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Multiple Choice
What concentration of the lead ion, Pb2+, must be exceeded to precipitate PbCl2 from a solution that is 1.00×10⁻² M in the chloride ion, Cl⁻? The Ksp for lead(II) chloride is 1.17×10⁻⁵.
A
1.00×10⁻² M
B
1.00×10⁻⁴ M
C
1.17×10⁻³ M
D
1.17×10⁻⁵ M
Verified step by step guidance
1
Understand the concept of solubility product constant (Ksp). The Ksp is the equilibrium constant for the dissolution of a sparingly soluble ionic compound. For PbCl2, the dissolution can be represented as: PbCl2(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq).
Write the expression for the solubility product constant (Ksp) for PbCl2. The Ksp expression is: . Here, [Pb²⁺] is the concentration of lead ions and [Cl⁻] is the concentration of chloride ions.
Substitute the given concentration of chloride ions into the Ksp expression. You are given [Cl⁻] = 1.00×10⁻² M. Substitute this value into the Ksp expression: .
Rearrange the Ksp expression to solve for the concentration of lead ions, [Pb²⁺]. This gives: .
Substitute the value of Ksp (1.17×10⁻⁵) into the rearranged expression to find the concentration of lead ions required to exceed the solubility product and precipitate PbCl2. Calculate: .