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

# Henderson equation; pH after NaOH added to acetic acid

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2072 Magh, 2+3 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-basic-naoh-acetic
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

Derive Henderson equation for basic buffer. Calculate the pH of a solution resulted by adding 0.1 gm of NaOH in to 100 ml of 0.1 M acetic acid solution. Given that the pKa for acid is 4.74.

## Short answer

For a basic buffer (weak base + salt), pOH = pKb + log([salt]/[base]), pH = 14 − pOH. Here, 0.1 g NaOH with 100 mL 0.1 M acetic acid forms an acidic buffer: moles NaOH 0.0025, acid left 0.0075, salt 0.0025; pH = 4.74 + log(0.0025/0.0075) = 4.26.

## Answer

#### Derivation of Henderson equation for basic buffer
A basic buffer has a weak base B and its salt BH⁺A⁻ (e.g. NH₄OH + NH₄Cl).

B + H₂O ⇌ BH⁺ + OH⁻
```text
Kb = [BH⁺][OH⁻]/[B]  →  [OH⁻] = Kb [base]/[salt]
(the salt is fully ionised, so [BH⁺] ≈ [salt]; the base is little ionised, so [B] ≈ [base])
−log[OH⁻] = −log Kb + log([salt]/[base])
pOH = pKb + log([salt]/[base])
pH  = 14 − pOH
```

#### Calculation
NaOH neutralises part of the acetic acid and forms the buffer CH₃COOH / CH₃COONa.
```text
Moles NaOH = 0.1/40 = 0.0025 mol
Moles CH₃COOH = 0.1 × 100/1000 = 0.010 mol
Acid left = 0.010 − 0.0025 = 0.0075 mol
Salt formed = 0.0025 mol

pH = pKa + log([salt]/[acid])
   = 4.74 + log(0.0025/0.0075)
   = 4.74 − 0.477 = 4.26
```
**Answer: pH ≈ 4.26**

## Check yourself

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

1. In the basic buffer derivation, [OH-] equals:
   A. Kb [salt]/[base]
   B. Kb x [base] x [salt]
   C. Kb/([base] + [salt])
   D. Kb [base]/[salt]
   Answer: D. From Kb = [BH+][OH-]/[B], with [BH+] = salt and [B] = base. (Key point: Derivation steps)

2. After finding pOH, how is pH obtained?
   A. pH = pOH
   B. pH = 7 + pOH
   C. pH = pOH - 14
   D. pH = 14 - pOH
   Answer: D. pH + pOH = 14 at 25°C. (Key point: Relating pOH to pH)

3. 0.1 g NaOH (M = 40) in 100 mL of 0.1 M acetic acid: acid left and salt formed?
   A. 0.0100 mol acid, 0.0025 mol salt
   B. 0.0025 mol acid, 0.0075 mol salt
   C. 0.0075 mol acid, 0.0075 mol salt
   D. 0.0075 mol acid, 0.0025 mol salt
   Answer: D. NaOH = 0.0025 mol; acid 0.010 - 0.0025 = 0.0075 mol; salt = 0.0025 mol. (Key point: Moles in the mixture)

4. pH = 4.74 + log(0.0025/0.0075) equals:
   A. 5.22
   B. 4.74
   C. 3.89
   D. 4.26
   Answer: D. log(1/3) = -0.477, so pH = 4.74 - 0.477 = 4.26. (Key point: Final pH)

## Related in Engineering Chemistry

- [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
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