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 %
