1

### NEET 2017

Concentration of the Ag+ ions in a saturated solution of Ag2C2O4 is 2.2 $\times$ 10$-$4 mol L$-$1. Solubility product of Ag2C2O4 is
A
2.66 $\times$ 10$-$12
B
4.5 $\times$ 10$-$11
C
5.3 $\times$ 10$-$12
D
2.42 $\times$ 10$-$8

## Explanation

Ag2C2O4(s) $\to$ 2Ag+ + C2O4 2–
S                  2S           S

Ksp = [Ag+] 2 [C2O4 2–] = [2S]2 [S]

[Ag+] = 2S = 2.2 × 10–4

or S = 1.1 × 10–4 M

$\therefore$ Ksp = [2.2 × 10–4 ]2 [1.1 × 10–4] = 5.3 × 10–12
2

### NEET 2017

The equilibrium constants of the following are

N2 + 3H2 $\rightleftharpoons$ 2NH3;     K1

N2 + O2 $\rightleftharpoons$ 2NO;     K2

H2 + ${1 \over 2}$O2 $\rightleftharpoons$ H2O;     K3

The equilibrium constant (K) of the reaction :

2NH3 + ${5 \over 2}$ O2 $\rightleftharpoons$ 2NO + 3H2O will be
A
K2K33/K1
B
K2K3/K1
C
K23K3/K1
D
K1K33/K2

## Explanation

2NH3 $\rightleftharpoons$ N2 + 3H2;     ${1 \over {{K_1}}}$

N2 + O2 $\rightleftharpoons$ 2NO;     K2

3H2 + ${3 \over 2}$O2 $\rightleftharpoons$ 3H2O;     (K3)3

By adding all equations, we get

2NH3 + ${5 \over 2}$ O2 $\rightleftharpoons$ 2NO + 3H2O

$\therefore$ K = ${{{K_2} \times {{\left( {{K_3}} \right)}^3}} \over {{K_1}}}$
3

### NEET 2016 Phase 1

MY and NY3, two nearly insoluble salts, have the same Ksp values of 6.2 $\times$ 10$-$13 at room temperature. Which statement would be true in regard to MY and NY3?
A
The salts MY and NY3 are more soluble in 0.5 M KY than in pure water.
B
The addition of the salt of KY to solution of MY and NY3 will have no effect on their solubilities.
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.

## Explanation

MY ⇌ M+ + Y
s        s       s

Ksp = s.s = s2

$\Rightarrow$ s = $\sqrt {6.2 \times {{10}^{ - 13}}}$

= 7.87 × 10–7 mol L–1

NY3 ⇌ N+ + 3Y
s        s       3s

Ksp = s.(3s)3 = 27s4

6.2 × 10–13 = 27s4

$\Rightarrow$ s = 3.89 × 10–4 mol L–1

Hence, molar solubility of MY in water is less than that of NY3.
4

### NEET 2016 Phase 2

The solubility of AgCl(s) with solubility product 1.6 $\times$ 10$-$10 in 0.1 M NaCl solution would be
A
1.26 $\times$ 10$-$5 M
B
1.6 $\times$ 10$-$9 M
C
1.6 $\times$ 10$-$11 M
D
zero

## Explanation

AgCl ⇌ Ag+ + Cl
s           s     s + 0.1

Concentration of Cl is (s + 0.1) mol L–1 because s mol L–1 from ionization of AgCl and 0.1 mol L–1 from ionization of 0.1 M NaCl.

Now, Ksp = [Ag+][Cl ]

$\Rightarrow$ 1.6 × 10–10 = s (s + 0.1)

$\Rightarrow$ 1.6 × 10–10 = s (0.1) {$\because$ s << 0.1}

$\Rightarrow$ s = 1.6 × 10–9 M