Question

$20$ mL of $0.1M{H}_{2}S{O}_4$ solution is added to $30$ mL of $0.2MN{H}_{4}OH$ solution. The pH of the resultant mixture is: $[p{K}_{b\left(N{H}_{4}OH\right)}=4.7]$.

• A. $9.4$
• B. $5.0$
• C. $9.0$
• D. $5.2$

Answer 1

The correct option is C $9.0$

$20$ ml $0.1$M $H_{2}S{O}_4\Rightarrow {\eta }_{H^{+}}=4mmol$
$30$ ml $0.2$M $N{H}_{4}OH\Rightarrow {\eta }_{N{H}_{4}OH=6}mmol$

$N{H}_{4}OH+H^{+}\rightleftharpoons N{H}^{⨁}+H_{2}O$
$\Rightarrow$ 6 4 0 0
$\Rightarrow$ 2 0 4 4

Solution is a basic buffer

$pOH=p{K}_b+log\frac{N{H}_4^{+}}{N{H}_{4}OH}$

$=4.7+log2$

$=4.7+0.3=5$

$pH=14-5=9$