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Q.
Ionisation of weak acid can be calculated by the formula
Equilibrium
Solution:
For weak acid dissociation equilibria, degree of dissociation $\alpha$ is given as :
$\alpha=\sqrt{\frac{K_a}{C}} \therefore \% \alpha=100 \sqrt{\frac{K_a}{C}}$
Also, $K_a=\frac{\left[ H ^{+}\right]\left[ A ^{-}\right]}{[ HA ]}=\frac{\left[ H ^{+}\right] c \alpha}{c(1-\alpha)}=\frac{\left[ H ^{+}\right] \alpha}{(1-\alpha)}$
$\log K_a=\log H ^{+}+\log \frac{\alpha}{1-\alpha}$
or $p K_a= pH +\log \frac{1-\alpha}{\alpha}$
$p K_a- pH =\log \frac{1-\alpha}{\alpha}$
$\frac{1-\alpha}{\alpha}=10^{ p K_a- pH }$
or $ \frac{1}{\alpha}=10^{ p K_a- pH }+1$
$\therefore \alpha=\frac{1}{\left[1+10^{ p K_a- pH }\right]}$
or $\% \alpha=\frac{100}{\left[1+10^{ p K_a- pH }\right]}$