Question

A solution contains 0.010 M Ba2+ and 0.010 M Ag+. If we add chromate (CrO42-) to...

A solution contains 0.010 M Ba2+ and 0.010 M Ag+. If we add chromate (CrO42-) to the solution:

Which ion starts to precipitate first?

Can 99.90% of either ion be precipitated by chromate without precipitating the other metal ion?

pKsp BaCrO4 = 9.67, pKsp Ag2CrO4=11.92

Homework Answers

Answer #1

Ksp of BaCrO4 is 2.14 x 10^-10

Ksp of Ag2CrO4 is 1.20 x 10^-12

As the Ksp of Ag2CrO4 is lower than the Ksp of BaCrO4, Ag+ would precipitate first from the solution

[CrO4^2-] needed for Ba2+ precipitation = 2.14 x 10^-10/0.01 = 2.14 x 10^-8 M

at this chromate concentration, [Ag+] = sq.rt.(1.20 x 10^-12/2.14 x 10^-8) = 7.5 x 10^-3 M

percentage of Ag+ precipitated = 7.5 x 10^-3 x 100/0.01 = 75%

So, precipitation of 99.9% Ag+ is not possible without precipitating the other metal ion

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