1

### AIPMT 2012 Prelims

pH of a saturated solution of Ba(OH)2 is 12. The value of solubility product (Ksp) of Ba(OH)2 is
A
3.3 $\times$ 10$-$7
B
5.0 $\times$ 10$-$7
C
4.0 $\times$ 10$-$6
D
5.0$\times$ 10$-$6

## Explanation

Ba(OH)2 Ba2+ + 2OH
At equilibrium x 2x

pH = – log[H+]

12 = – log [H+]

$\Rightarrow$ [H+] = 10–12

As, [H+][OH ] = 10–14

10–12 [OH ] = 10–14

$\Rightarrow$ [OH ] = 10–2

As [OH ] = 2x = 10–2 then x = 5.0 × 10–3

Now, Ksp = [Ba2+][OH ]2

Ksp = (5 × 10–3) (10–2)2 = 5.0 × 10–7
2

### AIPMT 2011 Mains

In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl$-$ concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium ?

(Ksp for AgCl = 1.8 $\times$ 10$-$10, Ksp for PbCl2 = 1.7 $\times$ 10$-$5)
A
[Ag+] = 1.8 $\times$ 10$-$7 M, [Pb2+] = 1.7 $\times$ 10$-$6 M
B
[Ag+] = 1.8 $\times$ 10$-$11 M, [Pb2+] = 8.5 $\times$ 10$-$5 M
C
[Ag+] = 1.8 $\times$ 10$-$9 M, [Pb2+] = 1.7 $\times$ 10$-$3 M
D
[Ag+] = 1.8 $\times$ 10$-$11 M, [Pb2+] = 1.7 $\times$ 10$-$4 M

## Explanation

Ksp[AgCl] = [Ag+][Cl-]

$\Rightarrow$ [Ag+] = ${{1.8 \times {{10}^{ - 10}}} \over {{{10}^{ - 1}}}}$ = 1.8 $\times$ 10$-$9 M

Ksp[PbCl2] = [Pb2+][Cl-]2

$\Rightarrow$ [Pb2+] = ${{1.7 \times {{10}^{ - 5}}} \over {{{10}^{ - 1}} \times {{10}^{ - 1}}}}$ = 1.7 $\times$ 10$-$3 M
3

### AIPMT 2011 Prelims

For the reaction, N2(g) + O2(g) $\rightleftharpoons$ 2NO(g), the equilibrium constant is K1. The equilibrium constant is K2 for the reaction,
2NO(g) + O2(g) $\rightleftharpoons$ 2NO2(g)
What is K for the reaction,
NO2(g) $\rightleftharpoons$ ${1 \over 2}$N2(g) + O2(g)
A
${1 \over {2{K_1}{K_2}}}$
B
${1 \over {4{K_1}{K_2}}}$
C
${\left[ {{1 \over {{K_1}{K_2}}}} \right]^{1/2}}$
D
${1 \over {{K_1}{K_2}}}$

## Explanation

N2(g) + ${1 \over 2}$O2(g) $\rightleftharpoons$ NO(g),     ${\left( {{K_1}} \right)^{{1 \over 2}}}$

NO(g) + ${1 \over 2}$O2(g) $\rightleftharpoons$ NO2(g),     ${\left( {{K_2}} \right)^{{1 \over 2}}}$
---------------------------------------------------
${1 \over 2}$N2(g) + O2(g) $\rightleftharpoons$ NO2(g)    ${\left( {{K_1}{K_2}} \right)^{{1 \over 2}}}$

So, for the reverse reaction which is the desired one the value of K will be reciprocal of this value. i.e.

K = ${\left[ {{1 \over {{K_1}{K_2}}}} \right]^{1/2}}$
4

### AIPMT 2011 Prelims

Which of the following is least likely to behave as Lewis base?
A
H2O
B
NH3
C
BF3
D
OH$-$

## Explanation

The species which have a lone pair of electrons to donate or a negative charge on it can act as a Lewis base. Here, N atom in NH3 and O atom in H2O have lone pair of electrons available for donation. In OH the negative charge on it results in to behave it as a Lewis base. But BF3 is an electron deficient species thus, it is least likely to behave as Lewis base.