1

### 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
2

### AIPMT 2008

If the concentration of OH$-$ ions in the reaction
Fe(OH)3(s) $\rightleftharpoons$ Fe3+(aq) + 3OH$-$(aq)
is decreased by 1/4 times, then equilibrium concentration of Fe3+ will increase by
A
64 times
B
4 times
C
8 times
D
16 times

## Explanation

Fe(OH)3(s) $\rightleftharpoons$ Fe3+(aq) + 3OH$-$(aq)

Equilibrium constant, Kc = ${{\left[ {F{e^{3 + }}} \right]{{\left[ {O{H^ - }} \right]}^3}} \over {\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}$

We know, equilibrium constant remains same at constant temperature.

Now, let the increase in concentration of Fe3+ be x times.

$\therefore$ Kc = ${{\left[ {x \times F{e^{3 + }}} \right]{{\left[ {{1 \over 4} \times O{H^ - }} \right]}^3}} \over {\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}$

= ${x \over {64}}{{\left[ {F{e^{3 + }}} \right]{{\left[ {O{H^ - }} \right]}^3}} \over {\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}$

$\Rightarrow$ Kc = ${x \over {64}}$Kc

$\Rightarrow$ ${x \over {64}}$ = 1

$\Rightarrow$ x = 64
3

### AIPMT 2008

The dissociation equilibrium of a gass AB2 can be represented as :
2AB2(g) $\rightleftharpoons$ 2AB(g) + B2(g)
The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure P is
A
(2Kp/P)1/2
B
(Kp/P)
C
(2Kp/P)
D
(2Kp/P)1/3

## Explanation

2AB2(g) &#8652; 2AB(g) + B2(g)
Initial mole 2 0 0
At equilibrium 2(1 - x) 2x x

Amount of moles at equilibrium = 2(1 – x) + 2x + x = 2 + x

${K_p} = {{{{\left[ {{p_{AB}}} \right]}^2}\left[ {{p_{{B_2}}}} \right]} \over {{{\left[ {{p_{A{B_2}}}} \right]}^2}}}$

= ${{{{\left[ {{{2x} \over {2 + x}} \times P} \right]}^2}\left[ {{x \over {2 + x}} \times P} \right]} \over {{{\left[ {{{2\left( {1 - x} \right)} \over {2 + x}} \times P} \right]}^2}}}$

= ${{\left[ {{{4{x^3}} \over {2 + x}} \times P} \right]} \over {4{{\left( {1 - x} \right)}^2}}}$

$\Rightarrow$ Kp = ${{{4{x^3} \times P} \over 2} \times {1 \over 4}}$

($\because$ 1 – x ≈ 1 and 2 + x ≈ 2)

$\Rightarrow$ x = ${\left( {{{8{K_p}} \over {4P}}} \right)^{{1 \over 3}}}$

$\Rightarrow$ x = ${\left( {{{2{K_p}} \over P}} \right)^{{1 \over 3}}}$
4

### AIPMT 2007

A weak acid, HA, has a Ka of 1.00 $\times$ 10$-$5. If 0.100 mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closest to
A
1.00%
B
99.9%
C
0.100%
D
99.0%

## Explanation

For weak acid degree of dissociation,

$\alpha$ = $\sqrt {{{{K_a}} \over C}}$

= $\sqrt {{{1 \times {{10}^{ - 5}}} \over {0.1}}} = {10^{ - 2}}$ = 1.00 %