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title: "A student tests whether chloride ions in rinse water will form a precipitate when treated with silver ions. The student mixes equal volumes of \\(6.0 \\times 10^{-5}\\text{ M AgNO}_3\\) and \\(6.0 \\times 10^{-6}\\text{ M NaCl}\\) at \\(25^\\circ\\text{C}\\). The volumes are assumed to be additive.  \\(\\text{AgCl(s)} \\rightleftharpoons \\text{Ag}^+\\text{(aq)}+\\text{Cl}^-\\text{(aq)}\\qquad K_{sp}=1.8\\times10^{-10}\\)  Which of the following correctly predicts whether a precipitate forms immediately after the solutions are mixed?"
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url: "https://nerd-notes.com/ubq/119354/"
date_modified: "2026-08-19T12:39:57+00:00"
---

# A student tests whether chloride ions in rinse water will form a precipitate when treated with silver ions. The student mixes equal volumes of \(6.0 \times 10^{-5}\text{ M AgNO}_3\) and \(6.0 \times 10^{-6}\text{ M NaCl}\) at \(25^\circ\text{C}\). The volumes are assumed to be additive.

\(\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)}+\text{Cl}^-\text{(aq)}\qquad K_{sp}=1.8\times10^{-10}\)

Which of the following correctly predicts whether a precipitate forms immediately after the solutions are mixed?

A student tests whether chloride ions in rinse water will form a precipitate when treated with silver ions. The student mixes equal volumes of \(6.0 \times 10^{-5}\text{ M AgNO}_3\) and \(6.0 \times 10^{-6}\text{ M NaCl}\) at \(25^\circ\text{C}\). The volumes are assumed to be additive.

\(\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)}+\text{Cl}^-\text{(aq)}\qquad K_{sp}=1.8\times10^{-10}\)

Which of the following correctly predicts whether a precipitate forms immediately after the solutions are mixed?

- **A.** A precipitate forms because \(Q_{sp}=9.0\times10^{-11}<K_{sp}\), indicating that the concentration of dissolved ions is too large.
- **B.** A precipitate forms because \(Q_{sp}=(6.0\times10^{-5})(6.0\times10^{-6})=3.6\times10^{-10}>K_{sp}\).
- **C.** No precipitate forms because mixing equal volumes makes each ion concentration one-fourth its initial value, giving \(Q_{sp}=2.25\times10^{-11}<K_{sp}\).
- **D.** No precipitate forms because mixing equal volumes halves each ion concentration, giving \(Q_{sp}=9.0\times10^{-11}<K_{sp}\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119354/*