Question

In an experiment to determine the solubility of lead chloride (PbCl2), 5.6g of (NH4)2SO4 was added to a
250ml solution containing an unknown amount of dissolved lead chloride (PbCl2) resulting in the formation of
Lead sulfate precipitate. The mass of the lead sulfate precipitate was found to be 1.645g. From the
information provided, calculate the mass of PbCl2 that dissolved in the given volume of solvent (250 ml),
hence the solubility and the solubility product constant (Ksp)

Answer 1

g PbSO4 = 1.645
mols PbSO4 = 1.645/molar mass = estimated 0.00542 = estd 0.00542mols Pb = estd 0.00542 mols PbCl2.

g PbCl2 = mols PbCl2 x molar mass PbCl2 and that is g/250mL. That x 4 = g/L
Then g/L divided by molar mass PbCl2 = M = (PbCl2) = (Pb^2+).
(Cl^-) = 2 x (Pb^2+).
Then Ksp = (Pb^2+)(Cl^-)^2 = ?

Answer 2

85g

Answer 3

11.676g pbcl2 and ksp= 0.642

Answer 4

Please the answers