Silver chromate is sparingly soluble in aqueous solutions. The Ksp of Ag2CrO4 is 1.12× 10^–12. What is the solubility (in mol/L) of silver chromate in 1.50 M potassium chromate aq solution? In 1.50 M silver nitrate aq solution? In pure water?

I'm assuming I need to write the solubility expression. If I have this right, Ksp=[2Ag]^2[CrO4] which becomes 1.12x10^-12=4s^3. But how to I then incorporate the questions it's asking for? Where would I plug in potassium chromate if there is no potassium in the original expression?

1 Answer
May 11, 2018

in potassium chromate s=4,3×107 mol/L
in silver nitrate s=5×1013 mol/L
in distilled waters=6.5×105 mol/L

Explanation:

Ag2CrO4=2Ag++CrO24
if s are the mol of salt that solubilize you have solubilized 2s mol of Ag+ and s mol of chromate
Kps=[Ag+]2×[CrO24]=(2s)2×s=4s3=1,12×1012
hence s=3Kps4=0.65×104=6.5×105 that is the solubility in mol/L in pure water

in a solution of one same ion you have also the ions of the new salt so you have for a solution of potassium chromate 1,50 M
Kps=[Ag+]2×[CrO24]=(2s)2×(s+1,5)=4s3=1,12×1012
since s is smallcompared with 1,5 you can write:
Kps=[Ag+]2×[CrO24]=(2s)2×1,5)=6s2=1,12×1012
hence s=2Kps6=0.43×106=4.3×107 mol/L that is about 100 times less soluble of the solution in pure water.

in a solution of silver nitrate 1,5 M you have
Kps=[Ag+]2×[CrO24]=(2s+1.5)2×s=4s3=1,12×1012
but s << 1.5, so you can write:
Kps=[Ag+]2×[CrO24]=(1.5)2×s=2.25s=1,12×1012
s=0.5×1012=5×1013 mol/L that is about 1 milion times less soluble than in pure water