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The Ksp of PbBr2 is 6.60×10−6. What is the molar solubility of PbBr2 in pure water?
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18. Aqueous Equilibrium
Solubility Product Constant: Ksp
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What is the molar solubility of PbBr2 in pure water? Ksp for PbBr2 is 4.67 × 10−6.
A
1.05 × 10−2
B
1.67 × 10−2
C
1.08 × 10−3
D
2.18 × 10−3
E
4.67 × 10−6
Verified step by step guidance1
Write the dissolution equation for PbBr2: PbBr2(s) ⇌ Pb²⁺(aq) + 2Br⁻(aq).
Define the molar solubility of PbBr2 as 's'. This means that at equilibrium, the concentration of Pb²⁺ will be 's' and the concentration of Br⁻ will be '2s'.
Write the expression for the solubility product constant (Ksp) for PbBr2: Ksp = [Pb²⁺][Br⁻]².
Substitute the equilibrium concentrations into the Ksp expression: Ksp = (s)(2s)² = 4s³.
Set the Ksp expression equal to the given Ksp value and solve for 's': 4s³ = 4.67 × 10⁻⁶. Solve for 's' to find the molar solubility.
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