1

### AIPMT 2009

The ionization constant of ammonium hydroxide is 1.77 $\times$ 10$-$5 at 298 K. Hydrolysis constant of ammonium chloride is
A
6.50 $\times$ 10$-$12
B
5.65 $\times$ 10$-$13
C
5.65 $\times$ 10$-$12
D
5.65 $\times$ 10$-$10

## Explanation

Ammonium chloride is a salt of weak base and strong acid. In this case hydrolysis constant Kh can be calculated as

Kh = ${{{K_w}} \over {{K_b}}}$ = ${{{{10}^{ - 14}}} \over {1.77 \times {{10}^{ - 5}}}}$ = 5.65 $\times$ 10$-$10
2

### AIPMT 2009

The dissociation constants for acetic acid and HCN at 25oC are 1.5 $\times$ 10$-$5 and 4.5 $\times$ 10$-$10 respectively. The equilibrium constant for the equilibrium
CN$-$ + CH3COOH $\rightleftharpoons$ HCN + CH3COO$-$ would be
A
3.0 $\times$ 10$-$5
B
3.0 $\times$ 10$-$4
C
3.0 $\times$ 104
D
3.0 $\times$ 105

## Explanation

CH3COOH $\rightleftharpoons$ CH3COO + H+,   K1 = 1.5 $\times$ 10$-$5

$\Rightarrow$ K1 = ${{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]} \over {\left[ {C{H_3}COOH} \right]}}$ = 1.5 $\times$ 10$-$5

HCN $\rightleftharpoons$ CN + H+,     K2 = 4.5 × 10–10

K2 = ${{\left[ {C{N^ - }} \right]\left[ {{H^ + }} \right]} \over {\left[ {HCN} \right]}}$ = 4.5 × 10–10

CN$-$ + CH3COOH $\rightleftharpoons$ HCN + CH3COO$-$

K = ${{\left[ {HCN} \right]\left[ {C{H_3}CO{O^ - }} \right]} \over {\left[ {C{N^ - }} \right]\left[ {C{H_3}COOH} \right]}}$

$\Rightarrow$ K = ${{{K_1}} \over {{K_2}}}$ = ${{1.5 \times {{10}^{ - 5}}} \over {4.5 \times {{10}^{ - 10}}}}$

= 3.33 × 104

$\simeq$ 3.0 × 104
3

### AIPMT 2008

Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H+ ion concentration in the mixture?
A
3.7 $\times$ 10$-$3 M
B
1.11 $\times$ 10$-$3 M
C
1.11 $\times$ 10$-$4 M
D
3.7 $\times$ 10$-$4 M

## Explanation

We know, pH = – log[H+]

For pH = 3, 3 = – log [H+]

$\Rightarrow$ [H+] = 10–3 M

For pH = 4 , 4 = – log [H+]

$\Rightarrow$ [H+] = 10–4 M

For pH = 5, 5 = – log [H+]

$\Rightarrow$ [H+] = 10–5 M

Total concentration of [H+]

M = ${{{{10}^{ - 3}}\left( {1 + 0.1 + 0.01} \right)} \over 3}$

M(V1 +V2 + V3) = M1V1 + M2V2 + M3V3

As V1 = V2 = V3 = V

$\Rightarrow$M(3V) = (M1 + M2 + M3)V

$\Rightarrow$ 3M = (10–3 + 10–4 + 10–5)

= ${{1.11 \times {{10}^{ - 3}}} \over 3}$

= 3.7 $\times$ 10$-$4 M
4

### AIPMT 2008

The values of for the reactions,

X $\rightleftharpoons$ Y + Z      . . . .(i)
A $\rightleftharpoons$ 2B       . . . .(ii)

are in the ratio 9 : 1. If degree of dissociation of X and A be equal, then total pressure at equilibrium (i) and (ii) are in the ratio
A
36 : 1
B
1 : 1
C
3 : 1
D
1 : 9

## Explanation

Given

X $\rightleftharpoons$ Y + Z      . . . .(i)

A $\rightleftharpoons$ 2B       . . . .(ii)

Let the total pressure for reaction (i) and (ii) be P1 and P2 respectively, then

${{{K_{{P_1}}}} \over {{K_{{P_2}}}}} = {9 \over 1}$

X Y + Z
Initial mole 1 0 0
At equilibrium 1 - $\alpha$ $\alpha$ $\alpha$

Total number of moles at equilibrium

= 1 - $\alpha$ + $\alpha$ + $\alpha$ = 1 + $\alpha$

$\therefore$ KP1 = ${{{P_Y} \times {P_Z}} \over {{P_X}}}$ = ${{{\alpha \over {1 + \alpha }} \times {P_1} \times {\alpha \over {1 + \alpha }} \times {P_1}} \over {{{1 - \alpha } \over {1 + \alpha }} \times {P_1}}}$

A 2B
Initial mole 1 0
At equilibrium 1 - $\alpha$ 2$\alpha$

Total number of moles at equilibrium

= 1 - $\alpha$ + 2$\alpha$ = 1 + $\alpha$

$\therefore$ KP2 = ${{{{\left( {{P_B}} \right)}^2}} \over {{P_A}}}$ = ${{{{\left( {{{2\alpha } \over {1 + \alpha }} \times {P_2}} \right)}^2}} \over {{{1 - \alpha } \over {1 + \alpha }} \times {P_2}}}$

$\therefore$ ${{{K_{{P_1}}}} \over {{K_{{P_2}}}}} = {{{P_1}} \over {4{P_2}}}$

$\Rightarrow$ ${{{P_1}} \over {4{P_2}}} = {9 \over 1}$

$\Rightarrow$ ${{{P_1}} \over {{P_2}}} = {{36} \over 1}$