Question

What is the \(\mathrm{pH}\) of a solution made by dissolving \(5.36\) grams of calcium fluoride in enough water to make \(680 \mathrm{~mL}\) of solution? The \(\mathrm{K}_{\mathrm{a}}\) for

HF is \(6.8 \times 10^{-4}\)

Answer 1

Moles of CaF2 = mass/molecular weight

= 5.36g / 78.07 g/mol

= 0.06865 mol

Moles of F- = 2 x 0.06865 mol = 0.1373 mol

[CaF2] = 0.1373 mol / 0.680 L = 0.2019 M

Consider the reaction

F- + H2O = HF+ + OH-

Kb = [HF+] [OH-] /[F-]

[HF+] = [OH-] = x

Kb = Kw/Ka = 10^-14 / 6.8*10^-4 = 1.47 x 10^-11

1.47 x 10^-11 = x² / (0.2019 - x)

x = 1.723 x 10^-6 = [OH-]

pOH = - log (1.723 x 10^-6) = 5.763

pH = 14 - pOH = 14 - 5.763 = 8.237