A solution contains 5.0 times 10^-2 M Ag^+ and 2.5 times 10^-2 M Mg^2+. Pb^2+ io

Question

A solution contains 5.0 times 10^-2 M Ag^+ and 2.5 times 10^-2 M Mg^2+. Pb^2+ ion in the form of Pb(NO_3)_2 was added incrementally to a solution that is initially 2.0 times 10^-3 M CrO_42^+ and 4.0 times 10^-2 M F^-. If Ksp(PbCrO_4) = 1.8 time 10^-14 and K_sp(PbF_2) = 3.6 time 10^-8, which of these two anions will form a precipitate with Pb^2+ first. Show all of your work. Calculate the % of the first anion remaining in solution at the point when the other anion begins to precipitate with Pb^2+.

Explanation / Answer

a)

Calculate solubility of each

Ksp PbCrO4 = [Pb+2][CrO4-2]

1.8*10^-14 = (Pb+2)(2*10^-3)

Pb+2 = (1.8*10^-14)/(2*10^-3)) = 9*10^-12 M

for

Ksp PbF2 = [Pb+2][F-]^2

3.6*10^-8 = (Pb+2)(4*10^-2)^2

[Pb+2] = (3.6*10^-8)((4*10^-2)^2) = 5.76*10^-11

so.. clearly, the PbF2 solution will precipitate first

b)

when the second precipitates...

[Pb+2] =  9*10^-12 M will be the concentration of the 2nd preicptate

so:

when

3.6*10^-8 = (9*10^-12)[F-]^2

[F-]= sqrt((3.6*10^-8)/((9*10^-12)) --> 63.24

which is mainly 0% left, since plenty of PbF2 has precipitated

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