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

Answer: (C)

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).
2. Write the expression for the solubility product constant (Ksp) for PbCl2. The Ksp expression is: $K_{\mathrm{sp}}=\; [Pb²⁺][Cl⁻]²$. Here, [Pb²⁺] is the concentration of lead ions and [Cl⁻] is the concentration of chloride ions.
3. 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: $K_{\mathrm{sp}}=\; [Pb²⁺](1.00\times 10⁻²)²$.
4. Rearrange the Ksp expression to solve for the concentration of lead ions, [Pb²⁺]. This gives: $[Pb²⁺]\; =\frac{K_{\mathrm{sp}}}{{1.00}^{-2}}$.
5. 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: $[Pb²⁺]\; =\frac{\mathrm{1.17\times 10⁻⁵}}{\mathrm{1.00\times 10⁻⁴}}$.