The ionization constant of $ClCH_{2}COOH$ is $1.35 \times 10^{-3}$. What will be the $pH$ of $0.1 \ M$ acid and its $0.1 \ M$ sodium salt solution?
For the weak acid $ClCH_{2}COOH$:
$K_{a} = 1.35 \times 10^{-3}$,$c = 0.1 \ M$.
$[H^{+}] = \sqrt{K_{a} \times c} = \sqrt{1.35 \times 10^{-3} \times 0.1} = \sqrt{1.35 \times 10^{-4}} = 0.0116 \ M$.
$pH = -\log(0.0116) = 1.935 \approx 1.94$.
For the salt $ClCH_{2}COONa$ (salt of weak acid and strong base):
$K_{h} = \frac{K_{w}}{K_{a}} = \frac{10^{-14}}{1.35 \times 10^{-3}} = 7.407 \times 10^{-12}$.
$[OH^{-}] = \sqrt{K_{h} \times c} = \sqrt{7.407 \times 10^{-12} \times 0.1} = \sqrt{7.407 \times 10^{-13}} = 8.606 \times 10^{-7} \ M$.
$pOH = -\log(8.606 \times 10^{-7}) = 7 - 0.935 = 6.065$.
$pH = 14 - pOH = 14 - 6.065 = 7.935 \approx 7.94$.
$K_{a} = 1.35 \times 10^{-3}$,$c = 0.1 \ M$.
$[H^{+}] = \sqrt{K_{a} \times c} = \sqrt{1.35 \times 10^{-3} \times 0.1} = \sqrt{1.35 \times 10^{-4}} = 0.0116 \ M$.
$pH = -\log(0.0116) = 1.935 \approx 1.94$.
For the salt $ClCH_{2}COONa$ (salt of weak acid and strong base):
$K_{h} = \frac{K_{w}}{K_{a}} = \frac{10^{-14}}{1.35 \times 10^{-3}} = 7.407 \times 10^{-12}$.
$[OH^{-}] = \sqrt{K_{h} \times c} = \sqrt{7.407 \times 10^{-12} \times 0.1} = \sqrt{7.407 \times 10^{-13}} = 8.606 \times 10^{-7} \ M$.
$pOH = -\log(8.606 \times 10^{-7}) = 7 - 0.935 = 6.065$.
$pH = 14 - pOH = 14 - 6.065 = 7.935 \approx 7.94$.
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