Dimethyl amine $(CH_3)_2NH$ is a weak base and its ionization constant is $5.4 \times 10^{-5}$. Calculate $[OH^{-}]$,$[H_3O^{+}]$,$pOH$ and $pH$ of its $0.2 \ M$ solution at equilibrium.
For a weak base,the equilibrium is: $(CH_3)_2NH + H_2O \rightleftharpoons (CH_3)_2NH_2^{+} + OH^{-}$.
Using the approximation $[OH^{-}] = \sqrt{K_b \times C}$,where $K_b = 5.4 \times 10^{-5}$ and $C = 0.2 \ M$.
$[OH^{-}] = \sqrt{5.4 \times 10^{-5} \times 0.2} = \sqrt{1.08 \times 10^{-5}} = \sqrt{10.8 \times 10^{-6}} \approx 3.286 \times 10^{-3} \ M$.
$pOH = -\log[OH^{-}] = -\log(3.286 \times 10^{-3}) \approx 2.48$.
$pH = 14 - pOH = 14 - 2.48 = 11.52$.
$[H_3O^{+}] = 10^{-pH} = 10^{-11.52} \approx 3.02 \times 10^{-12} \ M$.
Using the approximation $[OH^{-}] = \sqrt{K_b \times C}$,where $K_b = 5.4 \times 10^{-5}$ and $C = 0.2 \ M$.
$[OH^{-}] = \sqrt{5.4 \times 10^{-5} \times 0.2} = \sqrt{1.08 \times 10^{-5}} = \sqrt{10.8 \times 10^{-6}} \approx 3.286 \times 10^{-3} \ M$.
$pOH = -\log[OH^{-}] = -\log(3.286 \times 10^{-3}) \approx 2.48$.
$pH = 14 - pOH = 14 - 2.48 = 11.52$.
$[H_3O^{+}] = 10^{-pH} = 10^{-11.52} \approx 3.02 \times 10^{-12} \ M$.
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