Posted by **ChemMaster98** on Monday, March 31, 2008 at 7:18am.

The Ksp of Ag2CrO4(s) is 1.12 x 10^-12. Calculate the molar solubility of AgCrO4(s)

a)in pure water

b) in a solution of 0.10 mol/L sodium chromate, Na2CrO4(s)

- chemistry -
**DrBob222**, Monday, March 31, 2008 at 12:49pm
Ag2CrO4 ==> 2Ag^+ + CrO4^=

Ksp = (Ag^+)^2 (CrO4^=) = 1.12 x 10^-12

solubility Ag2CrO4 = S

(Ag^+) = 2S

(CrO4^=) = S

Plug into Ksp and solve for S.

For b.

Everything stays the same except (CrO4^=) = S + 0.1. Substitute into Ksp and solve for S. You can make the simplifying assumption that S + 0.1 = 0.1. If you do, to avoid solving a quadratic equation, then check at the end to make sure S + 0.10 = 0.1. If it isn't, then you will need to regroup and solve the quadratic or use successive approximations. I THINK S will be negligible when compared to 0.1 M.

Check my thinking. Check my work. Post your work if you get stuck.

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