What is the pH of a solution made by mixing 0.050 mol of NaCN with enough water to make a liter of solution

pH = 11

Explanation:

The equilibrium of a weak base as NaCN in water is:

NaCN(aq) + H₂O(l) ⇄ OH⁻(aq) + Na⁺(aq) + HCN(aq)

And kb, the equilibrium constant, is:

Kb = [OH⁻] [HCN] / [NaCN]

Where Kb of NaCN is 2.04x10⁻⁵

In the beginning, the [NaCN] is 0.050mol / L = 0.050M.

Both [OH⁻] and [HCN] are produced from this equilibrium, and its concentration is X, that is:

2.04x10⁻⁵ = [X] [X] / [0.050M]

1.02x10⁻⁶ = X²

X = 1x10⁻³ = [OH⁻]

As pOH = - log [OH⁻]

pOH = 3.00

And pH = 14 - pOH