Ionic EquilibriumHard
Question
The solubility of PbCI2 in water is 0.01 M 25oC. Its maximum concentration in 0.1 M NaCI will be:
Options
A.2 × 10-3 M
B.1 × 10-4 M
C.1.6 × 10-2 M
D.4 × 10-4 M
Solution
KSP of PbCI2 = 4s3 = 4 × (0.01)3 = 4 × 10-6
In NaCI solution for PbCI2; KSP = [Pb2+] [CI-]2
or 4 × 10-6 = [Pb2+] [0.1]2 ∴ [Pb2+] = 4 × 10-4 M
In NaCI solution for PbCI2; KSP = [Pb2+] [CI-]2
or 4 × 10-6 = [Pb2+] [0.1]2 ∴ [Pb2+] = 4 × 10-4 M
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