Question

% ionisation of weak acid can be calculated as:

• A. $\sqrt[100]{100\frac{K_a}{c}}$
• B. $\frac{100}{1+{10}^{(p{K}_a-pH)}}$
• C. Both are correct
• D. None is correct

Answer 1

Answer: (C)

Both are correct

$HA\leftrightharpoons H^{+}+A^{-}$

Initially, $\left[HA\right]=C\left[H^{+}\right]=0\left[A^{-}\right]=0$

$\left[HA\right]=C(1-\alpha )\left[H^{+}\right]=C\alpha \left[A^{-}\right]=C\alpha$

Ionic product of acid HA, $K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}\approx C{\alpha }^2$

Since, $1-\alpha \approx 1$

$\alpha =\sqrt{\frac{K_a}{C}}$ and $C=\frac{K_a}{{\alpha }^2}$

If $\alpha$ is not very small

$K_a=\frac{C{\alpha }^2}{1-\alpha }=\frac{\left[H^{+}\right]\alpha }{1-\alpha }$

$\frac{1-\alpha }{\alpha }=\frac{\left[H^{+}\right]}{K_a}$

$\frac{1}{\alpha }-1=\frac{\left[H^{+}\right]}{K_a}$

$\frac{1}{\alpha }=\frac{\left[H^{+}\right]}{K_a}+1$

$\frac{1}{\alpha }=\frac{{10}^{-pH}}{{10}^{-p{K}_a}}+1$

$\frac{1}{\alpha }={10}^{-p{K}_a-pH}+1$

$\frac{1}{\alpha }=\frac{1}{{10}^{-p{K}_a-pH}+1}$

$\therefore \%\alpha =\frac{100}{{10}^{p{K}_a-pH}+1}$

Hence ionisation of weak acid can be calculated as

$100\sqrt{\frac{K_a}{C}}$ and $\frac{100}{{10}^{p{K}_a-pH}+1}$