--- title: "pH of acetic acid and NaOH mixture" subject: "Engineering Chemistry" chapter: 1 key: "semester_2/engineering_chemistry:ph-acetic-naoh" source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/ph-acetic-naoh site: BCT Engineering --- # pH of acetic acid and NaOH mixture Engineering Chemistry · Chapter 1: Electro-chemistry and Buffer - Asked in: 2075 Baisakh, 2 marks - Video: [pH of a solution by mixing 100ml of 0.2M acetic acid and NaOH](https://www.youtube.com/watch?v=AeklVXoRcZA) — Ramesh Teppani - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/ph-acetic-naoh - Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-1 ## Question 60 mL of 0.5 M acetic acid is mixed with 40 mL of 0.25 M sodium hydroxide solution. What will be the pH of the mixture? (Given Ka = 1.85×10⁻⁵) ## Short answer Moles of CH₃COOH initially 30 mmol, NaOH 10 mmol; reaction leaves 20 mmol acid and 10 mmol salt, forming a buffer. pH = pKa + log(10/20) with pKa = 4.733 gives 4.43. ## Answer Acetic acid is a weak acid and NaOH is a strong base. NaOH reacts with part of the acid, so the mixture contains the acid and its salt: an acidic buffer. ```text Moles CH₃COOH = 0.5 × 60/1000 = 0.030 mol (30 mmol) Moles NaOH = 0.25 × 40/1000 = 0.010 mol (10 mmol) CH₃COOH + NaOH → CH₃COONa + H₂O Acid left = 30 − 10 = 20 mmol Salt formed = 10 mmol ``` (The total volume is common to both, so mmol can be used directly in the ratio.) ```text pKa = −log(1.85×10⁻⁵) = 4.733 pH = pKa + log([salt]/[acid]) = 4.733 + log(10/20) = 4.733 − 0.301 = 4.43 ``` **Answer: pH ≈ 4.43** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does mixing 30 mmol CH3COOH with 10 mmol NaOH give a buffer? A. Acid is in excess, so the mixture has acid plus its salt (CH3COONa) B. NaOH is completely used up and only water remains C. NaOH is in excess, giving a basic buffer D. Both are strong so no buffer forms Answer: A. NaOH converts part of the weak acid into its salt, leaving acid and salt together. (Key point: Why a buffer forms) 2. After reaction, how much acid and salt are present? A. Acid 20 mmol, salt 10 mmol B. Acid 30 mmol, salt 10 mmol C. Acid 10 mmol, salt 20 mmol D. Acid 20 mmol, salt 20 mmol Answer: A. 10 mmol NaOH consumes 10 mmol acid (30 - 10 = 20 left) and forms 10 mmol salt. (Key point: Amounts after reaction) 3. pKa = 4.733 and [salt]/[acid] = 10/20. What is the pH? A. 4.43 B. 5.03 C. 4.73 D. 4.13 Answer: A. pH = 4.733 + log(0.5) = 4.733 - 0.301 = 4.43. (Key point: Applying Henderson's equation) ## Related in Engineering Chemistry - [Henderson equation; pH after adding NaOH to acetic acid](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Ahenderson-acidic-ph) — Engineering Chemistry, chapter 1 - [Henderson equation; pH after NaOH added to acetic acid](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Ahenderson-basic-naoh-acetic) — Engineering Chemistry, chapter 1 - [Corrosion; pH after adding NaOH to acetic acid](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Acorrosion-ph-naoh-acetic) — Engineering Chemistry, chapter 1 - [Buffer; pH and degree of ionization of acetic acid](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Aacetate-buffer-ionization) — Engineering Chemistry, chapter 1 - [Buffer solution; sodium formate for pH 3.80](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Abuffer-sodium-formate) — Engineering Chemistry, chapter 1 - [Buffer action; pH after adding KOH to acetic acid buffer](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Aacidic-buffer-koh-ph) — Engineering Chemistry, chapter 1 ## Related in other subjects - [pH of mixture: 700 ml pH 6 and 500 ml pH 8](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Aph-mix-c76) — Water Supply Engineering, chapter 4 - [pH of mixture: 400 ml pH 6 and 700 ml pH 8](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Aph-mix-c69) — Water Supply Engineering, chapter 4 - [pH of mixture of three samples](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Aph-mix-b66) — Water Supply Engineering, chapter 4 - [Arrester coordination for 40 kA discharge](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aarrester-132kv-40ka-coordination) — High Voltage Engineering, chapter 4, asked 2×