Lithium cyanide, LiCN, is a salt formed from by the reaction of the weak acid hydrocyanic acid, HCN, and the strong base lithium hydroxide. The Ka of hydrocyanic acid is 4.90×10−10.

Question

Lithium cyanide, LiCN, is a salt formed from by the reaction of the weak acid hydrocyanic acid, HCN, and the strong base lithium hydroxide. The Ka of hydrocyanic acid is 4.90×10−10.

Using this information, determine the pH of a 0.119 M solution of LiCN that is prepared at 25∘C.

Answer 1

Use the K_a value to calculate K_b for LiCN.

K_b = K_w / K_a = 1 x 10^-14 / 4.90 x 10^-10 = 2.04 x 10^-5

Since Li⁺ is a spectator ion, the net ionic equation for the reaction of CN^- with water is:

CN- + H₂O <==> HCN + OH^-

K_b = [HCN] [OH^-] / [CN^-] = 2.04 x 10^-5

Setup an ICE table to determine the equilibrium concentration of each species.

[HCN]_eq = x

[OH^-]_eq = x

[CN^-]_eq = 0.119 - x --> assuming the value of x is significantly smaller than 0.119 M, the K_b expression simplifies to

2.04 x 10^-5 = (x) (x) / 0.119

x² = (2.04 x 10^-5) x 0.119

x² = 2.43 x 10^-6

x = √2.43 x 10^-6 = 1.56 x 10^-3

x = [OH^-] = 1.56 x 10^-3

pOH = –log [OH^-] = –log 1.56 x 10^-3 = 2.81

pH = 14.00 - 2.81 = 11.19