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1

AIPMT 2009

MCQ (Single Correct Answer)
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
2

AIPMT 2008

MCQ (Single Correct Answer)
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
3

AIPMT 2008

MCQ (Single Correct Answer)
The value of equilibrium constant of the reaction
HI(g) $$\rightleftharpoons$$ $${1 \over 2}$$H2(g) + $${1 \over 2}$$I2(g)
is 8.0. The The equilibrium constant of the reaction
H2(g) + I2(g) $$\rightleftharpoons$$ 2HI(g) will be
A
16
B
1/8
C
1/16
D
1/64

Explanation

HI(g) $$\rightleftharpoons$$ $${1 \over 2}$$H2(g) + $${1 \over 2}$$I2(g),

For this reaction, K = 8.0

Reversing the equation,

$$ \therefore $$$${1 \over 2}$$H2(g) + $${1 \over 2}$$I2(g) $$\rightleftharpoons$$ HI(g) ......(1)

For this reaction, K1 = $${1 \over 8}$$

H2(g) + I2(g) $$\rightleftharpoons$$ 2HI(g)

We get this equation by multiplying equation (1) by 2,

$$ \therefore $$ For this reaction, K2 = $${\left( {{1 \over 8}} \right)^2}$$ = $${1 \over {64}}$$
4

AIPMT 2008

MCQ (Single Correct Answer)
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 = $${{{{\left( {{P_B}} \right)}^2}} \over {{P_A}}}$$ = $${{{{\left( {{{2\alpha } \over {1 + \alpha }} \times {P_2}} \right)}^2}} \over {{{1 - \alpha } \over {1 + \alpha }} \times {P_2}}}$$

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

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

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

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