Determine the pH of a 0.22 M NaF solution at 25°C. The Ka of HF is 6.8 × 10^-4.

Answer 1

NaF is the salt of a strong base (NaOH) and a weak acid (HF), so this salt will be basic (pH > 7) when dissolved. We want to look at the hydrolysis reaction of F- as follows:

F^- + H2O ==> HF + OH^- where F^- is acting as a base, so we need to use the Kb

Recall that KaKb = Kw

Kb = Kw / Ka = 1x10^-14 / 6.8x10^-4

Kb = 1.47x10^-11

Kb = 1.47x10^-11 = [HF][OH-] / [F-]

1.47x10^-11 = (x)(x) / 0.22 - x and assuming x is small relative to 0.22 we can ignore it in the denominator

1.47x10^-11 = x² / 0.22

x² = 3.23x10-¹²

x = 1.80x10^-6 M = [OH-] ... which is quite small relative to 0.22 so above assumption was valid

pH + pOH = 14

pH = 14 - pOH

pOH = -log 1.80x10^-6 = 5.75

pH = 14 - 5.75

pH = 8.25