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

${NH}_{4}OH\left(aq\right)$ + {H}^{+}\left(aq\right)$⇌ {NH}_{4}^{+}\left(aq\right)$ + {H}_{2}O\left(l\right)$</p><p>${K}_{1} = \dfrac{\left[\overset{+}{N}{H}_{4}\right]}{\left[{NH}_{4}OH\right]\left[{H}^{+}\right]} = 1 .8 \times {10}^{9}$</p><p>${NH}_{4}OH\left(aq\right) ⇌ {NH}_{4}^{+}\left(aq\right) + {OH}^{-}\left(aq\right)$</p><p>${K}_{2} = \dfrac{\left[{NH}_{4}^{+}\right]\left[{OH}^{-}\right]}{\left[{NH}_{4}OH\right]}$</p><p>Multiply this by \left[{H}^{+}\right]$ and divide also

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

$= 1 .8 \times {10}^{9} \times 1 .0 \times {10}^{- 14} = 18 \times {10}^{- 6}$