---
title: "Buffer action; pH change on adding KOH"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:buffer-koh-change-ph"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/buffer-koh-change-ph
site: BCT Engineering
---

# Buffer action; pH change on adding KOH

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2069 Poush, 2+3 marks
- Video: [Buffer solution pH calculations | Chemistry | Khan Academy](https://www.youtube.com/watch?v=lsHq5aqz4uQ) — Khan Academy Organic Chemistry
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/buffer-koh-change-ph
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

Define the term buffer and buffer action. If 3.0 millimoles of KOH is added to 100 ml of a solution which is 0.10 M in acetic acid and 0.10 M in sodium acetate. What will be the change in pH of the solution?

## Short answer

A buffer solution resists pH change. Adding KOH converts acetic acid to acetate: initially 10 mmol each, after adding 3 mmol KOH acid = 7 mmol, salt = 13 mmol. pH change = log(13/7) ≈ +0.27, from 4.74 to 5.01.

## Answer

#### Buffer and buffer action
A **buffer** is a solution (a weak acid with its salt, or a weak base with its salt) that resists a change in pH on adding a small amount of strong acid or base. The property of resisting the change is called **buffer action**. In CH₃COOH / CH₃COONa the acetate ion neutralises added H⁺ (CH₃COO⁻ + H⁺ → CH₃COOH) and the acid neutralises added OH⁻ (CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O).

#### Calculation
```text
Initial: acid = 0.10 M × 100 mL = 10 mmol ; salt = 10 mmol
Initial pH = pKa + log(10/10) = pKa = 4.74

KOH added = 3.0 mmol:  CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
Acid left = 10 − 3 = 7 mmol
Salt now  = 10 + 3 = 13 mmol

New pH = pKa + log(13/7) = 4.74 + 0.269 = 5.01
Change in pH = log(13/7) = +0.27
```
(The change does not depend on pKa.) **Answer: pH rises by about 0.27 units** (from 4.74 to 5.01, taking pKa = 4.74).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which neutralises added OH- in acetic acid/acetate buffer?
   A. CH3COOH
   B. CH3COO-
   C. Na+
   D. H2O
   Answer: A. CH3COOH + OH- -> CH3COO- + H2O. (Key point: Buffer action)

2. With equal acid and salt (10 mmol each), initial pH =
   A. pKa
   B. pKa + 1
   C. pKa - 1
   D. 7
   Answer: A. log(1) = 0 so pH = pKa = 4.74. (Key point: Initial pH)

3. After 3 mmol KOH, acid and salt are:
   A. 7 and 13 mmol
   B. 13 and 7 mmol
   C. 10 and 13 mmol
   D. 7 and 10 mmol
   Answer: A. Acid decreases by 3, salt increases by 3. (Key point: After KOH)

4. The change in pH =
   A. +0.27
   B. -0.27
   C. +3
   D. +0.77
   Answer: A. log(13/7) = 0.27, pH rises from 4.74 to 5.01. (Key point: Change in pH)

## Related in Engineering Chemistry

- [Buffer action; pH after adding KOH to acetic acid buffer](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Aacidic-buffer-koh-ph) — Engineering Chemistry, chapter 1
- [Buffer; pH and degree of ionization of acetic acid](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Aacetate-buffer-ionization) — Engineering Chemistry, chapter 1
- [Henderson equation; pH after NaOH added to acetic acid](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Ahenderson-basic-naoh-acetic) — Engineering Chemistry, chapter 1
- [Henderson equation; pH after adding NaOH to acetic acid](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Ahenderson-acidic-ph) — Engineering Chemistry, chapter 1
- [Corrosion; pH after adding NaOH to acetic acid](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Acorrosion-ph-naoh-acetic) — Engineering Chemistry, chapter 1
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