The salt formed by the reaction of the weak acid hydrocyanic acid, HCN, with the strong base potassium hydroxide is

potassium cyanide, KCN. What is the hydroxide ion concentration of a 0.255 M solution of potassium cyanide at 25 °C

given that the value of for hydrocyanic acid is 4.9 X 10-102

2.28 * 10^-3 mol/L

Explanation:

The equation for the equilibrium is

CN^ - + H2O ⇌ HCN + OH^-

Ka = 4.9 * 10^-10

KaKb = Kw

4.9 * 10^-10 Kb = 1.00 * 10^-14

Kb = (1.00 * 10^-14) / (4.9 * 10^-10) = 2.05 * 10^-5

Now, we can set up an ICE table

CN^ - + H2O ⇌ HCN + OH^-

I / (mol/L) 0.255 0 0

C / (mol/L) - x + x + x

E / (mol/L) 0.255 - x x x

Ka = x^2 / (0.255 - x) = 2.05 * 10^-5

Check for negligibility

0.255 / (2.05 * 10^-5) = 12 000 > 400. ∴ x ≪ 0.255

x^2 = 0.255 (2.05 * 10^-5) = 5.20 * 10^-6

x = sqrt (5.20 * 10^-6) = 2.28 * 10^-3

[OH^-] = x mol/L = 2.28 * 10^-3 mol/L