In: Chemistry
Calculate the molar solubility of AgI in a 0.20 M
CaI2 solution.
Ksp of AgI = 8.5*10^-17 check this value you book
CaI2 -----------------> Ca^2+ (aq) + 2I^- (aq)
0.2M 0.2M 2*0.2 M = 0.4M
AgI (s) ---------------------> Ag^+ (aq) + I^- (aq)
s s+ 0.4
Ksp = [Ag^+][I^-]
8.5*10^-17 = s*(s+0.4) [ s<<<<<0.4 , S+ 0.4 = 0.4M]
8.5*10^-17 = s*0.4
s = 8.5*10^-17/0.4 = 2.125*10^-16 M
solubility of AgI is 2.125*10^-16 M