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 + H2  H3O+  +  CN-  Ka = 4.8 X 10-10
as  water is present it also dissociates  H2O  + H2  H3O+  +  OH  Kw = 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


Ka= [H3O+] [CN-][HCN]= (0.01α) (0.01α)0.014.8 × 10-10 = 0.01 α2α2 = (4.8 × 10-10)(0.01) = 4.8 ×10-12α =4.8 ×10-12 = 2.2 ×10-6
 
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 =  cα = 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.

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