How many grams of PbF2 (molar mass = 245.2) will dissolve in 550 mL of 0.25 M NaF solution? The Ksp for PbF2 is 2.69e-08.

Answer 1

Common ion problem.

PbF₂(s) <==> Pb²⁺(aq) + 2F^-(aq)

Ksp = 2.69x10^-8 = [Pb²⁺][F^-]²

Since the solution contains the common ion F^- (from the NaF), we will take the [F^-] as that of the NaF (0.25 M).

Ksp = 2.69x10^-8 = [Pb²⁺][0.25]²

2.69x10^-8 = 0.0625[Pb²⁺]

[Pb²⁺] = 4.30x10^-7 M = solubility of PbF₂ in 0.25 M NaF

Since the question is asking for GRAMS of PbF₂ in 550 ml, we can now convert the above answer as follows:

4.30x10^-7 moles PbF₂ / L x 0.550 L x 245.2 g / mol = 1.92x10^-4 g PbF₂