Question

The amount of $(N{H}_2{)}_{4}S{O}_4$ in $\text{g}$, which must be added to $500\text{mL}$ of $0.2M$ $N{H}_3$ to yield a solution of $pH=9.35$., if $K_b$ for $N{H}_3$=1.6$\times$${10}^{-5}$ will be:( Give your answer upto two decimal only.)
(Given, $\log 1.6=0.2,\log 70=1.85$)

Answer 1

According to Henderson - Hasselbalch equation,
$pOH=-\log K_b+\log \frac{\left[salt\right]}{\left[base\right]}$
$\Rightarrow pOH=-\log K_b+log\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}\cdot \cdot \cdot \left(i\right)$
$\left[N{H}_4^{+}\right]$ is obtaind from salt $(N{H}_2{)}_{4}S{O}_4$
$\text{Given}pH=9.35$
$\therefore pOH=14-9.35=4.65$
Millimole of $N{H}_{4}OH$ in solution $=0.2\times 500=100$
Let millimole of $N{H}_4^{+}$ added in solution$=a$
$\therefore \left[N{H}_4^{+}\right]=\frac{a}{500};\left[N{H}_{4}OH\right]=\frac{100}{500}$
So putting the values in equation $\left(i\right)$ we get,
$4.65=-\log (1.6\times {10}^{-5})+\log \frac{a/500}{100/500}$
$\Rightarrow 4.65=4.8+\log \frac{a}{100}$
$\Rightarrow -0.15=\log \frac{a}{100}$
$\Rightarrow -0.15=\log a-2$
$\Rightarrow 1.85=\log a$
$\therefore a=70$
$\text{Millimoles of}(N{H}_4{)}_{2}S{O}_4\text{added}=\frac{a}{2}=\frac{70}{2}=35$
Let the mass of $\left[N{H}_4{)}_{2}S{O}_4\right]$ added be $w$
$\therefore \frac{w}{132}=0.035$
$\therefore w_{(N{H}_4{)}_{2}S{O}_4}=4.62\text{g}$