The ionization constant of $HF$ is $3.2 \times 10^{-4}$. Calculate the degree of dissociation of $HF$ in its $0.02 \ M$ solution. Calculate the concentration of all species present ($H_{3}O^{+}$,$F^{-}$ and $HF$) in the solution and its $pH$.

(N/A) The following proton transfer reactions are possible:
$1) HF + H_{2}O \rightleftharpoons H_{3}O^{+} + F^{-} \quad K_{a} = 3.2 \times 10^{-4}$
$2) H_{2}O + H_{2}O \rightleftharpoons H_{3}O^{+} + OH^{-} \quad K_{w} = 1.0 \times 10^{-14}$
Since $K_{a} \gg K_{w}$,the first reaction is the principal reaction.
Let $\alpha$ be the degree of dissociation.

Species	$HF$	$H_{3}O^{+}$	$F^{-}$
Initial concentration $(M)$	$0.02$	$0$	$0$
Change $(M)$	$-0.02\alpha$	$+0.02\alpha$	$+0.02\alpha$
Equilibrium concentration $(M)$	$0.02(1-\alpha)$	$0.02\alpha$	$0.02\alpha$

Substituting into the equilibrium expression:
$K_{a} = \frac{[H_{3}O^{+}][F^{-}]}{[HF]} = \frac{(0.02\alpha)^{2}}{0.02(1-\alpha)} = \frac{0.02\alpha^{2}}{1-\alpha} = 3.2 \times 10^{-4}$
$0.02\alpha^{2} = 3.2 \times 10^{-4} - 3.2 \times 10^{-4}\alpha$
$\alpha^{2} + 0.016\alpha - 0.016 = 0$
Using the quadratic formula $\alpha = \frac{-b + \sqrt{b^{2} - 4ac}}{2a}$:
$\alpha = \frac{-0.016 + \sqrt{(0.016)^{2} - 4(1)(-0.016)}}{2} \approx 0.119 \approx 0.12$
Concentrations:
$[H_{3}O^{+}] = [F^{-}] = 0.02 \times 0.12 = 2.4 \times 10^{-3} \ M$
$[HF] = 0.02(1 - 0.12) = 0.0176 \ M = 1.76 \times 10^{-2} \ M$
$pH = -\log[H_{3}O^{+}] = -\log(2.4 \times 10^{-3}) = 3 - \log(2.4) \approx 3 - 0.38 = 2.62$