Lithium fluoride, LiF, is a salt formed from the neutralization

Question

Lithium fluoride, LiF, is a salt formed from the neutralization of the weak acid hydrofluoric acid, HF, with the strong base lithium hydroxide. Given that the value of Ka for hydrofluoric acid is 3.5×10−4, what is the pH of a 0.187 M solution of lithium fluoride at 25∘C

Answer 1

To find the answer to questions like this, you should look at the hydrolysis of the conjugate. In this case, the F^- ion.

F^- + H₂O ==> HF + OH^- In this case, F^- is the conjugate base, and we can find the Kb from Kw = KaKb

1x10^-14 = (3.5x10^-4)(Kb) and Kb = 2.86x10^-11

Kb = 2.86x10^-11 = [HF][OH^-]/[F^-]

2.86x10^-11 = (x)(x)/(0.187-x) and if we assume x is small, we can neglect it in the denominator

2.86x10^{-11 =} x^2/0.187

x² = 5.34x10^-2

x = [OH^-] = 2.31x10^-6 M (note: this value is small compared to 0.187 M so assumption was valid)

pOH = -log [OH^-] = -log 2.31x10^-6 = 5.64

pH = 14 - pOH

pH = 8.36

(note: since LiF is the salt of a strong base + weak acid, the pH should be > 7)