---
title: "A student adds excess solid \\(\\text{PbCl}_2\\) to a \\(0.10\\text{ M }\\text{NaCl}\\) solution at \\(25^\\circ\\text{C}\\). The dissolution equilibrium is represented below.  \\[ \\text{PbCl}_2(s) \\rightleftharpoons \\text{Pb}^{2+}(aq) + 2\\,\\text{Cl}^-(aq) \\]  At \\(25^\\circ\\text{C}\\), the solubility product constant, \\(K_{sp}\\), for \\(\\text{PbCl}_2\\) is \\(1.6 \\times 10^{-5}\\). Assuming that the contribution of \\(\\text{Cl}^-\\) from the dissolution of \\(\\text{PbCl}_2\\) is negligible, what is the molar solubility of \\(\\text{PbCl}_2\\) in this solution?"
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url: "https://nerd-notes.com/ubq/119698/"
date_modified: "2026-08-21T08:12:05+00:00"
---

# A student adds excess solid \(\text{PbCl}_2\) to a \(0.10\text{ M }\text{NaCl}\) solution at \(25^\circ\text{C}\). The dissolution equilibrium is represented below.

\[ \text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\,\text{Cl}^-(aq) \]

At \(25^\circ\text{C}\), the solubility product constant, \(K_{sp}\), for \(\text{PbCl}_2\) is \(1.6 \times 10^{-5}\). Assuming that the contribution of \(\text{Cl}^-\) from the dissolution of \(\text{PbCl}_2\) is negligible, what is the molar solubility of \(\text{PbCl}_2\) in this solution?

A student adds excess solid \(\text{PbCl}_2\) to a \(0.10\text{ M }\text{NaCl}\) solution at \(25^\circ\text{C}\). The dissolution equilibrium is represented below.

\[ \text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\,\text{Cl}^-(aq) \]

At \(25^\circ\text{C}\), the solubility product constant, \(K_{sp}\), for \(\text{PbCl}_2\) is \(1.6 \times 10^{-5}\). Assuming that the contribution of \(\text{Cl}^-\) from the dissolution of \(\text{PbCl}_2\) is negligible, what is the molar solubility of \(\text{PbCl}_2\) in this solution?

- **A.** \(1.6 \times 10^{-4}\text{ M}\)
- **B.** \(4.0 \times 10^{-4}\text{ M}\)
- **C.** \(1.6 \times 10^{-3}\text{ M}\)
- **D.** \(1.6 \times 10^{-2}\text{ M}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119698/*