---
title: "Buffer; pH and degree of ionization of acetic acid"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:acetate-buffer-ionization"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/acetate-buffer-ionization
site: BCT Engineering
---

# Buffer; pH and degree of ionization of acetic acid

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2067 Mangsir, 1+4 marks
- Video: [A-Level H2 Chemistry: How to calculate pH of Acidic Buffer](https://www.youtube.com/watch?v=jEo4x6DRyCk) — Winners Education - A Level H2 Chemistry 🇸🇬
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/acetate-buffer-ionization
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

What is a buffer? 1.64 g of anhydrous sodium acetate is added to 200 ml of 0.2M acetic acid. What is the pH of buffer? Calculate the degree of ionization of the acid in the solution. (Ka of acid = 1.8×10⁻⁵)

## Short answer

A buffer solution resists pH change on adding small amounts of strong acid/base; it contains a weak acid and its salt. For 200 mL 0.2 M CH3COOH with 1.64 g CH3COONa, [salt]=0.10 M, [acid]=0.2 M, pH=4.74+log(0.1/0.2)=4.44; degree of ionization α=[H⁺]/C=3.6×10⁻⁵/0.2=1.8×10⁻⁴.

## Answer

#### Buffer
A **buffer solution** resists a change in its pH when small amounts of a strong acid or base are added. It contains a weak acid and its salt (or a weak base and its salt).

#### pH of the buffer
```text
Moles CH₃COONa = 1.64/82 = 0.020 mol
[salt] = 0.020/0.200 = 0.10 M ;  [acid] = 0.2 M
pKa = −log(1.8×10⁻⁵) = 4.745
pH = pKa + log([salt]/[acid]) = 4.745 + log(0.10/0.20)
   = 4.745 − 0.301 = 4.44
```

#### Degree of ionisation of the acid
```text
[H⁺] = Ka [acid]/[salt] = 1.8×10⁻⁵ × 0.2/0.1 = 3.6×10⁻⁵ M
α = [H⁺]/C = 3.6×10⁻⁵/0.2 = 1.8×10⁻⁴
```
**Answer: pH = 4.44; degree of ionisation α = 1.8×10⁻⁴ (0.018 %)**. (Without the salt, α would be √(Ka/C) = 9.5×10⁻³; the common ion greatly suppresses ionisation.)

## Check yourself

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

1. 1.64 g CH3COONa (M = 82) in 200 mL: [salt] =
   A. 0.10 M
   B. 0.20 M
   C. 0.02 M
   D. 0.41 M
   Answer: A. 0.020 mol / 0.200 L = 0.10 M. (Key point: Salt concentration)

2. pKa 4.745, salt 0.10, acid 0.20: pH =
   A. 4.44
   B. 5.05
   C. 4.745
   D. 4.14
   Answer: A. 4.745 - 0.301 = 4.44. (Key point: pH)

3. [H+] = Ka [acid]/[salt] =
   A. 3.6x10^-5 M
   B. 9.0x10^-6 M
   C. 1.8x10^-5 M
   D. 7.2x10^-5 M
   Answer: A. 1.8x10^-5 x 0.2/0.1 = 3.6x10^-5. (Key point: [H+])

4. Compared with acid alone (alpha 9.5x10^-3), alpha in the buffer (1.8x10^-4) is:
   A. Much lower due to the common ion effect
   B. Higher
   C. Unchanged
   D. Zero
   Answer: A. Acetate from the salt suppresses ionisation. (Key point: Degree of ionisation)

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