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.
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).