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
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
Which of the following solutions would have same pH?...The solubility products of Al(OH)3 and Zn(OH)2 are 8.5 × 10-23 and 1.8 × 10-14 respectively. If NH4OH is added...The solubility of BaSO4 in water 2.42 × 103 gL–1 at 298 K. The value of solubility product (Ksp) will be (Gi...The species present in solution when CO2 is dissolved in water is...A solution contains a mixture of Ag+ (0.10 M) and Hg22+ (0.10 M), which are to be separated by selective precipitation. ...