Question

Calculate the weight of $(N{H}_4{)}_{2}S{O}_4$ which must be added to 500 mL of 0.2 M $N{H}_3$ to yield a solution of pH = 9.35, $K_b$ for $N{H}_3=1.78\times {10}^{-5}$ (in gm).

• A. 0.4
• B. 2
• C. 3.52
• D. 5.25

Answer 1

The correct option is D 5.25

$pOH=-log{K}_b+log\frac{\left[Salt\right]}{\left[Base\right]}$

$pOH=-log{K}_b+log\frac{\left[N{H}_4^{\oplus }\right]}{\left[N{H}_{4}OH\right]}$

$\because \left[N{H}_4^{\oplus }\right]$ is obtained from salt $(N{H}_4{)}_{2}S{O}_4$

$\because pH=9.35$, therefore, $pOH=14-9.35=4.65$

$\therefore$ mmol of $N{H}_{4}OH$ in solution $=0.2\times 500=100$

Let a millimoles of $N{H}_4^{\oplus }$ are added in solution.

$\left[N{H}_4^{\oplus }\right]=\frac{a}{500},\left[N{H}_{4}OH\right]=\frac{100}{500}$

$4.65=-log(1.78\times {10}^{-5})+log\left(\frac{a/500}{100/500}\right)$

$4.65=4.7796+log\left(\frac{a}{100}\right)\therefore a=79.51$

$\therefore$ mmol of $(N{H}_4{)}_{2}S{O}_4$ added $=\frac{a}{2}=\frac{79.51}{2}=39.755$

$\therefore \frac{W}{132}\times 1000=39.755\therefore W_{(N{H}_4{)}_{2}S{O}_4}=5.248g$