Homework Answers
Ksp of lead chromate is 2.0x10-16
PbCrO4(s) < -------> Pb+2 + CrO4-2 (aq)
a) solubility in 0.1M Na2CrO4 solution
PbCrO4(s) < -------> Pb+2 + CrO4-2 (aq)
0.1M initial due to common ion
x x 0.1 +x at equilibrium
Since x << 0.1 , we take [CrO4] = 0.1 M
Thus Ksp = [CrO4] [Pb+2] = 2.0x10-16 = x (0.1)
thus the solubility x = 2.0 x 10-15 M
b) solubility in pure water
PbCrO4(s) < -------> Pb+2 + CrO4-2 (aq)
s 0 0 initial
- s s equilibrium
Thus Ksp = s2 = 2.0 x 10-16
adn Solubility s = 1.414x10-8 M
c) in 0.001 M lead nitatre
PbCrO4(s) < -------> Pb+2 + CrO4-2 (aq)
0.001 0 inital
- 0.001+p p equilibrium
Again due to common ion effect p << 0.001
Ksp = 2.0x 10-16 = p (0.001)
or solubility in lead nitrate solution p = 2.0x 10-13 M
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