Calculate the pH of a 3.92×10-3 M solution of NaF, given that the Ka of HF = 6.80 x 10-4 at 25°C.

Question

Calculate the pH of a 3.92×10^-3 M solution of NaF, given that the K_a of HF = 6.80 x 10^-4 at 25°C.

Answer 1

This is very much like the previous question you posted about sodium acetate. In this case, the salt is NaF, and the conjugate base of HF is the F^- ion. So, as before, we will look at the hydrolysis of F- and use the Kb for F- to solve for pOH. Then convert to pH.

Hydrolysis of F^-:

F^- + H₂O ==> HF + OH^-

Write Kb expression:

Kb = [HF][OH^-] / [F^-] and Kb = 1x10^-14 / 6.80x10^-4 = 1.47x10^-11

1.47x10^-11 = (x)(x) / 3.92x10^-3 - x and assume x is small relative to 3.92x10^-3 and ignore

1.47x10^-11 = (x)(x) / 3.92x10^-3

x² = 5.76x10^-14

x = 2.40x10^-7 = [OH^-] (note: this is insignificant relative to 3.9x10^-3 M so above assumption was valid)

Now, depending on your level of Chemistry, because the [OH^-] is so low, we really should also consider the [OH^-] that comes from the H₂O, which is 1x10^-7 M. This would be the correct way to answer the question.

If you ignore the contribution from H₂O, we have...

pOH = -log 2.40x10^-7 = 6.62

pH = 14 - 6.62 = 7.38

If you consider the contribution from H₂O, we have...

[OH^-] = 2.40x10^-7 + 1x10^-7 = 3.40x10^-7 M

pOH = -log 3.40x10^-7 = 6.47

pH = 14 - 6.47 = 7.53