---
title: "A student titrates a \\(25.0 \\text{ mL}\\) sample of \\(0.100 \\text{ M } \\text{HC}_2\\text{H}_3\\text{O}_2\\text{(aq)}\\) (\\(K_a = 1.8 \\times 10^{-5}\\)) with \\(0.100 \\text{ M } \\text{NaOH(aq)}\\) at \\(25^\\circ\\text{C}\\). The titration reaction is represented by the following equation:  \\[ \\text{HC}_2\\text{H}_3\\text{O}_2\\text{(aq)} + \\text{OH}^-\\text{(aq)} \\rightarrow \\text{C}_2\\text{H}_3\\text{O}_2^-\\text{(aq)} + \\text{H}_2\\text{O(l)} \\]  Which of the following best predicts and justifies the \\(\\text{pH}\\) of the solution at the equivalence point of the titration?"
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url: "https://nerd-notes.com/ubq/121416/"
date_modified: "2026-08-23T05:02:09+00:00"
---

# A student titrates a \(25.0 \text{ mL}\) sample of \(0.100 \text{ M } \text{HC}_2\text{H}_3\text{O}_2\text{(aq)}\) (\(K_a = 1.8 \times 10^{-5}\)) with \(0.100 \text{ M } \text{NaOH(aq)}\) at \(25^\circ\text{C}\). The titration reaction is represented by the following equation:

\[
\text{HC}_2\text{H}_3\text{O}_2\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{C}_2\text{H}_3\text{O}_2^-\text{(aq)} + \text{H}_2\text{O(l)}
\]

Which of the following best predicts and justifies the \(\text{pH}\) of the solution at the equivalence point of the titration?

A student titrates a \(25.0 \text{ mL}\) sample of \(0.100 \text{ M } \text{HC}_2\text{H}_3\text{O}_2\text{(aq)}\) (\(K_a = 1.8 \times 10^{-5}\)) with \(0.100 \text{ M } \text{NaOH(aq)}\) at \(25^\circ\text{C}\). The titration reaction is represented by the following equation:

\[
\text{HC}_2\text{H}_3\text{O}_2\text{(aq)} + \text{OH}^-\text{(aq)} \rightarrow \text{C}_2\text{H}_3\text{O}_2^-\text{(aq)} + \text{H}_2\text{O(l)}
\]

Which of the following best predicts and justifies the \(\text{pH}\) of the solution at the equivalence point of the titration?

- **A.** \(\text{pH} = 7.00\), because the added moles of \(\text{OH}^-\text{(aq)}\) equal the initial moles of \(\text{HC}_2\text{H}_3\text{O}_2\text{(aq)}\), completely neutralizing all acidic and basic properties in the solution.
- **B.** \(\text{pH} > 7.00\), because all \(\text{HC}_2\text{H}_3\text{O}_2\text{(aq)}\) is converted to \(\text{C}_2\text{H}_3\text{O}_2^-\text{(aq)}\), which reacts with \(\text{H}_2\text{O(l)}\) according to \(\text{C}_2\text{H}_3\text{O}_2^-\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HC}_2\text{H}_3\text{O}_2\text{(aq)} + \text{OH}^-\text{(aq)}\).
- **C.** \(\text{pH} > 7.00\), because \(\text{NaOH}\) is a strong base that fully dissociates, leaving a net excess of unreacted \(\text{OH}^-\text{(aq)}\) ions at the stoichiometric point due to the weak dissociation of the acid.
- **D.** \(\text{pH} < 7.00\), because at the equivalence point \([\text{HC}_2\text{H}_3\text{O}_2] = [\text{C}_2\text{H}_3\text{O}_2^-]\), forming a buffer whose \(\text{pH}\) is equal to the \(\text{p}K_a\) of \(\text{HC}_2\text{H}_3\text{O}_2\) (\(\text{p}K_a \approx 4.74\)).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/121416/*