Chemistry • Year 12 • Module 6 • Lesson 13
Buffers: Mechanism, Calculations & Natural Systems
Apply buffer mechanism reasoning, the Henderson-Hasselbalch equation, and real data to explain buffer behaviour across biological and environmental contexts.
1. Interpret the buffer vs unbuffered pH response graph
The graph below shows how pH changes when increasing amounts of 1.00 mol/L HCl are added to two solutions: (i) 100 mL of pure water (initial pH 7.00) and (ii) 100 mL of an acetic acid / sodium acetate buffer with equal initial concentrations of 0.100 mol/L (initial pH = pKa = 4.74). Data are illustrative of typical experimental results. 9 marks
Figure 1. Illustrative pH response curves for pure water and 0.100 mol/L acetic acid / sodium acetate buffer on addition of 1.00 mol/L HCl. Buffer initial pH = pKa(CH₃COOH) = 4.74 (Ka = 1.8 × 10−5). Initial volume = 100 mL each. Volume change on HCl addition is negligible.
1.1 Using the graph, estimate the pH of pure water after 0.002 mol HCl is added to 100 mL. Show how you can check this estimate with a calculation. 2 marks
1.2 Describe the trend in pH of the buffer from 0 mol to 0.010 mol HCl added, and explain at the molecular level why the trend has this shape. 3 marks
1.3 The buffer curve drops steeply after approximately 0.010 mol HCl has been added. Using Henderson-Hasselbalch and the buffer composition, explain why this amount represents the buffer’s capacity limit. 2 marks
1.4 At 0.020 mol HCl added, the buffer pH has converged to a value close to that of pure water. Explain what has happened to the buffer components at this point. 2 marks
2. Cause-and-effect chain — respiratory acidosis in a patient with severe pneumonia
A patient with severe pneumonia cannot clear CO₂ from their lungs efficiently. Complete the cause-and-effect chain below to show how CO₂ accumulation leads to a fall in blood pH. Fill in each empty box with the correct chemical change or outcome. 5 marks
Each correct step earns 1 mark. The final “Overall outcome” earns 1 mark.
Equation: CO₂(aq) + H₂O →
3. Interpret buffer pH data — choosing the right weak acid for a target pH
A researcher at a Sydney university needs to prepare buffers at pH 4.00, pH 7.40 (simulating blood), and pH 8.30 (simulating ocean water). She has access to the three weak acids listed below. 6 marks
| Weak acid | Ka | pKa (calculated) | Effective buffer range |
|---|---|---|---|
| Acetic acid (CH₃COOH) | 1.8 × 10−5 | 4.74 | 3.74 – 5.74 |
| Dihydrogen phosphate (H₂PO₄⁻) | 6.2 × 10−8 | 7.21 | 6.21 – 8.21 |
| Boric acid (B(OH)₃) | 5.8 × 10−10 | 9.24 | 8.24 – 10.24 |
3.1 Identify the most suitable weak acid for each target pH and justify your choice by referring to the effective buffer range. 3 marks (1 per target pH)
3.2 Using Henderson-Hasselbalch, calculate the [A⁻]/[HA] ratio required to prepare a dihydrogen phosphate buffer at exactly pH 7.40. Show full working. 2 marks
3.3 Explain why the blood carbonate system (pKa = 6.10) can still maintain pH 7.40 even though 7.40 is well outside the pKa ± 1 effective range of a closed solution. 1 mark
4. Case study — NSW pool pH management with sodium bicarbonate
Public swimming pools in NSW (including those operated by councils across Sydney) maintain water pH between 7.2 and 7.6 for both bactericidal effectiveness of chlorine and swimmer safety. Pool managers add sodium bicarbonate (NaHCO₃) as an alkalinity increaser to build up the total dissolved carbonate, which acts as a buffer against the acidic by-products of chlorination. When excess acid builds up, HCO₃⁻ reacts with it: HCO₃⁻ + H⁺ → H₂CO₃ → CO₂ + H₂O. 5 marks
4.1 Identify the conjugate acid-base pair acting as the buffer in this pool system, and write the equation that shows how this buffer resists pH decrease from chlorination by-products (H⁺). 2 marks
4.2 A pool manager adds HCl to lower the pH of a pool that has become too alkaline. Explain why the pH changes slowly at first but then changes more rapidly once a large amount of HCl has been added. 2 marks
4.3 Predict and justify what effect evaporation of pool water (without replacing the pool chemicals) would have on the pH of a well-buffered pool over time. 1 mark
Q1.1 — Estimate pH of pure water after 0.002 mol HCl
From graph: pH appears to be approximately 2. Calculation check: n(HCl) = 0.002 mol added to 100 mL. [H⁺] = 0.002/0.100 = 0.020 mol/L. pH = −log(0.020) = 1.70. Accept graph reading of approximately 1.7–2.0.
Q1.2 — Trend description and molecular explanation
The buffer pH decreases gradually and almost linearly from pH 4.74 to approximately pH 4.60 as HCl is added from 0 to 0.010 mol (3 marks): (1) Trend: pH decreases only slightly as HCl is added, remaining close to pKa throughout this range. (2) Mechanism: each mole of H⁺ added reacts with CH₃COO⁻ (A⁻ + H⁺ → HA), reducing [A⁻] and increasing [HA] by equal amounts; because both were initially large (0.010 mol each), the fractional change in their ratio is small. (3) Henderson-Hasselbalch: the log([A⁻]/[HA]) term changes only slightly, so pH changes by a small amount.
Q1.3 — Why 0.010 mol represents capacity limit
n(CH₃COO⁻) initially = 0.100 mol/L × 0.100 L = 0.010 mol. When 0.010 mol HCl is added, all 0.010 mol of A⁻ is consumed (A⁻ + H⁺ → HA). With [A⁻] → 0, the Henderson-Hasselbalch term log([A⁻]/[HA]) → −∞, meaning the buffer can no longer resist acid addition and pH drops steeply.
Q1.4 — What has happened to buffer components at 0.020 mol HCl
All the conjugate base (CH₃COO⁻) has been consumed and converted to CH₃COOH. Now the solution is essentially a mixture of CH₃COOH (a weak acid) and excess HCl (a strong acid). The additional H⁺ from excess HCl is not consumed by any buffer component, so pH behaves like an unbuffered strong acid solution — close to pure water’s response. The buffer has been completely destroyed.
Q2 — Cause-and-effect chain
Step 1: [CO₂(aq)] increases; [H₂CO₃] increases. Equation: CO₂(aq) + H₂O → H₂CO₃.
Step 2: The ratio [HCO₃⁻]/[H₂CO₃] decreases (denominator increases, numerator unchanged or slightly decreasing), so log([HCO₃⁻]/[H₂CO₃]) decreases, and pH decreases.
Step 3: Blood pH falls below 7.35; classified as respiratory acidosis (pH is in the acidotic range relative to normal blood pH 7.35–7.45).
Overall outcome: If CO₂ retention continues, blood pH will continue to fall below 7.35 (respiratory acidosis), impairing enzyme function, oxygen delivery by haemoglobin, and cardiac function. Without treatment (assisted ventilation, bronchodilators), the condition is life-threatening.
Q3.1 — Matching acid to target pH
pH 4.00: Acetic acid (pKa = 4.74; range 3.74–5.74 includes 4.00). pH 7.40: Dihydrogen phosphate (pKa = 7.21; range 6.21–8.21 includes 7.40). pH 8.30: Dihydrogen phosphate is borderline (8.30 is within 6.21–8.21), but boric acid (pKa = 9.24; range 8.24–10.24) is more centred around 8.30 and provides better capacity. Accept dihydrogen phosphate with justification that 8.30 is within its range; award 1 mark for correct identification plus 1 mark for referring to the range.
Q3.2 — [A⁻]/[HA] ratio for phosphate buffer at pH 7.40
pH = pKa + log([A⁻]/[HA]) ⇒ log([HPO₄²⁻]/[H₂PO₄⁻]) = 7.40 − 7.21 = 0.19 ⇒ [HPO₄²⁻]/[H₂PO₄⁻] = 100.19 = 1.55. So approximately 1.55 parts HPO₄²⁻ for every 1 part H₂PO₄⁻ (or approximately 3:2 ratio).
Q3.3 — Blood system outside closed pKa ± 1 range
The blood carbonate system is an “open” system: CO₂ produced from H₂CO₃ decomposition is continuously exhaled by the lungs, which replenishes buffer capacity and maintains the [HCO₃⁻]/[H₂CO₃] ratio at ~20:1. In a closed solution, the buffer would be ineffective far from pKa, but the respiratory system continuously removes the acid component (H₂CO₃ via CO₂ expulsion), effectively regenerating the conjugate base ratio far beyond what a closed buffer could sustain.
Q4.1 — Pool buffer pair and equation
Conjugate pair: H₂CO₃ (weak acid) / HCO₃⁻ (conjugate base). Equation resisting pH decrease: HCO₃⁻ + H⁺ → H₂CO₃ → CO₂ + H₂O. (The conjugate base HCO₃⁻ consumes added H⁺, preventing pH from falling.)
Q4.2 — Why pH changes slowly then rapidly
Initially, added HCl is consumed by HCO₃⁻ (the conjugate base): HCO₃⁻ + H⁺ → H₂CO₃ → CO₂ + H₂O. While HCO₃⁻ remains in significant concentration, the buffer resists pH change and pH falls slowly. Once the HCO₃⁻ is exhausted (buffer capacity exceeded), no component remains to consume added H⁺, so each additional mole of HCl causes a large, rapid pH drop — the buffer has failed.
Q4.3 — Effect of evaporation on pool pH
The pH would remain essentially unchanged. Diluting or concentrating a buffer by evaporation (without adding or removing acid or base) does not change the [A⁻]/[HA] ratio, because both components concentrate proportionally. Henderson-Hasselbalch pH = pKa + log([A⁻]/[HA]); if both concentrations scale equally, the ratio and hence pH is unchanged.