Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}\leftrightharpoons \mathrm{N}{\mathrm{H}}_4^{+}+\mathrm{O}{\mathrm{H}}^{-}$

${\mathrm{k}}_{\mathrm{b}}=\frac{\left[\mathrm{N}{\mathrm{H}}_4^{+}\right]\mathrm{ }\left[\mathrm{O}{\mathrm{H}}^{-}\right]}{\left[\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}\right]}$

$\Rightarrow \mathrm{ }\mathrm{ }1.8\times {10}^{-5}=\frac{\left[\mathrm{N}{\mathrm{H}}_4^{+}\right]\mathrm{ }\left[0.01\right]}{\left[0.02\right]}$

{As$\mathrm{O}{\mathrm{H}}^{-}$will mainly come from$\mathrm{C}\mathrm{a}\left(\mathrm{O}{\mathrm{H}}_2\right)$only}

$\Rightarrow \left[\mathrm{N}{\mathrm{H}}_4^{+}\right]=36\times {10}^{-6}=\mathrm{a}\times {10}^{-6}$

$\Rightarrow \mathrm{ }\mathrm{ }\mathrm{a}=36$