Equilibrium constant for reaction N H _ 4 OH(aq) + H ^ + (aq) ⇌ N H...

$N{H}_{4}OH\left(aq\right)$ + {H}^{+}\left(aq\right)$⇌ N{H}_{4}^{+}\left(aq\right)$ + {H}_{2}O\left(l\right)$<br /><br />${K}_{1} = \dfrac{\left[N{H}_{4}^{+}\right]}{\left[N{H}_{4}OH\right]\left[{H}^{+}\right]} = 1.8 \times {10}^{9};$<br /><br />$N{H}_{4}OH\left(aq\right) ⇌ N{H}_{4}^{+}\left(aq\right) + O{H}^{-}\left(aq\right)$<br /><br />${K}_{2} = \dfrac{\left[N{H}_{4}^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}$<br /><br />Multiply this by \left[{H}^{+}\right]$ and divide also

${K}_{2} = \dfrac{\left[N{H}_{4}^{+}\right]\left[O\overset{-}{H}\right]\left[{H}^{+}\right]}{\left[N{H}_{4}OH\right]\left[{H}^{+}\right]} = {K}_{1} \times {K}_{w}$

$= 1.8 \times 1{0}^{9} \times 1.0 \times 1{0}^{- 14} = 18 \times 1{0}^{- 6}$