1

### AIPMT 2015

What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?
A
2.0
B
7.0
C
1.04
D
12.65

## Explanation

HCl + NaOH $\to$ NaCl + H2O
Initial  0.01       0.1            0          0
Final     0         0.09          0.01     0.01

As equal volumes of HCl and NaOH are added so the volume of resulting solution becomes double and the concentration of the solution becomes half.

$\therefore$ [OH-] = ${{0.09} \over 2}$ = 0.045 M

$\therefore$ pOH = – log[OH ] = –log [0.045] = 1.35

$\therefore$ pH = 14 – pOH = 14 – 1.35 = 12.65
2

### AIPMT 2014

For the reversible reaction,
N2(g) + 3H2(g) $\rightleftharpoons$ 2NH3(g) + heat

The equilibrium shifts in forward direction
A
by increasing the concentration of NH3(g)
B
by decreasing the pressure
C
by decreasing the concentrations of N2(g) and H2(g)
D
by increasing pressure and decreasiing temperature.

## Explanation

According to Le-Chatelier’s principle, the equilibrium shifts in that direction so as to oppose the applied change.

Given reaction is exothermic reaction. Hence according to Le-Chatelier's principle low temperature favours the forward reaction and on increasing pressure equilibrium will shift, towards lesser number of moles i.e. forward direction.
3

### AIPMT 2014

For a given exothermic reaction, Kp and K'p are the equilibrium constants at temperatures T1 and T2, respectively. Assuming that heat of reaction is constant in temperature range between T1 and T2, it is readily observed that
A
Kp > K'p
B
Kp < K'p
C
Kp = K'p
D
Kp = ${1 \over {k{'_p}}}$

## Explanation

log
K'p
Kp
= $- {{\Delta H} \over {2.303R}}\left[ {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right]$

For exothermic reaction, $\Delta$H = -ve means the temperature T2 is higher than T1.

$\therefore$ $\left[ {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right]$ is negative.

So log K'p - log Kp = -ve

$\Rightarrow$ log Kp > log K'p

$\Rightarrow$ Kp > K'p
4

### AIPMT 2014

Using the Gibb's energy change, $\Delta$Go = +63.3 kJ, for the following reaction,

Ag2CO3(s) $\rightleftharpoons$ 2 Ag+(aq) + CO32$-$ (aq)
the Ksp of Ag2CO3(s) in water at 25oC is
(R = 8.314 J K$-$1 mol$-$1)
A
3.2 $\times$ 10$-$26
B
8.0 $\times$ 10$-$12
C
2.9 $\times$ 10$-$3
D
7.9 $\times$ 10$-$2

## Explanation

We know, $\Delta$Go = – 2.303 RT log Ksp

$\therefore$ 63300 = – 2.303 × 8.314 × 298 log Ksp

$\Rightarrow$ log Ksp = -11.09

$\Rightarrow$ Ksp = 10-11.09

= 8.0 × 10–12