---
title: "Henderson equation; pH after adding NaOH to acetic acid"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:henderson-acidic-ph"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/henderson-acidic-ph
site: BCT Engineering
---

# Henderson equation; pH after adding NaOH to acetic acid

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2073 Magh, 1+2+2 marks
- Video: [Find the pH of a Buffer after adding NaOH](https://www.youtube.com/watch?v=-bhljuEApi8) — chemistNATE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/henderson-acidic-ph
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

Define Buffer. Derive Henderson's equation for acidic buffer. Calculate the pH of the solution formed by adding 0.4 g of NaOH on 500 mL 0.2 M acetic acid. pKa for acetic acid = 4.74.

## Short answer

A buffer resists pH changes. An acidic buffer (weak acid + its salt) follows pH = pKa + log([salt]/[acid]), derived from Ka. Adding 0.4 g NaOH to 500 mL 0.2 M acetic acid gives 0.01 mol salt and 0.09 mol acid; pH = 4.74 + log(0.01/0.09) = 3.79.

## Answer

#### Buffer
A buffer is a solution that resists a change in pH on adding small amounts of acid or base. An acidic buffer contains a weak acid (HA) and its salt (A⁻) with a strong base.

#### Derivation of Henderson's equation (acidic buffer)
For the weak acid HA: HA ⇌ H⁺ + A⁻
```text
Ka = [H⁺][A⁻]/[HA]
[H⁺] = Ka [HA]/[A⁻]
```
In the presence of the salt (fully ionised) the ionisation of HA is suppressed, so [HA] ≈ initial concentration of acid and [A⁻] ≈ concentration of salt:
```text
[H⁺] = Ka [acid]/[salt]
Taking −log on both sides:
−log[H⁺] = −log Ka − log([acid]/[salt])
pH = pKa + log([salt]/[acid])
```
This is **Henderson's equation**.

#### Calculation
```text
Moles NaOH = 0.4/40 = 0.01 mol
Moles CH₃COOH = 0.2 × 0.5 = 0.10 mol
CH₃COOH + NaOH → CH₃COONa + H₂O
Acid left = 0.10 − 0.01 = 0.09 mol
Salt formed = 0.01 mol

pH = 4.74 + log(0.01/0.09)
   = 4.74 − 0.954 = 3.79
```
**Answer: pH ≈ 3.79**

## Check yourself

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

1. In deriving Henderson's equation for an acidic buffer, why can [A-] be taken as the salt concentration?
   A. The acid is completely ionised
   B. The salt is fully ionised and suppresses the acid's ionisation
   C. The salt does not ionise
   D. Water supplies the A-
   Answer: B. Fully ionised salt provides nearly all A-, and common ion effect makes [HA] about equal to the initial acid concentration. (Key point: Assumptions in derivation)

2. Which is Henderson's equation for an acidic buffer?
   A. pH = pKa + log([acid]/[salt])
   B. pH = pKa + log([salt]/[acid])
   C. pH = 14 - pKa - log([salt]/[acid])
   D. pH = pKa - log([salt]/[acid])
   Answer: B. Taking -log of [H+] = Ka[acid]/[salt] gives pH = pKa + log([salt]/[acid]). (Key point: Henderson's equation)

3. 0.4 g NaOH (M = 40) is added to 500 mL of 0.2 M acetic acid. What are moles of NaOH and acid?
   A. 0.4 mol and 0.2 mol
   B. 0.01 mol and 0.10 mol
   C. 0.01 mol and 0.2 mol
   D. 0.10 mol and 0.01 mol
   Answer: B. NaOH = 0.4/40 = 0.01 mol; acid = 0.2 x 0.5 = 0.10 mol. (Key point: Moles of NaOH and acid)

4. Acid left 0.09 mol, salt 0.01 mol, pKa 4.74. What is the pH?
   A. 5.69
   B. 3.79
   C. 4.74
   D. 4.45
   Answer: B. pH = 4.74 + log(0.01/0.09) = 4.74 - 0.954 = 3.79. (Key point: Final pH)

## Related in Engineering Chemistry

- [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
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