Question

Consider PbI2 dissolving in water. PbI2 (s) <--> Pb2+ (aq) + 2I- (aq) Calculate Ksp if the max Pb2+ solubility = 0.00130M.

Answer 1

The solubility product constant (Ksp) for the PbI2 solution is approximately 5.093 x 10^-6 at room temperature.

Step-by-step explanation:

The solubility product constant, Ksp, describes the equilibrium constant for the dissolution of a sparingly soluble salt in water. For the reaction:

PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)

The expression for the Ksp is:

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

where [Pb2+] represents the molar concentration of Pb2+ ions in the solution and [I-] represents the molar concentration of I- ions in the solution.

Given that the molar solubility of PbI2 is 0.00230 M, we can assume that the concentration of Pb2+ ions is also 0.00230 M (since 1 mol of PbI2 produces 1 mol of Pb2+ ions). The concentration of I- ions will be twice that of Pb2+ ions, so [I-] = 2 * 0.00230 M = 0.00460 M.

Now, we can calculate the Ksp:

Ksp = (0.00230 M)(0.00460 M)^2

Ksp = 5.093 x 10^-6.

Answer 2

The Ksp of PbI2 is 8.79 × 10^-9.

Step-by-step explanation:

Ksp = [Pb²+][I¯]² = (1.30 × 10-3) (2.60 × 10-3)² = 8.79 × 10-9