---
title: "A student prepares four separate \\(100.0 \\text{ mL}\\) buffer solutions using a weak monoprotic acid, \\(\\text{HA}\\) (\\(\\text{p}K_a = 5.00\\)), and its conjugate base, \\(\\text{NaA}\\). To each solution, the student adds \\(0.0020 \\text{ mol}\\) of solid \\(\\text{NaOH}\\) without causing a significant volume change, and records the resulting \\(\\text{pH}\\) values in the table below.  | Buffer | \\([\\text{HA}]\\) (M) | \\([\\text{A}^-]\\) (M) | Initial \\(\\text{pH}\\) | Final \\(\\text{pH}\\) after addition of \\(0.0020 \\text{ mol NaOH}\\) | |—|—|—|—|—| | W | \\(0.50\\) | \\(0.50\\) | \\(5.00\\) | \\(5.03\\) | | X | \\(0.50\\) | \\(0.050\\) | \\(4.00\\) | \\(4.16\\) | | Y | \\(0.10\\) | \\(0.10\\) | \\(5.00\\) | \\(5.18\\) | | Z | \\(0.050\\) | \\(0.50\\) | \\(6.00\\) | \\(6.24\\) |  Based on the data, which of the following correctly ranks the buffer solutions from greatest to least buffer capacity against the added strong base, and provides the best justification?"
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url: "https://nerd-notes.com/ubq/123806/"
date_modified: "2026-09-28T12:30:23+00:00"
---

# A student prepares four separate \(100.0 \text{ mL}\) buffer solutions using a weak monoprotic acid, \(\text{HA}\) (\(\text{p}K_a = 5.00\)), and its conjugate base, \(\text{NaA}\). To each solution, the student adds \(0.0020 \text{ mol}\) of solid \(\text{NaOH}\) without causing a significant volume change, and records the resulting \(\text{pH}\) values in the table below.

| Buffer | \([\text{HA}]\) (M) | \([\text{A}^-]\) (M) | Initial \(\text{pH}\) | Final \(\text{pH}\) after addition of \(0.0020 \text{ mol NaOH}\) |
|—|—|—|—|—|
| W | \(0.50\) | \(0.50\) | \(5.00\) | \(5.03\) |
| X | \(0.50\) | \(0.050\) | \(4.00\) | \(4.16\) |
| Y | \(0.10\) | \(0.10\) | \(5.00\) | \(5.18\) |
| Z | \(0.050\) | \(0.50\) | \(6.00\) | \(6.24\) |

Based on the data, which of the following correctly ranks the buffer solutions from greatest to least buffer capacity against the added strong base, and provides the best justification?

A student prepares four separate \(100.0 \text{ mL}\) buffer solutions using a weak monoprotic acid, \(\text{HA}\) (\(\text{p}K_a = 5.00\)), and its conjugate base, \(\text{NaA}\). To each solution, the student adds \(0.0020 \text{ mol}\) of solid \(\text{NaOH}\) without causing a significant volume change, and records the resulting \(\text{pH}\) values in the table below.

| Buffer | \([\text{HA}]\) (M) | \([\text{A}^-]\) (M) | Initial \(\text{pH}\) | Final \(\text{pH}\) after addition of \(0.0020 \text{ mol NaOH}\) |
|---|---|---|---|---|
| W | \(0.50\) | \(0.50\) | \(5.00\) | \(5.03\) |
| X | \(0.50\) | \(0.050\) | \(4.00\) | \(4.16\) |
| Y | \(0.10\) | \(0.10\) | \(5.00\) | \(5.18\) |
| Z | \(0.050\) | \(0.50\) | \(6.00\) | \(6.24\) |

Based on the data, which of the following correctly ranks the buffer solutions from greatest to least buffer capacity against the added strong base, and provides the best justification?

- **A.** \(\text{W} > \text{Y} > \text{X} > \text{Z}\), because solutions with an equimolar ratio of weak acid to conjugate base always resist \(\text{pH}\) changes more effectively than solutions with unequal component ratios.
- **B.** \(\text{W} > \text{X} > \text{Y} > \text{Z}\), because the buffer with the smallest \(\Delta\text{pH}\) exhibits the greatest capacity, which is governed primarily by having a higher concentration of \(\text{HA}\) available to neutralize added \(\text{OH}^-\).
- **C.** \(\text{Z} > \text{Y} > \text{X} > \text{W}\), because solutions with higher initial \(\text{pH}\) values contain higher concentrations of conjugate base, maximizing their ability to absorb additional basic species.
- **D.** \(\text{W} > \text{X} > \text{Z} > \text{Y}\), because buffer capacity is determined solely by total buffer concentration (\([\text{HA}] + [\text{A}^-]\)), meaning solutions with a total concentration of \(0.55 \text{ M}\) resist base addition better than a solution with \(0.20 \text{ M}\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123806/*