Question

What is the pH of a solution prepared by mixing 50.00 mL of 0.30 M HF with 50.00 mL of 0.030 M NaF? Assume that the volume of the solutions are additive and that Ka = 1.72 × 10-4 for HF.

Answer 1

Given, Ka = 1.72 X 10^-4

Correponding pKa = -logKa = -log1.72 X 10^-4 = 3.764

Now, find the moles of HF and NaF in given solution:

Moles of HF = (0.05 L)(0.3 M) = 0.015 mol
Moles of NaF = (0.05 L)(0.03 M) = 0.0015 mol

These two react with a 1:1 stoichiometry,

And final volume of solution = 0.05 + 0.05 = 0.1 L

Now, Concentrations of both are:

[HF] = 0.015 mol / 0.1L = 0.15 M
[NaF] = 0.0015 mol / 0.1L = 0.015 M

pH = pka + log[NaF] / [HF]

pH = 3.764 + log [0.015] / [0.15]

pH = 3.764 + (-1)

pH = 2.764