The solubility product, Ksp, of PbBr2 is 8.9 × 10-6. Calculate the molar solubility in:?
c) The solubility product, Ksp, of PbBr2 is 8.9 × 10-6. Calculate the molar solubility in:
i. Pure water.
ii. 0.20 M KBr.
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PbBr2 ===> Pb2+ + 2Br-
So the molar solubility of PbBr2 = [Pb2+]
Ksp = [Pb2+][Br-]^2 = 8.9 x 10^-6
i. Let [Pb2+] = x. Then [Br-] = 2x
(x)(2x)^2 = 4x^3 = 8.9 x 10^-6
x^3 = 2.2 x 10^-6
x = 1.3 x 10^-2 M = [Pb2+] = molar solubility of PbBr2
ii. Let [Pb2+] = x. Then [Br-] = 0.20 + 2x. But because x is so small, we can neglect it, and [Br-] = 0.20.
(x)(0.20)^2 = 8.9 x 10^-6
x = 2.2 x 10^-4 M