ammonium iodide is NH4I
iodide ions will form
Ag+ + I- = AgI
Pb+2 + I- = PbI2
Get Ksp data:
AgI 8.52×10–17; PbI2 9.8×10–9
Therefore, Ksp << Ag than PbI2
then, most likely AgI precipitates first
b)
find [I-] when it begins to precipitate
Ksp = [Ag+][I-]
8.52*10^-17 = (6.56*10^-3)[I-]
[I-] = (8.52*10^-17) / (6.56*10^-3) = 1.2987*10^-14 M
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can u help plZ?
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