What is the K_sp of PbBr₂ if the concentration of PbBr₂(aq) in a saturated solution is 1.3 x 10^-2 moles/L?

Question

Quizplus · Accepted Answer

The dissolution of PbBr₂ in water can be represented as PbBr₂(s) ⇌ Pb²⁺(aq) + 2Br⁻(aq). If the concentration of PbBr₂ is 1.3 x 10⁻² M, then [Pb²⁺] = 1.3 x 10⁻² M and [Br⁻] = 2(1.3 x 10⁻²) = 2.6 x 10⁻² M. The Ksp = [Pb²⁺][Br⁻]² = (1.3 x 10⁻²)(2.6 x 10⁻²)² = 8.8 x 10⁻⁶.

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What Is the K_sp of PbBr₂ If the Concentration of PbBr₂(aq)