1

### AIPMT 2002

Solution of 0.1 N NH4OH and 0.1 N NH4Cl has pH 9.25. Then find out pKb of NH4OH
A
9.25
B
4.75
C
3.75
D
8.25

## Explanation

NH4OH and NH4Cl constitute to form a basic buffer.

pOH = pKb + log ${{\left[ {Salt} \right]} \over {\left[ {Base} \right]}}$

We know, pOH+ pH = 14 or pOH = 14 – pH

$\therefore$ 14 - pH - $\log {{\left[ {Salt} \right]} \over {\left[ {Base} \right]}}$ = pKb

$\Rightarrow$ 14 - 9.25 - $\log {{0.1} \over {0.1}}$ = pKb

$\Rightarrow$ 14 – 9.25 – 0 = pKb

$\Rightarrow$ pKb = 4.75
2

### AIPMT 2002

Which has highest pH?
A
CH3COOK
B
Na2CO3
C
NH4Cl
D
NaNO3

## Explanation

Na2CO3 is a salt of weak acid H2CO3 and strong base NaOH, therefore, its aqueous solution will be basic hence has pH more than 7.
3

### AIPMT 2002

Reaction BaO2(g) $\rightleftharpoons$ BaO(s) + O2(g); $\Delta$H = +ve. In equilibrium condition, pressure of O2 depends on
A
increase mass of BaO2
B
increase mass of BaO
C
increase temperature on equilibrium
D
increase mass of BaO2 and BaO both.

## Explanation

For the reaction

BaO2(g) $\rightleftharpoons$ BaO(s) + O2(g); H = +ve.

At equilibrium Kp = PO2

[For solid and liquids concentration term is taken as unity]

Hence, the value of equilibrium constant depends only upon partial pressure of O2. Further on increasing temperature formation of O2 increases as this is an endothermic reaction.
4

### AIPMT 2002

Solubility of MX2 type electrolytes is 0.5 $\times$ 10$-$4 mole/lit., then find out Ksp of electrolytes.
A
5 $\times$ 10$-$12
B
25 $\times$ 10$-$10
C
1 $\times$ 10$-$13
D
5 $\times$ 10$-$13

## Explanation

MX2 Ag2+ + 2X-
s s 2s

Ksp = [M2+] [X]2 = (S)(2S)2 = 4S3

$\Rightarrow$ Ksp = 4(0.5 × 10–4)3 = 5 × 10–13