What is the molar solubility of AgCl?

Question

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 x 10^-10 and the Kf for the complex ion is 2.9 x 10^13

Answer 1

Let's assume that the molar solubility of AgCl is x.
The Ag+ that comes from the dissolved AgCl will be in the bound form in the complex and also in free form (which takes part in 2 euilibrium reactions)
So the total amount of Ag+ is
[Ag+]total= [Ag+]+ [Ag(S2O3)2(-3)] = x
If we call [Ag(S2O3)2(-3)]=y, then the free Ag+ at equilibrium is

[Ag+]=x-y

.. .. .. .. .. AgCl <=> Ag+ +Cl-
Dissolve .. x
produce .. .. .. .. .. .. x .. .. ..x
At equil .. .. .. .. .. .. x-y .. .. x

Ksp= [Ag+][Cl-] =(x-y)x =1.8*10^-10 (1)

.. .. .. .. .. .. Ag+ + 2S2O3(-2) <=> Ag(S2O3)2(-3)
Initial .. .. .. x .. .. .. .. 1
React .. .. ..y .. .. .. .. 2y
Produce .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. y
At equil. .. x-y .. .. .. 1-2y .. .. .. .. .. .. .. y

Kf= [Ag(S2O3)2(-3)] / [Ag+][S2O3(-2)]^2 =>
y/[(x-y)(1-2y)^2] =2.9*10^13 (2)

You have a system of two equations. It is quite complex so it is best to solve it with a special calculator. There is a free on-line tool for this
from http://www.quickmath.com/ but it is temporarily down.
Otherwise, if you have a program like Mathematica, use that to solve the system

Then
x=
y=

If it is restored in time, I'll edit my post and include the solution.

[Edit] the server is back
You have 4 possible solutions

x1=4.433*10^-8, y1=0.500
x2=4.434*10^-8, y2=0.499
x3=y3=0.497
x4=y4=0.503

Now, since the solubility of AgCl alone is s=SQRT(Ksp)=
SQRT(1.8*10^-10) = 1.34*10^-5 M, when you can form the complex, solubility should increase. Since x1,x2 < 1.34*10^-5 M they are rejected.
x4 is rejected because for y4=0.503 you would have S2O3(-2) reacting 2*0.503=1.006 which is slightly more than what we initially had (1.000)
so the answer is x3=0.497 M