---
title: "A student prepares a saturated aqueous solution of \\(\\text{CaCO}_3\\) in a beaker at \\(25\\ ^\\circ\\text{C}\\) with a small amount of solid remaining at the bottom, as represented by the equilibrium equation below.  \\[ \\text{CaCO}_3(s) \\rightleftharpoons \\text{Ca}^{2+}(aq) + \\text{CO}_3^{2-}(aq) \\]  The student then adds a few drops of \\(6.0\\text{ M }\\text{HCl}(aq)\\) to the mixture at constant temperature, without significantly changing the total volume. Which of the following best predicts and explains the effect of adding the acid on the amount of solid \\(\\text{CaCO}_3\\)?"
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url: "https://nerd-notes.com/ubq/121400/"
date_modified: "2026-08-23T05:02:05+00:00"
---

# A student prepares a saturated aqueous solution of \(\text{CaCO}_3\) in a beaker at \(25\ ^\circ\text{C}\) with a small amount of solid remaining at the bottom, as represented by the equilibrium equation below.

\[ \text{CaCO}_3(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{CO}_3^{2-}(aq) \]

The student then adds a few drops of \(6.0\text{ M }\text{HCl}(aq)\) to the mixture at constant temperature, without significantly changing the total volume. Which of the following best predicts and explains the effect of adding the acid on the amount of solid \(\text{CaCO}_3\)?

A student prepares a saturated aqueous solution of \(\text{CaCO}_3\) in a beaker at \(25\ ^\circ\text{C}\) with a small amount of solid remaining at the bottom, as represented by the equilibrium equation below.

\[ \text{CaCO}_3(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{CO}_3^{2-}(aq) \]

The student then adds a few drops of \(6.0\text{ M }\text{HCl}(aq)\) to the mixture at constant temperature, without significantly changing the total volume. Which of the following best predicts and explains the effect of adding the acid on the amount of solid \(\text{CaCO}_3\)?

- **A.** The amount of solid \(\text{CaCO}_3\) decreases because \(\text{Cl}^-(aq)\) reacts with \(\text{Ca}^{2+}(aq)\) to form insoluble \(\text{CaCl}_2(s)\), shifting the equilibrium to the right.
- **B.** The amount of solid \(\text{CaCO}_3\) decreases because \(\text{H}^+(aq)\) reacts with \(\text{CO}_3^{2-}(aq)\) to form \(\text{HCO}_3^-(aq)\), shifting the equilibrium to the right.
- **C.** The amount of solid \(\text{CaCO}_3\) increases because \(\text{H}^+(aq)\) acts as a common cation with \(\text{Ca}^{2+}(aq)\), shifting the equilibrium to the left.
- **D.** The amount of solid \(\text{CaCO}_3\) increases because lowering the \(\text{pH}\) decreases the value of \(K_{sp}\) for \(\text{CaCO}_3\), shifting the equilibrium to the left.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/121400/*