The ka for hydrocyanic acid (hcn) is 4.9 * 10-10. what is [oh-] in a 0.075 m solution of sodium cyanide (nacn) ?

by using ICE table:

CN - + H2O ↔ HCN + OH-

initial 0.075 0 0

change - X + X + X

Equ (0.075-X) X X

when Kb = Kw/Ka

∴Kb = 1 x 10^-14 / 4.9 x 10^-10

= 2 x 10^-5

and when Kb = [HCN][OH]/[CN-]

by substitution:

2 x 10^-5 = X^2 / (0.075 - X) by solving for X

∴ x = 0.0012 M

∴[OH] = 0.0012 M