Question

the solubility of lead (II) sulfate is 8.0x10^-2 g/L. what is the solubility product constant for lead (II) sulfate?

Answer 1

Solution :-

Lets first calculate the molarity of the PbSO4

moles = mass / molar mass

           = 8.0*10^-2 g / 303.26 g per mol

           = 2.638*10^-4 mol

molarity = moles / volume in liter

             = 2.638*10^-4 mol / 1 L

             = 2.638*10^-4 M

PbSO4 ------ > Pb^2+ + SO4^2-

ksp = [Pb^2+][SO4^2-]

ksp = [2.638*10^-4] [2.638*10^-4]

ksp =6.96*10^-8

Therefore the solubility product constant ksp = 6.96*10^-8