Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\text{pOH}={\text{pK}}_{\text{b}}+\log \frac{\left[{\text{NH}}_4^{+}\right]}{\left[{\text{NH}}_3\right]}$

$5.74=4.74+\log \frac{\left[{\text{NH}}_4^{+}\right]}{0.1}$

or$\log \frac{\left[\mathrm{N}{\mathrm{H}}_4^{+}\right]}{0.1}=\mathrm{ }1$

or$\frac{\left[\mathrm{N}{\mathrm{H}}_4^{+}\right]}{0.1}=\mathrm{ }10$

or$\left[\mathrm{N}{\mathrm{H}}_4^{+}\right]=\mathrm{ }1\mathrm{ }\left(\mathrm{M}\right)$

$\therefore {\left({\text{NH}}_4^{+}\right)}_2{\text{SO}}_4$required$=0.5\text{ mole}$