Question

Which of the two solutions $0.1M$ $HCN$ $(K_a=4\times {10}^{-10})$ and $0.1M$ $HF$ $(K_a=6.7\times {10}^{-4})$ will have greater degree of ionisation and to what extent?

Answer 1

Degree of dissociation '$\alpha$' for weak acid=$\sqrt{\frac{K_a}{C}}$

For $0.1MHCN$

${\alpha }_1=\sqrt{\frac{K_a}{C}}=\sqrt{\frac{4\times {10}^{-10}}{0.1}}$

For $0.1MHF$

${\alpha }_2=\sqrt{\frac{K_a}{C}}=\sqrt{\frac{6.7\times {10}^{-4}}{0.1}}$

$\frac{{\alpha }_2}{{\alpha }_1}=\frac{\sqrt{K_{a2}}}{\sqrt{K_{a1}}}=1294$

So ${\alpha }_2=1294{\alpha }_1$.