Calculate the solubility of the following compound in moles per liter: PbI2, Ksp = 1.4 x 10^-8. A) 1.4 x 10^-8 M
B) 0.0019 M
C) 5.6 x 10^-8 M
D) 0.0024 M
E) 0.0015 M
Solution:
PbI2(s) <--> Pb^2+(aq) + 2I- (aq)
I 0 0
C X 2X
E X 2X
Ksp = 1.4 x 10^-8 = [Pb^2+][I-]^2
Ksp = 1.4 x 10^-8 = x(2x)^2 = 4x^3
X= 0.0015 M
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