What is the pH of a solution made by dissolving 3.56 grams of calcium fluoride in enough water to make 360 mL of solution?

Question

The K_a for HF is 6.80x10^–4.

Answer 1

CaF₂ is the salt of a weak acid (HF) and a strong base, Ca(OH)₂. Consequently, it will be a basic salt and the pH should be >7. We approach this problem by looking at the hydrolysis of the F- anion (which originates from CaF2 when dissolved in water)

F-(aq) + H2O(l) ==> HF(aq) + OH-(aq) (Calcium is a spectator ion and is not included)

molar mass CaF₂ = 78.07 g / mole

moles CaF₂ used = 3.56 g x 1 mol / 78.07 g = 0.04560 moles CaF₂

moles F- = 0.04560 mol CaF₂ x 2 mol F- / mol = 0.09120 mols F-

[F-] = 0.09120 mols / 360 ml x 1000 ml / L = 0.2533 M

F-(aq) + H2O(l) ==> HF(aq) + OH-(aq)

Since F- is acting as a base, we find the Kb as follows:

KaKb = 1x10^-14

Kb = 1x10^-14 / 6.80x10^-4 = 1.47x10^-11

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

1.47x10^-11 = (x)(x) / 0.2533 - x (ignore x in denominator since Kb is so small)

1.47x10^-11 = x² / 0.2533

x² = 3.73x10^-12

x = [OH-] = 1.93x10^-6 M

pOH = -log [OH-] = -log 1.93x10^-6

pOH = 5.71

pH = 14 - pOH

pH = 8.29 (basic as predicted)