Asked by ChemMaster98
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)
a)in pure water
b) in a solution of 0.10 mol/L sodium chromate, Na2CrO4(s)
Answers
Answered by
DrBob222
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.
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.
Answered by
Buster
yeah... its cool
Answered by
ERICK JOHN
Im not sure ...need more explanations please?
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