Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}\left(\mathrm{a}\mathrm{q}\right)\mathrm{ }+\mathrm{ }{\mathrm{H}}^{+}\left(\mathrm{a}\mathrm{q}\right)\rightleftharpoons \mathrm{N}{\mathrm{H}}_4^{+}\left(\mathrm{a}\mathrm{q}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

${\mathrm{K}}_1=\frac{\left[\mathrm{N}{\mathrm{H}}_4^{+}\right]}{\left[\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}\right]\left[{\mathrm{H}}^{+}\right]}=1.8\times {10}^9;$

$\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}\mathrm{ }\left(\mathrm{a}\mathrm{q}\right)\rightleftharpoons \mathrm{N}{\mathrm{H}}_4^{+}\left(\mathrm{a}\mathrm{q}\right)+\mathrm{O}{\mathrm{H}}^{-}\left(\mathrm{a}\mathrm{q}\right)$

${\mathrm{K}}_2=\frac{\left[\mathrm{N}{\mathrm{H}}_4^{+}\right]\left[\mathrm{O}{\mathrm{H}}^{-}\right]}{\left[\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}\right]}$

Multiply this by$\left[{\mathrm{H}}^{+}\right]$and divide also

${\mathrm{K}}_2=\frac{\left[\mathrm{N}{\mathrm{H}}_4^{+}\right]\left[\mathrm{O}\stackrel{-}{\mathrm{H}}\right]\left[{\mathrm{H}}^{+}\right]}{\left[\mathrm{N}{\mathrm{H}}_4\mathrm{O}\mathrm{H}\right]\left[{\mathrm{H}}^{+}\right]}={\mathrm{K}}_1\times {\mathrm{K}}_{\mathrm{w}}$

$=\mathrm{ }1.8\times 10^9\times 1.0\times 10^{-14}=18\times 10^{-6}$