What is the pH?

Question

What is the pH of a 0.340 M solution of NaCN (Ka of HCN is 4.9 × 10⁻¹⁰)?

Answer 1

CN^- is a conjugate base of a weak acid. It will hydrolyze (react with water) to produce hydroxide ions

CN^-(aq) + H₂O(l) ⇔ HCN(aq) + OH^-(aq)

0.340 0 0

-x +x +x

0.340-x x x

We drop x since Kb value is small and the concentration of products is much smaller than the concentration of the reactants

This is a base, we need to use Kb

Ka x Kb = Kw

Kb = Kw/Ka = (1 x 10^-14)/4.9 × 10^-10= 2.0 x 10^-5

Kb = x²/0.340

2.0 x 10^-5 = x²/0.340

x= [OH^-] =2.61 x 10^-3

pOH = log [OH^-] = 2.583 pH = 14 - pOH = 11.417

Note: the Ka and Kb constants should not be considered for significant figures. You want to look at the concentrations.