.35
pH = 14 - 1.35 = 12.65

10.

Addition of salt KY (having a common ion Y^-) decreases the solubility of MY and NY₃ due to common ion effect.
Option (a) and (b) are wrong
For salt MY, MY ⇌ M⁺ + Y^-
K_sp = (s) (s)
6.2 × 10^-13 = s²
\(\therefore\) s = \(\sqrt{6.2 \times 10^{-13}} = 10^{-7}\)
For salt NY₃,
NY₃⇌ N³⁺ + 3Y^-
K_sp = (s) (3s)³
K_sp = 27s⁴
\(s = (\frac{6.2 \times 10^{-13}}{27})^{1/4}\)
s = 10^-4
The molar solubility of MY in water is less than of NY₃

11.

(a)
10^-10

12.

PbI_2(s) ⇌ Pb²⁺_(aq) + 2I^-_(aq)
K_sp = (s) (2s)²
3.2 × 10^-8 = 4s³
s = (3.2 × 10^-8/4)^1/3
= ( 8 x 10^-9)^1/3
= 2 x 10^-3 M

13.

AgCl_(s) ⇌ Ag⁺_(aq) + Cl^-_(aq)
NaCl → Na⁺ +Cl^-
0.1M 0.1M 0.1M
K_sp = 1.6 × 10^-10
K_sp = [Ag⁺][ Cl^-]
K_sp = (s) (s + 0.1)
0.1 >>> s
∴ s + 0.1 = 0.1
∴ s = 1.6 × 10^-10/ 0.1 = 1.6 × 10^-9

14.

pH = -log₁₀ [H⁺]
∴ [H⁺] = 10^-pH
Let the volume be x ml.
V₁M₁ +V₂M₂ +V₃M₃ = VM
∴ x ml of 10^-1 M+ x ml of 10^-2 M + x ml of 10^-3 M
= 3 x ml of [H⁺]
\(∴ [H^+] = \frac{ x[0.1 + 0.01 + 0.001]}{3x}\)
\(= \frac{ [0.1 + 0.01 + 0.001]}{3}\)
\(= \frac{ [0.111]}{3}\)
= 0.037
= 3.7 x 10^-2

15.

C₅H₅N + H-OH ⇌ C₅H₅ ⁺NH + OH^-
\(\frac { { \alpha }^{ 2 }{ C } }{ 1-\alpha } =K_b\)
\(\alpha\)²C \(\approx\) K_b
\(\alpha = \sqrt {\frac{K_b} C} = \sqrt {\frac{1.7 \times 10^{-9}} {0.1}}\)
\(= \sqrt{1.7} \times 10^{-4}\)
Percentage of dissociation =\(= \sqrt{1.7} \times 10^{-4}\) x 100
= 1.3 x 10^-2 = 0.013%

16.

HCOONa + HOH ⇌ NaOH + H-COOH
strong base weak acid
Basic in nature.
C₆H₅NH₃Cl + HOH ⇌ H₃O⁺ + C₆H₅ - NH₂ + Cl^-
acidic
KCN + H - OH ⇌ KOH + HCN
Base strong base weak acid
basic, acidic, basic is correct.

17.

18.

BF₃ → electron deficient → Lewis acid
PF₃ → electron rich → Lewis base
CO → having lone pair of electron → Lewis base
F → unshared pair of electron → Lewis base

19.

BF₃ → electron deficient → Lewis acid
PF₃ → electron rich → Lewis base
CF₄ → neutral → neither Lewis acid nor base
SiF₄ → neutral → neither Lewis acid nor base

20.

Basic buffer is the solution which has weak base and its salt.
NH₄OH + HCI → NH₄CI +H₂O + NH₄OH
200 ml 100 ml salt 100 ml weak base

21.

H₂O + H₂O ⇌ H₃O⁺ + OH^-
acid 1 base 1 acid 2 base 2
HF + H₂O ⇌ H₃O⁺ + F^-
acid 1 base 1 acid 2 base 2
∴ Conjugate bases are OH^- and F^- respectively.

22.

Ca(OH)₂ ⇌ Ca²⁺ + 2OH^-
Given that pH = 9
pOH = 14 - 9 = 5
[pOH = - log₁₀ [OH]]
[OH^-] = 10^[pOH]
[OH] = 10^-5 M
Ksp = [Ca²⁺] [OH^-]
= 10^-5/2 x (10^-5)² = 0.5 x 10^-15

23.

BaSO₄ ⇌ Ba²⁺ + SO₄^2-
K_sp = (s) (s)
K_sp = (s)²
= (2.42 × 10^-3gL^-1)²
\(=\frac{2.42 \times 10^{-3} gL^{-1}}{233 \text{ g mol}^{-1}}\)
= (0.01038 x 10^-3)²
= (1.038 x x 10^-5)²
= 1.077 x 10^-10
= 1.08 × 10^-10mol²L^-2

24.

75 mL\(\frac{M}{5}\)HCl + 25mL\(\frac{M}{5}\)NaOH
No. of moles of HCI = 0.2 x 75 x 10^-3 = 15 x 10^-3
No. of moles of NaOH = 0.2 x 25 x 10^-3
= 5 x 10^-3
No. of moles of HCl after mixing
= 15 x 10^-3 - 5 x 10^-3
= 10 x 10^-3
\(\therefore\) Concentration of HCI = No. of moles of HCI/Vol in litre
= 10 x 10^-3/100 x 10^-3 = 0.1 M
For (iii) solution. pH of 0.1 M HCI = -log₁₀ (0.1) = 1

25.

\(\mathrm{Ag}_{2} \mathrm{C_2O_4} \rightleftharpoons 2 \mathrm{Ag}_{}^{+}+\mathrm{C_2O}_{4{}}^{2-}\)
\(\left[\mathrm{Ag}^{+}\right]=2 .24 \) ×10^-4mol L^-1
\(\mathrm{C_2O}_{4{}}^{2-} = \frac {2.24 \times 10 ^{-4}}{2}\) mol L^-1
= 1.12 ×10^-4mol L^-1
K_sp = [Ag]² [C₂O₄^2-]
= (2.24 ×10^-4mol L^-1) (1.12 ×10^-4mol L^-1)
= 5.619 × 10^-12mol³L^-3