Chemistry - Ionic Equilibrium Question with Solution | TestHub

It is a basic buffer
$\mathrm{p}\mathrm{H}=8.65$,
pOH = 14 – 8.65 = 5.35
$\mathrm{p}\mathrm{O}\mathrm{H}=\mathrm{p}{\mathrm{K}}_{\mathrm{b}}+\mathrm{l}\mathrm{o}\mathrm{g}\frac{\left[\stackrel{\oplus }{\mathrm{N}}{\mathrm{H}}_4\right]}{\left[\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}\right]}$
5.35 = 4.75 + log$\frac{\left[\stackrel{\oplus }{\mathrm{N}}{\mathrm{H}}_4\right]}{\left[\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}\right]}$
$\frac{\left[\stackrel{\oplus }{\mathrm{N}}{\mathrm{H}}_4\right]}{\left[\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}\right]}=$antilog (5.35 – 4.75)
= antilog (0.6)$\approx 4.0$
So$\frac{\left[\stackrel{\oplus }{\mathrm{N}}{\mathrm{H}}_4\right]}{\left[\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}\right]}=\frac{30\times 0.2}{0.3\times \mathrm{V}}\approx 4.0$
So V = 5 mL.