Posted by John L on Thursday, April 3, 2008 at 12:06pm.
AgCl <==> Ag^+ + Cl^-
NaCl ==> Na^+ + Cl^-
Ksp = (Ag^+)(Cl^-) = 1.8 x 10^-10
Let S = solubility of AgCl, then
(Ag^+) = S
(Cl^-) = S+0.01
Solve for S.
Note: A similar problem to this post (0.1 M NaCl instead of 0.01 M NaCl) was on a couple of days ago; the one who posted said that the answer came back incorrect. Check my work. Check myu thinking.
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