Question

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

Answer 1

2.28 × 10^-3 mol/L

Step-by-step 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