Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\underset{\mathrm{4 }\mathrm{m}\mathrm{m}\mathrm{o}\mathrm{l}}{\mathrm{H}\mathrm{N}{\mathrm{O}}_3}+\underset{\mathrm{6 }\mathrm{m}\mathrm{m}\mathrm{o}\mathrm{l}}{\mathrm{N}{\mathrm{H}}_3}\to \underset{\mathrm{4 }\mathrm{m}\mathrm{m}\mathrm{o}\mathrm{l}}{\mathrm{N}{\mathrm{H}}_4\mathrm{N}{\mathrm{O}}_3}$
[Since${\mathrm{W}}_{\mathrm{B}}\left(\mathrm{N}{\mathrm{H}}_3\right)$is left and salt$\left(\mathrm{N}{\mathrm{H}}_4\mathrm{N}{\mathrm{O}}_3\right)$is formed so it is a basic buffer]
Base left = 6 – 4 = 2 mmol
[Acid] =$40\times 0.1$= 4 mmol
[Base]$=20\times 0.3=6$
pOH$=\mathrm{p}{\mathrm{K}}_{\mathrm{b}}+\mathrm{l}\mathrm{o}{\mathrm{g}}_{10}\frac{\left[\mathrm{S}\mathrm{a}\mathrm{l}\mathrm{t}\right]}{\left[\mathrm{B}\mathrm{a}\mathrm{s}\mathrm{e}\right]}$
So, pOH$=4.7447+\mathrm{l}\mathrm{o}\mathrm{g}4/2=5.0457$
and pH = 14 – 5.0457 = 8.95.