---
title: "A student titrates a sample of a weak diprotic acid, \\(\\text{H}_2\\text{A}\\), with \\(0.10 \\text{ M NaOH}\\). The titration curve generated from the experimental data is shown below. Based on the titration curve, which of the following best identifies the value of \\(pK_{a2}\\) for \\(\\text{H}_2\\text{A}\\) and the reasoning used to determine it?"
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url: "https://nerd-notes.com/ubq/119359/"
date_modified: "2026-08-19T12:39:58+00:00"
---

# A student titrates a sample of a weak diprotic acid, \(\text{H}_2\text{A}\), with \(0.10 \text{ M NaOH}\). The titration curve generated from the experimental data is shown below. Based on the titration curve, which of the following best identifies the value of \(pK_{a2}\) for \(\text{H}_2\text{A}\) and the reasoning used to determine it?

A student titrates a sample of a weak diprotic acid, \(\text{H}_2\text{A}\), with \(0.10 \text{ M NaOH}\). The titration curve generated from the experimental data is shown below. Based on the titration curve, which of the following best identifies the value of \(pK_{a2}\) for \(\text{H}_2\text{A}\) and the reasoning used to determine it?

![A 2D line graph titled Titration of H2A with NaOH showing pH versus volume of titrant. The horizontal x-axis is labeled Volume of 0.10 M NaOH added (mL) with gridlines from 0 to 50 in increments of 10. The vertical y-axis is labeled pH with gridlines from 0 to 14 in increments of 2. A single solid black curve starts at (0, 2.0), passes through (10, 3.5), rises steeply through a vertical inflection at (20, 6.0), passes through (30, 7.2), rises steeply through a second vertical inflection at (40, 10.0), and levels off toward (50, 12.3). Points on the curve at x = 10, 20, 30, and 40 are marked with small solid black dots. No other curves, text, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787143198-jnlkzB.jpg)

- **A.** \(3.5\), because it is the pH at the first half-equivalence point at \(10.0 \text{ mL}\) of added \(\text{NaOH}\)
- **B.** \(6.0\), because it is the pH at the first equivalence point at \(20.0 \text{ mL}\) of added \(\text{NaOH}\)
- **C.** \(7.2\), because it is the pH at the second half-equivalence point at \(30.0 \text{ mL}\) of added \(\text{NaOH}\)
- **D.** \(10.0\), because it is the pH at the second equivalence point at \(40.0 \text{ mL}\) of added \(\text{NaOH}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/119359/*