# 7.What is the solubility of Pb(IO3)2 (a) in mol/L, and (b) in g/L ? Ksp of Pb(IO3)2 = 2.5 x 10^-13?

**Answer:**

The equilibrium is :

Pb(IO3)2 <----> Pb2+ + 2IO3-

Ksp = [Pb2+][ IO32-]^2

let x = moles/L Pb(IO3)2 that dissolve : we get x mol/L pb2+ and 2x mol/L IO3-

2.5 x 10^-13 = ( x)(2x)^2 = 4x^3

x = 3.97 x 10^-5 M (a)

Molecular weight = 557 g/mol

557 g/mol x 3.97 x 10^-5 mol/L = 0.0221 g/L (b)

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