Ksp of AgBr = 3.3 * 10^-13
AgBr (s)..................> Ag+ (aq) + Br- (aq)
Ksp = [Ag+][Br-]
or
3.3 * 10^-13 = [Ag+] * 0.0260
or
[Ag+] = 1.27 * 10^-11 M
thus
maximum amount of Ag+ ion = 1.27 * 10^-11 M
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