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 %