Question

The pH of solution by mixing 50 mL of 0.6 M $N{H}_{4}OH$ and 50 mL of 0.1M $H_{2}S{O}_4$ is : Given $K_a$ for $N{H}_4^{+}={10}^{-9}$

• A. 4.39
• B. 6.61
• C. 9.3
• D. 7.39

Answer 1

The correct option is C 9.3

$2N{H}_{4}OH+H_{2}S{O}_4\to (N{H}_4{)}_{2}S{O}_4+2H_{2}O$
$\begin{array}{ccccc}Meg.att=0& 30& 10& 0& 0\\ Meq.afterreaction& 20& 0& 10& 10\end{array}$
$\therefore$ $\left[\right(N{H}_4{)}_{2}S{O}_4]=\frac{1}{2\times 100}M$
$\therefore$ $\left[N{H}_4^{+}\right]=\frac{10\times 2}{2\times 100}M=0.1M$
$\left[N{H}_{4}OH\right]=\frac{20}{100}=0.2M$
$pOH=P{K}_b+log\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}={10}^{-5}=log\frac{0.1}{0.2}=5-0.3010=4.6989$
$\therefore$ $pH=9.3010$