---
title: "A student titrates a \\(50.0\\text{ mL}\\) sample of \\(0.10\\text{ M }\\text{CH}_3\\text{COOH(aq)}\\) with \\(0.10\\text{ M }\\text{NaOH(aq)}\\) at \\(25\\ ^\\circ\\text{C}\\) and continuously monitors the electrical conductivity of the solution as a function of titrant volume. Up to the equivalence point (\\(50.0\\text{ mL}\\) of titrant added), the electrical conductivity increases at a moderate, steady rate.  Which of the following best predicts and explains the behavior of the electrical conductivity as additional \\(\\text{NaOH(aq)}\\) is added past the \\(50.0\\text{ mL}\\) equivalence point?"
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url: "https://nerd-notes.com/ubq/123649/"
date_modified: "2026-09-28T12:01:53+00:00"
---

# A student titrates a \(50.0\text{ mL}\) sample of \(0.10\text{ M }\text{CH}_3\text{COOH(aq)}\) with \(0.10\text{ M }\text{NaOH(aq)}\) at \(25\ ^\circ\text{C}\) and continuously monitors the electrical conductivity of the solution as a function of titrant volume. Up to the equivalence point (\(50.0\text{ mL}\) of titrant added), the electrical conductivity increases at a moderate, steady rate.

Which of the following best predicts and explains the behavior of the electrical conductivity as additional \(\text{NaOH(aq)}\) is added past the \(50.0\text{ mL}\) equivalence point?

A student titrates a \(50.0\text{ mL}\) sample of \(0.10\text{ M }\text{CH}_3\text{COOH(aq)}\) with \(0.10\text{ M }\text{NaOH(aq)}\) at \(25\ ^\circ\text{C}\) and continuously monitors the electrical conductivity of the solution as a function of titrant volume. Up to the equivalence point (\(50.0\text{ mL}\) of titrant added), the electrical conductivity increases at a moderate, steady rate.

Which of the following best predicts and explains the behavior of the electrical conductivity as additional \(\text{NaOH(aq)}\) is added past the \(50.0\text{ mL}\) equivalence point?

- **A.** The conductivity decreases because excess \(\text{OH}^-\text{(aq)}\) ions suppress the hydrolysis of \(\text{CH}_3\text{COO}^-\text{(aq)}\) via the common-ion effect, reducing the total concentration of dissolved ions.
- **B.** The conductivity remains constant because the neutralization reaction has reached completion, so no additional charge-carrying species are produced in the mixture.
- **C.** The conductivity increases at a greater rate because additional \(\text{Na}^+\text{(aq)}\) and highly conductive, unreacted \(\text{OH}^-\text{(aq)}\) ions accumulate directly in solution rather than reacting to produce \(\text{CH}_3\text{COO}^-\text{(aq)}\) ions.
- **D.** The conductivity increases at a slower rate because dilution from the added titrant volume offsets the increase in the total number of dissolved ions.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123649/*