Calculate the solubility of AgCl in:

Ksp = 1.6x10^-10
a) 100 ml of 4.00 x 10^-3 M Calcium chloride
b) 100 ml of 4.00 x 10^-3 M Calcium Nitrate

all i know is 1.6x10^-10=[Ag+][Cl-]
which is 1.6x10^-10=x^2
x=1.3x10^-5, giving the M of Ag and Cl. Don't know how to go further

3 answers

What you know to do, and very well I might add, is to calculate Ag and Cl in a saturated solution of AgCl.
These two problems are slightly different. Basically they illustate the common ion effect. The first one has CaCl2 in the solution (the common ion is Cl^-) and the second one is calcium nitrate (no common ion). Use your method to calculate the solubility in Ca(NO3)2 since there is no common ion (technically the solubility is increased with a salt BUT I suspect you haven't covered that yet).
The first part s done this way.
AgCl(s) ==> Ag^+ + Cl^-
Ksp = (Ag^+)(Cl^-) = 1.6E-10
And
CaCl2 ==> Ca^2+ + 2Cl^-

Now substitute into the Ksp expression.
(Ag^+) = x
(Cl^-) = x from the AgCl and 0.008M for a total of x+0.008. (Where did the 0.008 come from? That is 0.004 M CaCl2 so the chloride is just twice that.) Solve for x and that will be the molarity. That is moles/L; you want grams/100 mL. So moles/L x 0.1 L (from the 100 mL) gives you moles/100 mL and that x molar mass = grams AgCl.

.000032

3 pi