What is the solubility product constant, K _sp, for PbI₂ if a saturated solution is formed after 0.54 g of PbI₂ dissolves in 1.00 liter of water at 25 ° C?
Molar Mass(PbI₂) = 461.0 g/mol

Question

What is the solubility product constant, K _sp, for PbI₂ if a saturated solution is formed after 0.54 g of PbI₂ dissolves in 1.00 liter of water at 25 ° C? Molar Mass(PbI₂) = 461.0 g/mol

Quizplus · Accepted Answer

First, we need to calculate the concentration of PbI₂ in the saturated solution. This can be done using the mass of PbI₂ dissolved and its molar mass: n = m/M = 0.54 g / 461.0 g/mol = 1.17 × 10^-3 mol Assuming all of the PbI₂ dissolves and ionizes completely in water, the concentration of Pb2+ and I- in the saturated solution are also 1.17 × 10^-3 mol/L. Therefore, the solubility product constant can be calculated as follows: K _sp = [Pb2+][I-]^2 = (1.17 × 10^-3)^1(1.17 × 10^-3)^2 = 1.61 × 10^-9 The closest answer choice is C, which matches this value.

What Is the Solubility Product Constant, K sp, for PbI2

What Is the Solubility Product Constant, K _sp, for PbI₂