Calculate the solubility product of a lead sulphate solution if 3.03 x 10-2 g of the product is dissolved in 1000 mL of water at 25oC.

Answer 1

To calculate the solubility product (Ksp) of lead sulphate (PbSO4), we need to determine the concentration of the dissolved product in moles per liter (mol/L).

First, we need to convert the given mass of PbSO4 into moles:
Mass of PbSO4 = 3.03 x 10^(-2) g
Molar mass of PbSO4 = 207.2 g/mol (Pb: 207.2 g/mol, S: 32.1 g/mol, O: 16.0 g/mol)

Number of moles of PbSO4 = Mass / Molar mass = (3.03 x 10^(-2) g) / (207.2 g/mol) = 1.464 x 10^(-4) mol

Next, we need to calculate the concentration of Pb2+ and SO42- ions in the solution. Lead sulphate dissociates according to the following equation:
PbSO4(s) ⇌ Pb2+(aq) + SO42-(aq)

Since the stoichiometric ratio between PbSO4 and Pb2+ is 1:1, the concentration of Pb2+ ions will be equal to the concentration of PbSO4. Thus, [Pb2+] = (1.464 x 10^(-4)) mol/L.

The same applies to the concentration of SO42- ions, so [SO42-] = (1.464 x 10^(-4)) mol/L.

Finally, we use the concentrations of Pb2+ and SO42- ions to calculate the solubility product (Ksp) of PbSO4:
Ksp = [Pb2+][SO42-] = (1.464 x 10^(-4))² = 2.1456 x 10^(-8)

Therefore, the solubility product (Ksp) of the lead sulphate solution at 25°C is 2.1456 x 10^(-8).