1

AIPMT 2009

What is the [OH$-$] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)2?
A
0.40 M
B
0.0050 M
C
0.12 M
D
0.10 M

Explanation

Number of equivalents of H+ = 20.0 × 0.050 milliequivalents

= 1.0 milliequivalents

Number of equivalents of OH = 2 × 30.0 × 0.10

= 6.0 milliequivalents

$\therefore$ Equivalents of OH- left after neutralization

= 6 – 1 = 5 milliequivalents

Total volume after neutralization

= 20.0 + 30.0 mL

= 50 mL

[OH-] = ${5 \over {50}}$ = 0.1 M
2

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
3

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
4

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