Calculate the minimum number of moles of Pb(NO3)2 required to start precipitation in 50.0mL of 0.15M ZnI2

Answer 1

PbI2 ==> Pb^2+ + 2I^-
Ksp = (Pb^2+)(I^-)^2
(I^-) = 0.15M*2 = 0.30M
Substitute and solve for (Pb^2+); the unit is mols/L. You want mols in 50 mL (0.050L) and mols = M x L = ?

Answer 2

.0225

Answer 3

0.0225?

Answer 4

got 1.275x10^-10 here

Answer 5

I got 4.72x10^-9