The solubility of PbBr₂ is 0.427 g per 100.0 mL of solution at 25°C.Determine the value of the solubility product constant for this strong electrolyte.Lead(II)bromide does not react with water.

Question

Quizplus · Accepted Answer

The solubility product constant (Ksp) for PbBr2 can be calculated using the solubility and the dissociation equation: PbBr2 ⇌ Pb^2+ + 2Br^-. First, convert the solubility from grams to moles: 0.427 g/100 mL = 0.00427 g/mL. The molar mass of PbBr2 is approximately 367 g/mol, so the molarity is 0.00427 g/mL / 367 g/mol = 0.0116 M. The concentration of Pb^2+ is 0.0116 M, and the concentration of Br^- is 2 * 0.0116 M = 0.0232 M. Ksp = [Pb^2+][Br^-]^2 = (0.0116)(0.0232)^2 = 6.3 x 10^-6.

The Solubility of PbBr2 Is 0 ×\times× 10-4 B)2 ×\times× 10-4 C)3 ×\times×

The Solubility of PbBr₂ Is 0 $\times$ 10^-⁴
B)2 $\times$ 10^-⁴
C)3 $\times$