---
title: "A student titrates \\(50.0 \\text{ mL}\\) of a \\(0.100 \\text{ M}\\) weak monoprotic acid, \\(\\text{HA}\\ (\\text{p}K_a = 4.80)\\), with \\(0.100 \\text{ M}\\ \\text{NaOH}\\). The equivalence point is reached after adding \\(50.0 \\text{ mL}\\) of \\(\\text{NaOH}\\). The student records the magnitude of the \\(\\text{pH}\\) change, \\(|\\Delta\\text{pH}|\\), resulting from the addition of a \\(0.50 \\text{ mL}\\) increment of \\(\\text{NaOH}\\) at two different stages: Stage 1 at \\(V_{\\text{NaOH}} = 25.0 \\text{ mL}\\), and Stage 2 at \\(V_{\\text{NaOH}} = 48.0 \\text{ mL}\\).  Which of the following correctly compares \\(|\\Delta\\text{pH}|\\) at Stage 1 with \\(|\\Delta\\text{pH}|\\) at Stage 2 and provides the best justification?"
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url: "https://nerd-notes.com/ubq/123779/"
date_modified: "2026-09-28T12:30:17+00:00"
---

# A student titrates \(50.0 \text{ mL}\) of a \(0.100 \text{ M}\) weak monoprotic acid, \(\text{HA}\ (\text{p}K_a = 4.80)\), with \(0.100 \text{ M}\ \text{NaOH}\). The equivalence point is reached after adding \(50.0 \text{ mL}\) of \(\text{NaOH}\). The student records the magnitude of the \(\text{pH}\) change, \(|\Delta\text{pH}|\), resulting from the addition of a \(0.50 \text{ mL}\) increment of \(\text{NaOH}\) at two different stages: Stage 1 at \(V_{\text{NaOH}} = 25.0 \text{ mL}\), and Stage 2 at \(V_{\text{NaOH}} = 48.0 \text{ mL}\).

Which of the following correctly compares \(|\Delta\text{pH}|\) at Stage 1 with \(|\Delta\text{pH}|\) at Stage 2 and provides the best justification?

A student titrates \(50.0 \text{ mL}\) of a \(0.100 \text{ M}\) weak monoprotic acid, \(\text{HA}\ (\text{p}K_a = 4.80)\), with \(0.100 \text{ M}\ \text{NaOH}\). The equivalence point is reached after adding \(50.0 \text{ mL}\) of \(\text{NaOH}\). The student records the magnitude of the \(\text{pH}\) change, \(|\Delta\text{pH}|\), resulting from the addition of a \(0.50 \text{ mL}\) increment of \(\text{NaOH}\) at two different stages: Stage 1 at \(V_{\text{NaOH}} = 25.0 \text{ mL}\), and Stage 2 at \(V_{\text{NaOH}} = 48.0 \text{ mL}\).

Which of the following correctly compares \(|\Delta\text{pH}|\) at Stage 1 with \(|\Delta\text{pH}|\) at Stage 2 and provides the best justification?

- **A.** \(|\Delta\text{pH}|\) is greater at Stage 1 than at Stage 2, because the higher concentration of unreacted \(\text{HA}\) at Stage 1 drives a greater extent of neutralization.
- **B.** \(|\Delta\text{pH}|\) is smaller at Stage 1 than at Stage 2, because \([\text{HA}] \approx [\text{A}^-]\) at Stage 1, which minimizes the fractional change in the ratio \(\dfrac{[\text{A}^-]}{[\text{HA}]}\) upon adding base.
- **C.** \(|\Delta\text{pH}|\) is smaller at Stage 1 than at Stage 2, because the neutralization reaction between \(\text{HA}\) and \(\text{OH}^-\) is reversible at Stage 1 but becomes irreversible near the equivalence point.
- **D.** \(|\Delta\text{pH}|\) is greater at Stage 1 than at Stage 2, because the high concentration of conjugate base \(\text{A}^-\) at Stage 2 suppresses further changes in \(\text{pH}\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123779/*