Chemistry - Chemical Equilibrium Question with Solution | TestHub

${\mathrm{NH}}_4\mathrm{OH}\left(\mathrm{aq}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{NH}}_4^{+}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

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

${\mathrm{NH}}_4\mathrm{OH}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{NH}}_4^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

${\mathrm{K}}_2=\frac{\left[{\mathrm{NH}}_4^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{NH}}_4\mathrm{OH}\right]}$

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

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

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

value of x = 18.