A solution contains 0.10 M iodide, I^-, and 0.10 M carbonate ion, CO^2-_3 A stud

Question

A solution contains 0.10 M iodide, I^-, and 0.10 M carbonate ion, CO^2-_3 A student wants to know if 99% of one cation can be removed from the other by adding aqueous Pb(NO_3)_2. (a) When solid Pb(NO_3)_2 is added to the solution, which salt precipitates first, PbI_2 or PbCO_3? (K_sp for PbI_2 is 8.7 times 10^-9; K_sp = for PbCO_3 is 1.5 times 10^-13) (b) What is the concentration of the first ion that precipitates (I^- or CO^2-_3) when the second, more soluble salt begins to precipitate? (c) Can 99% of one cation be removed from the other in this fashion?

Explanation / Answer

a)

Ksp of PbI2 = 8.7 x 10^-9

Ksp of PbCO3 = 1.5 x 10^-13

That one with lower Ksp precipitates first, so PbCO3 precipitates first.

b)

When second one starts to precipitaes

that is PbCO3 precipitated before PbI2.

[Pb^+] required to precipitate [I-] is

PbI2 <===>Pb2+ + 2I-

Ksp(PbI2) = [Pb2+][I-]2

[Pb2+] = Ksp/[I-]^2 = (8.7 x 10^-9) / (0.10 M)^2 = 8.7 x 10^-7

[CO3^2-] at this stage = Ksp(PbCO3) / [Pb2+] = (1.5 x 10^-13) / (8.7 x 10^-7) = 1.7 x 10^-7

[CO3^2-] = 1.7 x 10^-7 M

c)

Yes.

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