Answer: $\frac{100}{10^{pK_a-pH}+1}$
$HA \rightleftharpoons H^+ + A^-$
Initially, [HA] = C, [$H^+$] = 0, [$A^-$] = 0
$[HA] = C(1-\alpha), [H^+] = C\alpha , [A^-] = C\alpha$
Ionic Product of the acid HA, $K_a = \frac{[H^+][A^-]}{[HA]} \approx C\alpha^2$ Since $1-\alpha \approx 1$
$\Rightarrow \alpha = \sqrt{\frac{K_a}{C}}$ and $C = \frac{K_a}{\alpha^2}$
% $\alpha = 100 \sqrt{\frac{K_a}{C}}$
If $\alpha$ is not very small
$K_a = \frac{C \alpha^2}{1-\alpha} = \frac{[H^+] \alpha}{1 - \alpha}$
$\Rightarrow \frac{1-\alpha}{\alpha} = \frac{[H^+]}{K_a}$
$\Rightarrow \frac{1}{\alpha} - 1 = \frac{H^+}{K_a}$
$\Rightarrow \frac{1}{\alpha} = \frac{H^+}{K_a} + 1$
$\Rightarrow \frac{1}{\alpha} = \frac{10^{-pH}}{10^{-pK_a}} + 1$
$\Rightarrow \frac{1}{\alpha} = 10^{pK_a - pH} + 1$
$\Rightarrow \frac{1}{\alpha} = \frac{1}{10^{pK_a-pH} +1}$
$\therefore$ % $\alpha = \frac{100}{10^{pK_a-pH}+1}$