The solubility product of lead chloride (PbCl2) is 1.2 * 10^-5 at 25°C. This means that at equilibrium, the concentration of lead ions (Pb2+) and chloride ions (Cl-) in a saturated solution of lead chloride is 1.2 * 10^-5 moles per liter (mol/L).

The solubility product is a measure of the solubility of a compound in water. It is defined as the product of the concentrations of the ions that make up the compound, each raised to the power of its stoichiometric coefficient in the balanced chemical equation for the dissolution reaction.

For lead chloride, the dissolution reaction is:

PbCl2(s) <=> Pb2+(aq) + 2Cl-(aq)

The solubility product expression for lead chloride is:

Ksp = [Pb2+][Cl-]^2

where Ksp is the solubility product, [Pb2+] is the concentration of lead ions in mol/L, and [Cl-] is the concentration of chloride ions in mol/L.

At equilibrium, the concentrations of lead ions and chloride ions in a saturated solution of lead chloride are equal, so we can substitute [Pb2+] for [Cl-] in the solubility product expression:

Ksp = [Pb2+]^3

Solving for [Pb2+], we get:

[Pb2+] = (Ksp)^(1/3) = (1.2 * 10^-5)^(1/3) = 1.08 * 10^-2 mol/L

Therefore, the solubility of lead chloride in water at 25°C is 1.08 * 10^-2 mol/L.