---
title: "Propionic acid buffer pH; effect of HCl"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:propionic-buffer-hcl"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/propionic-buffer-hcl
site: BCT Engineering
---

# Propionic acid buffer pH; effect of HCl

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2068 Magh, 1+4 marks
- Video: [Calculate the pH Change if We Add Acid to a Buffer Solution of Propanoic acid and sodium propanoate](https://www.youtube.com/watch?v=aDEYLTcdFOU) — Dr. Masi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/propionic-buffer-hcl
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

What is a buffer solution? The Ka of propionic acid is 1.34×10⁻⁵. What is the pH of a solution containing 0.5M propionic acid and 0.25M sodium propionate? What will be the pH of this solution after 0.1M HCl has been added to the buffer?

## Short answer

A buffer resists pH changes. For propionic acid buffer, pH = pKa + log([salt]/[acid]) = 4.873 + log(0.25/0.5) = 4.57. After adding 0.1 M HCl, salt decreases to 0.15 M, acid increases to 0.60 M, giving pH = 4.27.

## Answer

#### Buffer solution
A solution that resists a change of pH on the addition of small amounts of acid or alkali. Propionic acid + sodium propionate is an acidic buffer.

#### pH of the buffer
```text
pKa = −log(1.34×10⁻⁵) = 4.873
pH = pKa + log([salt]/[acid])
   = 4.873 + log(0.25/0.5)
   = 4.873 − 0.301 = 4.572
```

#### pH after adding 0.1 M HCl
HCl (strong acid) reacts with propionate: C₂H₅COO⁻ + H⁺ → C₂H₅COOH. Taking 0.1 mol per litre and no change of volume:
```text
Salt   = 0.25 − 0.10 = 0.15 M
Acid   = 0.50 + 0.10 = 0.60 M
pH = 4.873 + log(0.15/0.60) = 4.873 − 0.602 = 4.271
```
**Answer: pH of buffer = 4.57; after adding HCl, pH = 4.27** (a fall of only 0.30 unit, whereas 0.1 M HCl in pure water would give pH 1).

## Check yourself

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

1. Ka = 1.34x10^-5, acid 0.5 M, salt 0.25 M: pH =
   A. 4.57
   B. 4.87
   C. 5.17
   D. 4.27
   Answer: A. 4.873 + log(0.5) = 4.573. (Key point: Initial pH)

2. When HCl is added to the buffer:
   A. Propionate converts to propionic acid
   B. Propionic acid converts to propionate
   C. Chloride ions react
   D. pH becomes 1
   Answer: A. C2H5COO- + H+ -> C2H5COOH. (Key point: Effect of HCl)

3. After 0.1 M HCl: salt and acid are:
   A. 0.15 M and 0.60 M
   B. 0.35 M and 0.40 M
   C. 0.25 M and 0.50 M
   D. 0.15 M and 0.40 M
   Answer: A. Salt falls by 0.10, acid rises by 0.10. (Key point: New composition)

4. New pH =
   A. 4.27
   B. 4.57
   C. 1.00
   D. 4.87
   Answer: A. 4.873 + log(0.15/0.60) = 4.271. (Key point: New pH)

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