1

### AIPMT 2012 Mains

Given that the equilibrium constant for the reaction,
2SO2(g) + O2(g) $\rightleftharpoons$ 2SO3(g)
has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature ?
SO3(g) $\rightleftharpoons$ SO2(g) + ${1 \over 2}$ O2(g)
A
1.8 $\times$ 10$-$3
B
3.6 $\times$ 10$-$3
C
6.0 $\times$ 10$-$2
D
1.3 $\times$ 10$-$5

## Explanation

2SO2(g) + O2(g) $\rightleftharpoons$ 2SO3(g),    K

SO3(g) $\rightleftharpoons$ SO2(g) + ${1 \over 2}$ O2(g),     K' = $\sqrt {{1 \over K}}$

$\therefore$ K' = $\sqrt {{1 \over {278}}} =$ 6.0 $\times$ 10$-$2
2

### AIPMT 2012 Prelims

Buffer solutions have constant acidity and alkalinity because
A
these give unionised acid or base on reaction with added acid or alkali
B
acids and alkalies in these solutions are shielded from attack by other ions
C
they have large excess of H+ or OH$-$ ions
D
they have fixed value of pH

## Explanation

For this example,

CH3COOH ⇌ CH3COO + H+ ;

CH3COONa ⇌ CH3COO + Na+

when few drops of HCl are added to this buffer, the H+ of HCl immediatly combine with CH3COO ions to form undissociated acetic acid molecules. Thus there will be no H+ ions to combine with CH3COO ions to form undissociated acetic acid molecules. Thus there will be no appreciable change in its pH value. Like wise if few drops of NaOH are added, the OH ions will combine with H+ ions to form unionised water molecule. Thus pH of solution will remain constant.
3

### AIPMT 2012 Prelims

Equimolar solutions of the following substances were prepared separately. Which one of these will record the highest pH value ?
A
BaCl2
B
AlCl3
C
LiCl
D
BeCl2

## Explanation

All of the given salts have same anion i.e., Cl which on hydrolysis gives HCl which is a strong acid.

Now, among the salts which have cation that gives a strongest base on hydrolysis of salt have the highest pH value. As Ba form Ba(OH)2 which is a stronger base thus, it results in the highest pH value.
4

### 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