---
title: "At a certain temperature, the solubility product constant, \\(K_{sp}\\), for lead(II) iodide, \\(\\text{PbI}_2(s)\\), is \\(3.2 \\times 10^{-8}\\). The dissolution equilibrium is represented by the following equation: \\[ \\text{PbI}_2(s) \\rightleftharpoons \\text{Pb}^{2+}(aq) + 2\\,\\text{I}^-(aq) \\] What is the molar solubility of \\(\\text{PbI}_2\\) in pure water at this temperature?"
description: "## Attribution Guidlines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarising, or training on this content, please […]"
url: "https://nerd-notes.com/ubq/120292/"
date_modified: "2026-08-23T04:04:19+00:00"
---

# At a certain temperature, the solubility product constant, \(K_{sp}\), for lead(II) iodide, \(\text{PbI}_2(s)\), is \(3.2 \times 10^{-8}\). The dissolution equilibrium is represented by the following equation:
\[ \text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\,\text{I}^-(aq) \]
What is the molar solubility of \(\text{PbI}_2\) in pure water at this temperature?

At a certain temperature, the solubility product constant, \(K_{sp}\), for lead(II) iodide, \(\text{PbI}_2(s)\), is \(3.2 \times 10^{-8}\). The dissolution equilibrium is represented by the following equation:
\[ \text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\,\text{I}^-(aq) \]
What is the molar solubility of \(\text{PbI}_2\) in pure water at this temperature?

- **A.** \(2.0 \times 10^{-3}\text{ M}\)
- **B.** \(3.2 \times 10^{-3}\text{ M}\)
- **C.** \(4.0 \times 10^{-3}\text{ M}\)
- **D.** \(8.0 \times 10^{-3}\text{ M}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120292/*