Question

The solubility of PbI2 is measured and found to be 0.609 g/L. Use this information to calculate a Ksp value for lead iodide. Ksp ?

Answer 1

Ksp  = 9.2 × 10-9 .

Answer:- I Given - solubility of Pb Iz = 0.609 g/L _mal set of Pb Iz = 4610) gl'mol Molarity of Pb I2 is = 0.609 GIL 1,32 x 103 md/ 461.01 g/mol They equilibrium is Pb I2 Pb2+ + 21 i Ksp = [Pb2+] (I°12 Since-one mode of Pb Iz glves. one mole of Pb2t and two moles of I- Therefore [Pb 2+] = 1. 3x10 3 mol/L and [I] = 2X1:32X163 mol/L ..LI) = 2.64 X163 mol/L Now, Ksp = [Pb2+] [172 | = (1:32X163) (264X163) is ksp = 9.2 x 1011 Therebene Ksp of lead jodide is Ksp = 9.2x10?