Question

Suppose 208.3 mg of PbCl2 was added to 15.0 mL of water in a flask, and the solution was

allowed to reach equilibrium at 20.0 °C. Some solute remained at the bottom of the flask after
equilibrium, and the solution was filtered to collect the remaining PbCl2, which had a mass of 53.3
mg. What is the solubility of PbCl₂ (in g/L)?
Express the concentration in grams per liter to three significant figures.

Answer 1

The solubility of PbCl₂ in g/L is 0.180 g/L. This can be calculated by first solving for the moles of PbCl₂ in the solution at equilibrium. Since the mass of the PbCl₂ is 208.3 mg, the moles can be calculated as 0.002083 mol. The volume of the solution is 15.0 mL, and since 1 L = 1000 mL, the volume in liters is 0.015 L. Therefore, the solubility of PbCl₂ is 0.002083 mol/0.015 L = 0.180 g/L.