Question

20 $mL$ of $0.1M$ $H_{2}S{O}_4$ solution is added to $30mL$ of $0.2M$ $N{H}_{4}OH$ solution. The $pH$ of the resultant mixture is:
$[P{k}_{b}ofN{H}_{4}OH=4.11]$

• A. 9.4
• B. 5
• C. 9
• D. 5.2

Answer 1

The correct option is C 9

Moles equivalent of $H_{2}S{O}_4$= $20\times 0.1=2$
Moles equivalent of $N{H}_{4}OH$= $30\times 0.2=6$

${\mathrm{NH}}_4\mathrm{OH}+H^{+}\rightleftharpoons {{\mathrm{NH}}_4}^{+}+H_{2}O64002044\mathrm{Solution}\mathrm{is}\mathrm{basic}\mathrm{buffer}\mathrm{pOH}={\mathrm{pk}}_b+\log \frac{{{\mathrm{NH}}_4}^{+}}{{\mathrm{NH}}_4\mathrm{OH}}=4.7+\log \frac{4}{2}=4.7+\log 2=4.7+0.3=514=\mathrm{pOH}+\mathrm{pH}\mathrm{pH}=14-5=9$