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.
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.
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