Posted by Stacy on Wednesday, November 16, 2011 at 10:45pm.
........AgCl(s) ==> Ag^+(aq) + Cl^-(aq)
..........x...........x.........x
How many moles KCl were added. That is 0.200g/molar mass KCl = about 0.003 but you need to be more accurate. In 1L that will be 0.003 M.
..........KCl ==> K^+ + Cl^-
initial..0.003....0......0
equil......0....0.003....0.003
Now substitute into the Ksp expression for AgCl.
1.8E-10 = (Ag^+)(Cl^-)
(Ag^+) = x from AgCl.
(Cl^-) = x from AgCl + 0.003 from KCl. You have one unknown, solve for that which is the solubility of AgCl under these conditions.
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