Calculate the amount of ${\left({\text{NH}}_4\right)}_2{\text{SO}}_4$ in grams which must be added to 500 ml of 0.2 M ${\text{NH}}_3$ to yield a solution of pH = 9, ${\mathrm{K}}_{\mathrm{b}} \mathrm{for} {\mathrm{NH}}_3 = 2 \times {10}^{-5}$

${\text{P}}^{\mathrm{OH}}= -\log {\mathrm{K}}_{\mathrm{b}} + \log \frac{\left[{\mathrm{NH}}_4^{+}\right]}{\left[{\text{NH}}_4\text{OH}\right]}$

Let ‘a’ millimoles of ${\text{NH}}_4^{+}$ is added to a solution  

millimoles of ${\mathrm{NH}}_4\mathrm{OH} = 500 \times 0.2 = 100$

$\therefore \left[{\mathrm{NH}}_4^{+}\right] = \left[\mathrm{salt}\right] = \mathrm{a}$

Given ${\text{P}}^{\mathrm{H}}{\text{= 9 ; P}}^{\mathrm{OH}}\text{= 5}$ 

$\therefore 5 = - \log 2 \times {10}^{-5} + \log \frac{\text{a}}{100}$

$\therefore \text{a} = 200 \mathrm{millimoles} = 0.2 \mathrm{moles}$

Moles of ${\left({\text{NH}}_4\right)}_2{\text{SO}}_4$ added = $\frac{a}{2}=0.1 \mathrm{mol}$

$\therefore W {\left(N{H}_4\right)}_{2}S{O}_4 = 0.1 \times 132 = 13.2$g