---
title: "Henderson's equation for basic buffer"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:henderson-basic"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/henderson-basic
site: BCT Engineering
---

# Henderson's equation for basic buffer

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2072 Asoj, 2 marks
- Video: [Henderson Hasselbalch Equation for Basic Buffer Solution - Chemical Equilibrium - Chemistry Class 11](https://www.youtube.com/watch?v=V8nxmdjCjNg) — Ekeeda
- 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
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

Derive Henderson's equation for basic buffer solution.

## Short answer

A basic buffer contains a weak base and its salt. From Kb = [BH⁺][OH⁻]/[B], with [BH⁺]≈[salt] and [B]≈[base], we get pOH = pKb + log([salt]/[base]), and pH = 14 − pOH.

## Answer

A basic buffer contains a weak base (B) and its salt with a strong acid (BH⁺ A⁻), for example NH₄OH + NH₄Cl.

**Derivation**

For the weak base: B + H₂O ⇌ BH⁺ + OH⁻
```text
Kb = [BH⁺][OH⁻]/[B]
[OH⁻] = Kb [B]/[BH⁺]
```
The salt is fully ionised and its common ion BH⁺ suppresses the ionisation of the base, so [B] ≈ [base] and [BH⁺] ≈ [salt]:
```text
[OH⁻] = Kb [base]/[salt]
−log[OH⁻] = −log Kb − log([base]/[salt])
pOH = pKb + log([salt]/[base])
```
This is Henderson's equation for a basic buffer. Since pH + pOH = 14:

pH = 14 − pKb − log([salt]/[base])

## Check yourself

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

1. A basic buffer consists of:
   A. Weak acid and its salt
   B. Weak base and its salt with a strong acid
   C. Strong base and strong acid
   D. Weak base and a weak acid
   Answer: B. For example NH4OH with NH4Cl. (Key point: Basic buffer composition)

2. In the derivation, [BH+] is replaced by [salt] because:
   A. The base is completely ionised
   B. The salt is fully ionised and common ion suppresses the base's ionisation
   C. BH+ is a gas
   D. The salt is insoluble
   Answer: B. The salt supplies nearly all BH+, and the base stays mostly unionised, so [B] = [base]. (Key point: Key assumption)

3. Which equation is Henderson's equation for a basic buffer?
   A. pOH = pKb + log([base]/[salt])
   B. pOH = pKb + log([salt]/[base])
   C. pH = pKb + log([salt]/[base])
   D. pOH = pKb - log([salt]/[base])
   Answer: B. From [OH-] = Kb[base]/[salt], pOH = pKb + log([salt]/[base]); pH = 14 - pOH. (Key point: Final form)

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