Determine the percent ionization of a 0.145 m hcn solution.

When Ka of HCN  = 6.2x10^-10 

              HCN       H3O+      CN-
initial    0.145          0            0
change -X              +X          +X
final     (0.145-X)     +X          +X

Ka= [H3O+][CN]- / [HCN]
by substitution:
we can assume [HCN] = 0.145
6.2x10^-10 = X*X / (0.145) by solving this equation,
X = 9.5 x 10^-6 
∴ [ H3O+] = 9.5 X 10^-6 
∴percent ionization = [H3O+]/[HCN] *100
                               = 9.5X10^-6 / 0.145 *100
                               = 0.0066 %

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