Calculate the degree of ionization of 0.01M solution of HCN.Ka = 4.8 X 10-10.Also calculate H⁺ ion concentration of the solution

Assume the degree of dissociation to be α. The acid dissociation reaction will be 
  HCN + H₂O $\iff$  H₃O⁺  +  CN^-  K_a = 4.8 X 10^-10
as  water is present it also dissociates  H₂O  + H₂O $\iff$  H₃O⁺  +  OH^-   K_w = 1 X 10^-14

The initial and equilibrium concentration of the reactants and products for HCN, H₃O⁺ and CN^- will be

Initial conc  0.01M  0  0
Change      -0.01α  0.01α  0.01α
Equilibrium conc  0.01-0.01α or 0.01(1-α)  cα  cα

As HCN is a weak acid as evidenced by low Ka value,  α value is negligible and 1-α = 1 and 0.01(1-α) = 0.01


$$\begin{gathered} K_a= \frac{\left[H3O^{+}\right] \left[C{N}^{-}\right]}{\left[HCN\right]}= \frac{(0.01\alpha ) (0.01\alpha )}{0.01} \\ 4.8 \times {10}^{-10} = 0.01 {\alpha }^2 \\ {\alpha }^2 = (4.8 \times {10}^{-10})(0.01) = 4.8 \times {10}^{-12} \\ \alpha =\sqrt{4.8 \times {10}^{-12}} = 2.2 \times {10}^{-6} \end{gathered}$$
 
degree of dissociation is 2.2 X 10^-6

For pH, we need the concentration of H₃O⁺ which is contributed by HCN and H₂O dissociation 9as HCn is a weak acid, this cannot be ignored)

[H₃O⁺] i.e. [H⁺]  form HCN dissociation =  c$\alpha$ = 0.01 X ( 2.2 X 10^-6 ) = 2.2 X 10^-8 M

[H₃O⁺] i.e. [H⁺]  form water dissociation = 1X 10^-7 M

Total [H⁺] = (2.2 X 10^-8) + (1 X 10^-7 ) = 1.22 X 10^-7 M

pH = -log [H+] = -log (1.22 X 10^-7) = - (-6.9137) = 6.91.