Calculate the solubility of Ag2CrO4 (Ksp = 9.0 × 10–12) in a 0.021 M AgNO3 solution.
Concentration of Ag2CrO4 = 0.21 M
Ksp of Ag2CrO4 = 9.0*10-12
Solubility of Ag2CrO4 =?
Ag2CrO4 --------- Ag2+ + CrO42-
Ksp = [Ag+]2 [CrO42-]
Let, S is the solubility of CrO42- in
the solution, so;
Ksp= S2 * S
9.0*10-12 = S3
S = 2.0800*10-4 M
The solubility for CrO42- will be;
= 2.0800*10-4 M
And for Ag+ will be;
[Ag+] = 2 * 2.0800*10-4
= 4.160*10-4 M
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