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Calculate the minimum number of moles of Pb(NO3)2 required to start precipitation in 50.0mL of 0.15M ZnI2Asked by May
Calculate the minimum number of moles of Pb(NO3)2 required to start precipitation in 50.0mL of 0.15M ZnI2
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Answered by
DrBob222
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 = ?
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 = ?
Answered by
Anonymous
.0225
Answered by
buddy
0.0225?
Answered by
buddy
got 1.275x10^-10 here
Answered by
Anonymous
I got 4.72x10^-9
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