At $298 \ K$ temperature,calculate the $pH$ of a $0.25 \ M$ solution of $(CH_3)_2NH$ given that its $K_b$ is $5.4 \times 10^{-4}$.
(A) For a weak base $(CH_3)_2NH$,the dissociation is: $(CH_3)_2NH + H_2O \rightleftharpoons (CH_3)_2NH_2^+ + OH^-$.
The concentration of $OH^-$ is given by $[OH^-] = \sqrt{K_b \times C}$.
$[OH^-] = \sqrt{5.4 \times 10^{-4} \times 0.25} = \sqrt{1.35 \times 10^{-4}} = 1.16 \times 10^{-2} \ M$.
$pOH = -\log[OH^-] = -\log(1.16 \times 10^{-2}) = 2 - 0.064 = 1.936$.
Since $pH + pOH = 14$,$pH = 14 - 1.936 = 12.064$.
The concentration of $OH^-$ is given by $[OH^-] = \sqrt{K_b \times C}$.
$[OH^-] = \sqrt{5.4 \times 10^{-4} \times 0.25} = \sqrt{1.35 \times 10^{-4}} = 1.16 \times 10^{-2} \ M$.
$pOH = -\log[OH^-] = -\log(1.16 \times 10^{-2}) = 2 - 0.064 = 1.936$.
Since $pH + pOH = 14$,$pH = 14 - 1.936 = 12.064$.
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