Question
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.
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.
Answers
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.
........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.
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