What is the molar solubility of AgCl in 1.0 M K2S2O3 if the complex ion Ag(S2O3)2^3-forms? The Ksp for AgCl is 1.8 × 10^-10 and the Kf for Ag(S2O3)2^3- is 2.9 × 10^13.

Answer 1

AgCl(s) ==> Ag^+ + Cl^-
Ag^+ + 2[S2O3]^3- ==> [Ag(S2O3)2]^3-
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AgCl(s) + 2[S2O3]^3- ==> [Ag(S2O3)2]^3- + Cl^-
I...solid...........1.0.......................0........................0
C..................-2S........................S........................S
E...................1.0-2S...................S........................S
Krxn = Kf*Ksp = (S)(S)/(1.0-2S) ^2
Solve for S = solubility.
Post your work if you get stuck.

Answer 2

So I was able to set this up like you did, However the answer key stated that the answer was 0.5 M which makes no sense with the set up we both got as the denominator would be 0 and wouldnt equal the reaction of the ion formation constant and the solubility product constant

Answer 3

You're right. If 0.5 M is S then the denominator is 0 and that can't be. However, I think you should have solved the equation. I did and the equation solution is two roots, both 0.5M. How can that be? Because the quadratic is 5.22E3 -2.088E4 S + 2.088E4 S^2 - 1S^2. So 1S^2 subtracted from 2.088E4 S^2 is for all practical purposes 2.088E4 S^2. The solution of that equation is 0.5M but it's that VERY SMALL fraction less than 0.5 M that would make the denominator not quite zero. In other words making the approximation that 1S^2 subtracted from 2.088E4 S^2 = 2.088E4 S^2 is why the answer is 0.5 M and not (Probably something like) 0.49999 or whatever. Hope this helps.

Answer 4

Yeah sorry, I wasn't thinking straight when doing that problem. Thank you very much for your help