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s AgCl dalam larutan CaCl2 0,05 M
AgCl <-> Ag^+ + Cl^-
s s s
CaCl2 -> Ca^2+ + 2Cl^-
0,05 M 0,05 M 0,1M
[Cl] total = 0,1M
Ksp = s2 = (10,5)^2 = 10^-10
Ksp AgCl = [Ag+][Cl-]
10^-10 = s x 0,1
s = 10^-10 dibagi 10^-1 = 10^-9
Jadi, kelarutam AgCl dalam CaCl2 0,05 M adalah 10^-9