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.
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.
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
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.
Yeah sorry, I wasn't thinking straight when doing that problem. Thank you very much for your help
To find the molar solubility of AgCl in 1.0 M K2S2O3, we need to consider the equilibrium between AgCl dissolving and the formation of the complex ion Ag(S2O3)2^3-.
Let's assume that x mol/L of AgCl dissolves in the presence of 1.0 M K2S2O3. Therefore, the concentration of Ag+(aq) ions will also be x mol/L.
The balanced equation for the dissolution of AgCl is:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
The equilibrium constant for this reaction is the solubility product constant (Ksp) for AgCl, which is given as 1.8 × 10^-10.
Ksp = [Ag+(aq)][Cl-(aq)]
Since the concentration of Ag+(aq) equals x and the concentration of Cl-(aq) also equals x, the equation becomes:
Ksp = x * x
Ksp = x^2
Now, let's consider the formation of the complex ion Ag(S2O3)2^3-. The balanced equation for this reaction is:
Ag+(aq) + 2S2O3^2-(aq) ⇌ Ag(S2O3)2^3-(aq)
The equilibrium constant for this reaction is the formation constant (Kf) for Ag(S2O3)2^3-, which is given as 2.9 × 10^13.
Kf = [Ag(S2O3)2^3-(aq)] / ([Ag+(aq)] * [S2O3^2-(aq)]^2)
Since the concentration of Ag+(aq) is x and the concentration of S2O3^2-(aq) is 1.0 M (from the K2S2O3 solution), the equation becomes:
Kf = [Ag(S2O3)2^3-(aq)] / (x * (1.0)^2)
Now, we can rearrange the equation for Kf to isolate the concentration of Ag(S2O3)2^3-(aq):
[Ag(S2O3)2^3-(aq)] = Kf * x
Since the concentration of Ag(S2O3)2^3-(aq) is equal to x, we can substitute it into the expression for the equilibrium constant Ksp:
Ksp = [Ag+(aq)][Cl-(aq)]
Ksp = x * [Cl-(aq)]
Now, we can derive an equation combining Ksp and Kf:
x^2 = Ksp * (1.0)^2 / (Kf * x)
x^2 = Ksp / Kf
Finally, we can calculate the value of x, which represents the molar solubility of AgCl in 1.0 M K2S2O3:
x = √(Ksp / Kf)
Substituting the given values of Ksp and Kf into the equation, we find:
x = √(1.8 × 10^-10 / 2.9 × 10^13)
Calculating this expression, we can find the molar solubility of AgCl in 1.0 M K2S2O3.