---
title: "A student adds excess solid \\(\\text{AgCl(s)}\\) to separate beakers containing pure water and \\(0.10\\text{ M }\\text{NaCl(aq)}\\) at \\(25^\\circ\\text{C}\\). After equilibrium is established, undissolved solid remains in both beakers. Assume that no silver-containing complex ions form.  \\[\\text{AgCl(s)}\\rightleftharpoons\\text{Ag}^+\\text{(aq)}+\\text{Cl}^-\\text{(aq)}\\qquad K_{sp}=1.8\\times10^{-10}\\]  Which statement best compares the molar solubility of \\(\\text{AgCl(s)}\\) in the two beakers and explains the difference?"
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url: "https://nerd-notes.com/ubq/119529/"
date_modified: "2026-08-19T12:41:37+00:00"
---

# A student adds excess solid \(\text{AgCl(s)}\) to separate beakers containing pure water and \(0.10\text{ M }\text{NaCl(aq)}\) at \(25^\circ\text{C}\). After equilibrium is established, undissolved solid remains in both beakers. Assume that no silver-containing complex ions form.

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

Which statement best compares the molar solubility of \(\text{AgCl(s)}\) in the two beakers and explains the difference?

A student adds excess solid \(\text{AgCl(s)}\) to separate beakers containing pure water and \(0.10\text{ M }\text{NaCl(aq)}\) at \(25^\circ\text{C}\). After equilibrium is established, undissolved solid remains in both beakers. Assume that no silver-containing complex ions form.

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

Which statement best compares the molar solubility of \(\text{AgCl(s)}\) in the two beakers and explains the difference?

- **A.** The molar solubility is lower in \(0.10\text{ M }\text{NaCl(aq)}\), because the \(\text{Na}^+\text{(aq)}\) ions strengthen the ionic attractions within the \(\text{AgCl(s)}\) crystal lattice.
- **B.** The molar solubility is lower in \(0.10\text{ M }\text{NaCl(aq)}\), because \(\text{Cl}^-\text{(aq)}\) is a common ion and the fixed value of \(K_{sp}\) requires a smaller equilibrium \([\text{Ag}^+]\) when \([\text{Cl}^-]\) is larger.
- **C.** The molar solubility is higher in \(0.10\text{ M }\text{NaCl(aq)}\), because added \(\text{Cl}^-\text{(aq)}\) shifts the dissolution equilibrium toward products to consume the added ion.
- **D.** The molar solubility is the same in both liquids, because \(K_{sp}\) has the same value for both systems at the same temperature.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119529/*