---
title: "Corrosion; pH after adding NaOH to acetic acid"
subject: "Engineering Chemistry"
chapter: 1
key: "semester_2/engineering_chemistry:corrosion-ph-naoh-acetic"
source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/corrosion-ph-naoh-acetic
site: BCT Engineering
---

# Corrosion; pH after adding NaOH to acetic acid

Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer

- Asked in: 2067 Chaitra, 1+4 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/corrosion-ph-naoh-acetic
- Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1

## Question

What is corrosion? Calculate the pH of a resulted solution when 0.1 gm of NaOH is added to 200ml of 0.1M acetic acid solution. (pKa = 4.74).

## Short answer

Corrosion is the slow destruction of metals by chemical/electrochemical attack of the environment, e.g., rusting of iron. Adding 0.1 g NaOH to 200 mL 0.1 M CH3COOH yields a buffer with 0.0175 mol acid and 0.0025 mol salt; pH = 4.74 + log(0.0025/0.0175) = 3.90.

## Answer

#### Corrosion
The slow destruction of a metal by chemical or electrochemical attack of its surroundings (air, moisture, gases) is called **corrosion**. Example: rusting of iron, 4Fe + 3O₂ + 2xH₂O → 2Fe₂O₃·xH₂O (hydrated ferric oxide); formation of green patina on copper.

#### Calculation
NaOH converts part of the acetic acid into sodium acetate, giving an acidic buffer.
```text
Moles NaOH = 0.1/40 = 0.0025 mol
Moles CH₃COOH = 0.1 × 200/1000 = 0.020 mol
Acid left = 0.020 − 0.0025 = 0.0175 mol
Salt formed = 0.0025 mol

pH = pKa + log([salt]/[acid])
   = 4.74 + log(0.0025/0.0175)
   = 4.74 − 0.845 = 3.895
```
**Answer: pH ≈ 3.90**

## Check yourself

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

1. Corrosion is:
   A. Slow destruction of a metal by chemical or electrochemical attack of surroundings
   B. Melting of a metal
   C. Plating of a metal
   D. Alloy formation
   Answer: A. Such as rusting of iron. (Key point: Corrosion)

2. Rust is mainly:
   A. Hydrated ferric oxide, Fe2O3.xH2O
   B. FeCl3
   C. Fe(OH)2 only
   D. Fe metal
   Answer: A. 4Fe + 3O2 + 2xH2O -> 2Fe2O3.xH2O. (Key point: Rust)

3. 0.1 g NaOH in 200 mL of 0.1 M acid: acid left and salt:
   A. 0.0175 mol and 0.0025 mol
   B. 0.020 mol and 0.0025 mol
   C. 0.0025 mol and 0.0175 mol
   D. 0.0175 mol and 0.0175 mol
   Answer: A. Acid 0.020 - 0.0025 = 0.0175; salt 0.0025. (Key point: Moles)

4. pH = 4.74 + log(0.0025/0.0175) =
   A. 3.90
   B. 4.74
   C. 5.59
   D. 3.40
   Answer: A. log(1/7) = -0.845; 4.74 - 0.845 = 3.895. (Key point: pH)

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