A solution is 1.0x10-4 M in NaF, Na2S, and Na3PO4. What would be the order of precipitation as a source of Pb2+ is added gradually to the solution? Assume negligible volume change.

Question

Given: Ksp of PbF₂= 4.0x10^-8, Ksp of PbS= 7.0x10^-29, Ksp of Pb₃(PO₄)₂= 1.0x10^-54 .

Show your complete solutions.

Answer 1

PbF₂(s) <==> Pb²⁺(aq) + 2F^-(aq)

Ksp = 4.0x10^-8 = [Pb²⁺][F-]² = [Pb2+][2x10^-4)²

[Pb²⁺] = 1 M

PbS(s) <==> Pb²⁺(aq) + S^2-(aq)

Ksp = 7x10^-29 = [Pb²⁺][S^2-] = [Pb²⁺][1.0x10^-4)

[Pb²⁺] = 7x10^-25 M

Pb₃(PO₄)₂(s) <==> 3Pb²⁺(aq) + 2PO₄^3-(aq)

Ksp = 1x10^-54 = [Pb²⁺]³[PO₄^3-)² = [Pb²⁺]³[2x10^-4]²

[Pb²⁺] = 2.9x10^-16 M

Order of precipitation will be

PbS

Pb₃(PO₄)₂

PbF₂