Type: MCQ

20 mL of 0.1 M $\text{H}_2\text{SO}_4$ solution is added to 30 mL of 0.2 M $\text{NH}_4\text{OH}$ solution. The pH of the resultant mixture is:

[$\text{pK}_b$ of $\text{NH}_4\text{OH} = 4.7$]

(a) 5.2
(b) 9.0
(c) 5.0
(d) 9.4

Show Answer

mmol of $\text{H}_2\text{SO}_4 = 20 \times 0.1 = 2$

mmol of $\text{NH}_4\text{OH} = 30 \times 0.2 = 6$

$\text{H}_2\text{SO}_4 + 2\text{NH}_4\text{OH} \rightarrow (\text{NH}_4)_2\text{SO}_4 + 2\text{H}_2\text{O}$

Initial: 2, 6; Final: 0, $(6 - 2 \times 2) = 2$; $(\text{NH}_4)_2\text{SO}_4 = 2$ mmol

$[\text{NH}4\text{OH}]{left} = 2$ mmol

$[\text{NH}_4^+] = 2 \times 2 = 4$ mmol

Total Volume $= 30 + 20 = 50$ mL

$\text{pOH} = \text{pK}_b + \log \frac{[\text{Salt}]}{[\text{Base}]} = 4.7 + \log \frac{4/50}{2/50} = 4.7 + \log 2 = 5$

$\text{pH} = 14 - \text{pOH} = 14 - 5 = 9$