A Level Chemistry - How to Solve: Buffer Solutions (pH Calculation, Henderson-Hasselbalch, Buffer Action)

How to Solve: Buffer Solutions (pH Calculation, Henderson-Hasselbalch, Buffer Action)

Complete Guide For GCSE/A-Level Chemistry (Edexcel, AQA, OCR) – Worth 6-10 marks in exams

Introduction

"Mastering buffer solutions unlocks 6-10 marks in your exam—enough to boost your grade from a C to an A. Hospitals use them to keep IV drips safe, your blood uses them to stay at pH 7.4, and even your shampoo relies on them. Today, you’ll learn the exact steps to calculate pH, predict buffer action, and avoid the traps that cost students marks."

WHAT YOU NEED TO KNOW FIRST

Before starting, you must understand: 1. pH and pKa: pH = -log[H⁺], pKa = -log(Ka). Lower pKa = stronger acid. 2. Weak acids/bases: They partially dissociate (e.g., CH₃COOH ⇌ CH₃COO⁻ + H⁺). 3. Equilibrium expressions: Ka = [H⁺][A⁻]/[HA] for a weak acid HA.

If any of these are unclear, pause and review them first.

KEY TERMS & FORMULAS

Key Terms

Formulas

1. Henderson-Hasselbalch Equation (for weak acid buffers)
   pH = pKa + log([A⁻]/[HA])
2. pH: pH of the buffer solution.
3. pKa: -log(Ka) of the weak acid.
4. [A⁻]: Concentration of conjugate base (mol/dm³).
5. [HA]: Concentration of weak acid (mol/dm³).

6. MEMORISE THIS (not always given on exam sheets).

7. Ka Expression (for weak acids)
   Ka = [H⁺][A⁻]/[HA]

8. Given on exam sheet (but you must know how to rearrange it).

9. pH from [H⁺]
   pH = -log[H⁺]

10. Given on exam sheet.

STEP-BY-STEP METHOD

Follow these 5 steps for every buffer pH calculation:

1. Identify the weak acid and its conjugate base

2. Example: CH₃COOH (weak acid) and CH₃COO⁻ (conjugate base).

3. Write down the given values

4. pKa of the weak acid.

5. Concentrations of [HA] and [A⁻] (or moles if volumes are given).

6. Check if you need to convert moles to concentrations

7. If given moles and volume, use: concentration = moles/volume (dm³).

8. Plug into the Henderson-Hasselbalch equation

9. pH = pKa + log([A⁻]/[HA]).

10. Tip: If [A⁻] = [HA], pH = pKa (log(1) = 0).

11. Calculate and round to 2 decimal places

12. Use a calculator. Log(2) ≈ 0.30, log(0.5) ≈ -0.30.

Worked Example Using the Steps

Question: A buffer contains 0.10 mol/dm³ CH₃COOH and 0.20 mol/dm³ CH₃COO⁻. pKa of CH₃COOH = 4.76. Calculate the pH.

Step 1: Weak acid = CH₃COOH, conjugate base = CH₃COO⁻. Step 2: [HA] = 0.10 mol/dm³, [A⁻] = 0.20 mol/dm³, pKa = 4.76. Step 3: Concentrations already given (no conversion needed). Step 4: pH = 4.76 + log(0.20/0.10) = 4.76 + log(2) = 4.76 + 0.30. Step 5: pH = 5.06.

What we did and why: We used the Henderson-Hasselbalch equation because the solution is a buffer (weak acid + conjugate base). The ratio [A⁻]/[HA] determines the pH shift from pKa.

WORKED EXAMPLES

Example 1 – Basic

Question: A buffer is made from 0.050 mol/dm³ HCOOH (pKa = 3.75) and 0.025 mol/dm³ HCOO⁻. Calculate its pH.

Solution: 1. Weak acid = HCOOH, conjugate base = HCOO⁻. 2. [HA] = 0.050, [A⁻] = 0.025, pKa = 3.75. 3. No conversion needed. 4. pH = 3.75 + log(0.025/0.050) = 3.75 + log(0.5) = 3.75 - 0.30. 5. pH = 3.45.

What we did and why: The buffer ratio [A⁻]/[HA] < 1, so pH < pKa. We used the Henderson-Hasselbalch equation directly.

Example 2 – Medium (Moles Given)

Question: 0.020 moles of NH₃ (weak base) and 0.010 moles of NH₄Cl (conjugate acid) are dissolved in 500 cm³ of water. pKb of NH₃ = 4.75. Calculate the pH.

Solution: 1. Weak base = NH₃, conjugate acid = NH₄⁺ (from NH₄Cl). 2. Moles: NH₃ = 0.020, NH₄⁺ = 0.010. Volume = 0.500 dm³. 3. Convert moles to concentrations:
- [NH₃] = 0.020/0.500 = 0.040 mol/dm³.
- [NH₄⁺] = 0.010/0.500 = 0.020 mol/dm³. 4. For a weak base buffer, use pOH = pKb + log([BH⁺]/[B]).
- pOH = 4.75 + log(0.020/0.040) = 4.75 - 0.30 = 4.45. 5. pH = 14 - pOH = 14 - 4.45 = 9.55.

What we did and why: We converted moles to concentrations first. For a weak base buffer, we used pOH instead of pH, then converted to pH.

Example 3 – Exam-Style (Disguised)

Question: A student mixes 25 cm³ of 0.10 mol/dm³ CH₃COOH with 15 cm³ of 0.10 mol/dm³ NaOH. pKa of CH₃COOH = 4.76. Calculate the pH of the resulting solution.

Solution: 1. Identify the reaction: CH₃COOH + NaOH → CH₃COO⁻ + H₂O. 2. Calculate moles:
- CH₃COOH: 0.025 dm³ × 0.10 mol/dm³ = 0.0025 mol.
- NaOH: 0.015 dm³ × 0.10 mol/dm³ = 0.0015 mol. 3. Determine remaining species:
- NaOH is limiting. It reacts with CH₃COOH to form CH₃COO⁻.
- Moles of CH₃COOH left = 0.0025 - 0.0015 = 0.0010 mol.
- Moles of CH₃COO⁻ formed = 0.0015 mol. 4. Total volume = 25 + 15 = 40 cm³ = 0.040 dm³. 5. Concentrations:
- [HA] = 0.0010/0.040 = 0.025 mol/dm³.
- [A⁻] = 0.0015/0.040 = 0.0375 mol/dm³. 6. Henderson-Hasselbalch:
- pH = 4.76 + log(0.0375/0.025) = 4.76 + log(1.5) = 4.76 + 0.18. 7. pH = 4.94.

What we did and why: This is a partial neutralisation question. We calculated moles first, then used the reaction to find the buffer components. The key was recognising that NaOH converts some CH₃COOH to CH₃COO⁻.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP

"Here’s what you need to remember the night before your exam: 1. Buffers resist pH change because they contain a weak acid + conjugate base (or weak base + conjugate acid). 2. Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). Memorise this! 3. For weak base buffers, use pOH = pKb + log([BH⁺]/[B]), then pH = 14 - pOH. 4. Always check units: Convert cm³ to dm³ if needed. 5. Partial neutralisation? Calculate moles first, then find remaining acid/base. 6. Common traps: Ignoring volume changes, mixing up pKa/pKb, and misplacing the log ratio. 7. Buffer action explanation: Added H⁺ reacts with A⁻, added OH⁻ reacts with HA—equilibrium shifts to remove them.

You’ve got this. Now go ace that exam!"