Calculate the $pH$ of $0.02 \ M$ $ClCH_2COOH$. Given its $K_a = 1.36 \times 10^{-3}$,calculate its $pK_b$.

(N/A) For a weak acid $ClCH_2COOH$,the concentration $C = 0.02 \ M$ and $K_a = 1.36 \times 10^{-3}$.
First,calculate the degree of dissociation $\alpha = \sqrt{K_a / C} = \sqrt{1.36 \times 10^{-3} / 0.02} = \sqrt{0.068} \approx 0.2608$.
$[H^+] = C \times \alpha = 0.02 \times 0.2608 = 0.005216 \ M$.
$pH = -\log[H^+] = -\log(0.005216) \approx 2.283$.
To find $pK_b$,first calculate $pK_a = -\log(K_a) = -\log(1.36 \times 10^{-3}) \approx 2.866$.
Using the relation $pK_a + pK_b = 14$,we get $pK_b = 14 - 2.866 = 11.134$.