Question

Calculate the amount of $(N{H}_4{)}_{2}S{O}_2$ in grams which must be added to 500 ml of 0.2 M $N{H}_{4}OH$ to yield a solution of $pH=9$. $K_b\left(N{H}_3\right)=2\times {10}^{-5}$

• A. 3.24 g
• B. 4.24 g
• C. 13.2 g
• D. 6.24 g

Answer 1

The correct option is C 13.2 g

Given, $pH=9,K_b\left(N{H}_3\right)=2\times {10}^{-5},\left[N{H}_3\right]=0.02M,V=500ml$
let 'a' millimoles of $N{H}_4^{+}$ is added to a solution having millimoles
$N{H}_{4}OH=500\times 0.2=100\therefore \left[N{H}_4^{+}\right]=\left[salt\right]=\frac{a}{500}$ and $\left[N{H}_{4}OH\right]=\left[Base\right]=\frac{100}{500}$
$pH+pOH=14\Rightarrow pOH=14-9=5$
Using Henderson equation,
$pOH=-\log K_b+\log \frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}$ $5=-\log 2\times {10}^{-5}+\log \frac{\frac{a}{500}}{\frac{100}{500}}$
$5=5-\log 2+\log \frac{\frac{a}{500}}{\frac{100}{500}}$
$log2=\log \frac{\frac{a}{500}}{\frac{100}{500}}$
$\therefore a=200$ millimoles = 0.2 mol
$(N{H}_4{)}_{2}S{O}_4\text{added}=\frac{a}{2}=0.1$ mol $\therefore W_{(N{H}_4{)}_{2}S{O}_4}=\text{number of mole}\times \text{molar mass of}(N{H}_4{)}_{2}S{O}_4$
$=0.1\times 132=13.2g$