The solubility of PbBr₂ is 0.427 g per 100 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

To find the solubility product constant (Ksp) for PbBr2, first convert the solubility from grams to moles. The molar mass of PbBr2 is approximately 367 g/mol. Solubility in moles/L = 0.427 g/100 mL * (1 mol/367 g) * (1000 mL/1 L) = 0.0116 mol/L. PbBr2 dissociates into Pb^2+ and 2 Br^-, so [Pb^2+] = 0.0116 M and [Br^-] = 2 * 0.0116 M = 0.0232 M. Ksp = [Pb^2+][Br^-]^2 = (0.0116)(0.0232)^2 = 6.3 * 10^-6.

The Solubility of PbBr2 Is 0

The Solubility of PbBr₂ Is 0