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

Let x be the molarity of Pb²⁺ added and R the solubility quotient

R(PbF₂) = x[F^-]² = x(1x10^-4)²

R(PbS) = x[S^2-] = x(1x10^-4)

R(Pb₃(PO₄)₂ = x³[PO₄^3-]² = x³[1x10^-4]²

For precipitation, you need R > K_sp ^.Each of these equations can be used to solve for x required for precipitation.. Whichever one has the lowest x will precipitate first.

I'll do the first one x = 4 x 10^-8/(1 x 10^-4)² = 4