---
title: "Buffering capacity; NH₄Cl for pH 9 buffer"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:buffer-nh4cl-ph9"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/buffer-nh4cl-ph9
site: BCT Engineering
---

# Buffering capacity; NH₄Cl for pH 9 buffer

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2070 Bhadra, 1+4 marks
- Video: [, A buffer solution with pH 9 is to be prepared by mixing NH_4Cl and NH_4OH. Calculate the number...](https://www.youtube.com/watch?v=ngqgp6pe2v0) — PW Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/buffer-nh4cl-ph9
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

What is meant by buffering capacity? A research student desires to prepare one litre of a solution buffered at pH 9.00. How many grams of ammonium chloride have to be added to one litre of 0.20 M NH₃ to make such a buffer? [pKb value of ammonia is 4.75]

## Short answer

Buffering capacity is moles of strong acid or base needed to change pH of 1 L buffer by one unit. For pH 9, pOH=5, [NH4Cl]=0.20×10^(0.25)=0.356 M. Mass of NH4Cl = 0.356 × 53.49 = 19.0 g.

## Answer

#### Buffering capacity
The number of moles of strong acid or base required to change the pH of one litre of the buffer by one unit. It is greatest when [salt] = [base] (or acid).

#### Calculation
Basic buffer: NH₃ + NH₄Cl, pKb = 4.75.
```text
pOH = 14 − 9.00 = 5.00
pOH = pKb + log([NH₄Cl]/[NH₃])
5.00 = 4.75 + log([NH₄Cl]/0.20)
log([NH₄Cl]/0.20) = 0.25  →  ratio = 1.778
[NH₄Cl] = 0.20 × 1.778 = 0.3557 M

Moles in 1 L = 0.3557 mol
Molar mass NH₄Cl = 53.49 g/mol
Mass = 0.3557 × 53.49 = 19.0 g
```
**Answer: about 19.0 g of NH₄Cl.**

## Check yourself

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

1. Buffering capacity is greatest when:
   A. [salt] is far less than [base]
   B. pH = 14
   C. only salt is present
   D. [salt] = [base]
   Answer: D. Capacity is maximum with equal concentrations of the two components. (Key point: Buffering capacity)

2. For NH3/NH4Cl buffer, pOH =
   A. pKb + log([NH3]/[NH4Cl])
   B. pKb - log([NH4Cl]/[NH3])
   C. 14 + pKb
   D. pKb + log([NH4Cl]/[NH3])
   Answer: D. pOH = pKb + log([salt]/[base]). (Key point: Equation for basic buffer)

3. pOH = 5.00, pKb = 4.75. [NH4Cl]/[NH3] =
   A. 0.56
   B. 1.0
   C. 0.25
   D. 1.778
   Answer: D. log ratio = 0.25, ratio = 10^0.25 = 1.778. (Key point: Ratio)

4. [NH4Cl] = 0.3557 M in 1 L, M = 53.49 g/mol. Mass =
   A. 9.5 g
   B. 0.36 g
   C. 38 g
   D. 19.0 g
   Answer: D. 0.3557 x 53.49 = 19.0 g. (Key point: Mass)

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