Posted by Anonymous on Friday, May 4, 2012 at 4:46pm.
...........AgCl ==> Ag^+ + Cl^-
I............x........0........0
C...........-x.........x........x
E...........-x..........x......x
........NaCl ==> Na^+ + Cl^-
I.......0.15M.....0......0
C.......-0.15M....0.15....0.15
E........0.........0.15...0.15
Ksp + (Ag^+)(Cl^-)
1.8E-10 = (x)(x+0.15) note:(Cl^-) = x+ 0.15 or x from the AgCl + 0.15 from the NaCl. This a problem illustrating the common ion effect. The solubility of AgCl is DECREASED significantly by the addition of a common ion.
Solve for x.
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