What is the molar solubility of AgCl, Ksp= 1.60x^-10, in a solution that also contains 0.0250 M KCl?

The answer is 1.60x^-10.

But I am not sure why this is. Does it have to do with the solubility rules? Please Help.

No. It has to do with the common ion effect in which Cl is the common ion. That decreases the solubility of AgCl from what it would normally be. But I can tell you the answer is NOT 1.60E-10 for that is Ksp and not the solubility.

........AgCl ==> Ag^+ + Cl^-
I.......solid....0.......0
C.......-x.......x.......x
E......solid-x...x.......x

KCl is completely ionized and completely soluble.
Then KCl ==> K^+ + Cl^-
I....0.025....0.....0
C....-0.025..0.025..0.025
E.....0.....0.025...0.025

Ksp = (Ag^+)(Cl^-)
(Ag^+) = solubility = x from above.
(Cl^-) = x from AgCl + 0.025 from KCl for total of x+0.025
Solve for x = solubility
It should be approx 6E-9 M.

To determine the molar solubility of AgCl in a solution containing 0.0250 M KCl, you need to consider the common ion effect. The common ion effect states that the solubility of a salt is reduced when a common ion is present in the solution.

In this case, both AgCl and KCl contain Cl- as the common ion. When AgCl dissolves in water, it dissociates into Ag+ and Cl- ions. Similarly, KCl dissociates into K+ and Cl- ions. So, the presence of KCl in the solution increases the concentration of Cl- ions.

To find the molar solubility of AgCl in the presence of KCl, you can use the Ksp expression for AgCl: Ksp = [Ag+][Cl-].

Let's assume that the molar solubility of AgCl is "s". Then, the concentration of Ag+ ions will also be "s" since one Ag+ ion is formed for every one AgCl molecule that dissolves. However, the concentration of Cl- ions will not only come from the dissociation of AgCl but also from the KCl solution.

The concentration of Cl- ions from AgCl can be calculated as "s", whereas the concentration of Cl- ions from KCl is given as 0.0250 M. Therefore, the total concentration of Cl- ions in the solution will be 0.0250 M + "s".

Since AgCl is sparingly soluble, we can assume that the concentration of Ag+ ions contributed by it will be negligible compared to the concentration of Ag+ ions supplied by the Ksp of AgCl. Hence, we can ignore the contribution of Ag+ ions from AgCl.

Finally, substituting the concentrations into the Ksp expression:
Ksp = [Ag+][Cl-] = (s)(0.0250 M + s)

From the problem statement, we know that Ksp = 1.60 x 10^-10. So, we can solve the equation:
1.60 x 10^-10 = s(0.0250 M + s)

As the quadratic equation involves a squared term, we can further assume that the value of "s" is small compared to 0.0250 M. Consequently, we can neglect "s" in comparison to 0.0250 M, simplifying the equation:
1.60 x 10^-10 ≈ s(0.0250 M)

Now, solving for "s":
s = (1.60 x 10^-10) / (0.0250 M) ≈ 6.40 x 10^-9 M

Hence, the molar solubility of AgCl in a solution containing 0.0250 M KCl is approximately 6.40 x 10^-9 M.