In: Chemistry
The Ksp of PbI2 is 1.4 x 10-8. Calculate the molar solubility of PbI2 in 0.01 M NaI.
1.5 x 10-3 M |
1.4 x 10-4 M |
1.4 x 10-6 M |
5.6 x 10-2 M |
answer : 1.4 x 10^-4 M
concentration of [I-] = 0.01 M
PbI2 <-------------> Pb+2 + 2 I-
S 2S
S 2S + 0.01
S 0.01
Ksp = [Pb+2][I-]^2
1.4 x 10^-8 = S x (0.01)^2
S = 1.4 x 10^-4 M
solubility of PbI2 = 1.4 x 10^-4 M