Chemistry • Year 12 • Module 6 • Lesson 8
pH and pOH: Calculations for Strong Acids & Bases
Build HSC Band 5–6 technique on pH and pOH calculations — evaluating multi-step scenarios, critiquing sources, and writing extended chemical reasoning.
1. Data + scenario — multi-step mixing with evaluation (Band 5–6)
8 marks Band 5–6
Scenario. A research student is preparing a pH-buffered solution for a cell biology experiment. She begins by mixing two stock solutions:
- Solution A: 80.0 mL of 0.120 mol/L HCl (a strong acid)
- Solution B: 60.0 mL of 0.100 mol/L Ba(OH)₂ (a strong diprotic base)
She records the result and then makes two errors in a second attempt: (i) she uses only V = 80.0 mL (the acid volume) as the total volume; (ii) she forgets the diprotic factor for Ba(OH)₂.
Data table (student recordings from a calibrated pH meter):
| Trial | V(acid, mL) | V(base, mL) | pH measured | pH predicted? |
|---|---|---|---|---|
| 1 (correct) | 80.0 | 60.0 | 12.23 | |
| 2 (error i) | 80.0 | 60.0 | 12.23 | Student predicts 12.48 |
| 3 (error ii) | 80.0 | 60.0 | 12.23 | Student predicts 1.59 |
Q1. In your response you must:
- Calculate the correct pH of the Trial 1 mixture, showing all steps.
- Show how error (i) produces pH = 12.48 and explain the conceptual mistake.
- Show how error (ii) produces pH = 1.59 and explain the conceptual mistake.
- Evaluate which error (i or ii) produces a larger absolute pH deviation from the correct value and justify your evaluation.
- Explain why pH = 7.00 is not achieved in this mixture despite both being strong, and state what volumes would need to be used to achieve pH 7.00.
2. Source critique — evaluate a scientific claim (Band 5–6)
7 marks Band 5–6
“Ocean pH has dropped from 8.2 to 8.1 since the Industrial Revolution — a decrease of only 0.1 units, which is barely measurable. Scientists are exaggerating the threat of ocean acidification; a change this small cannot meaningfully alter [H⁺] enough to affect marine life.”
— Comment posted on an Australian science news article, 2024.
Q2. Critically evaluate this claim. In your response you must:
- Calculate [H⁺] at pH 8.2 and at pH 8.1 and determine the factor by which [H⁺] has increased.
- Identify the specific scientific flaw in the commenter’s reasoning.
- Explain the correct chemistry of why a 0.1 pH unit change is significant.
- Describe one experimental measurement that could demonstrate this significance to a sceptic.
- Assess whether the claim is “exaggeration” or a “misunderstanding of scale”, with justification.
- Identify one named marine organism and explain the mechanism by which elevated [H⁺] threatens it.
Q1 — Multi-step mixing with evaluation
Step 1 — Calculate moles:
n(H⁺) from HCl = 0.120 mol/L × 0.0800 L = 9.60 × 10−3 mol.
n(OH⁻) from Ba(OH)₂ = 0.100 mol/L × 0.0600 L × 2 = 1.20 × 10−2 mol.
Step 2 — Find excess:
OH⁻ is in excess: n(excess OH⁻) = 1.20 × 10−2 − 9.60 × 10−3 = 2.40 × 10−3 mol.
Step 3 — Correct pH:
V(total) = 80.0 + 60.0 = 140.0 mL = 0.1400 L.
c(OH⁻) = 2.40 × 10−3 / 0.1400 = 0.01714 mol/L.
pOH = −log(0.01714) = 1.766. pH = 14 − 1.766 = 12.23.
Error (i) — V = 80.0 mL only:
c(OH⁻)(wrong) = 2.40 × 10−3 / 0.0800 = 0.0300 mol/L.
pOH = −log(0.0300) = 1.52. pH = 14 − 1.52 = 12.48. Conceptual mistake: the excess species is diluted into both solution volumes, not just the acid volume. Using only V(acid) overestimates [OH⁻] and gives an inflated pH.
Error (ii) — forgetting diprotic factor:
If diprotic factor forgotten: n(OH⁻) = 0.100 × 0.0600 = 6.00 × 10−3 mol (half the correct value). Now n(H⁺) = 9.60 × 10−3 mol; n(OH⁻) = 6.00 × 10−3 mol. H⁺ is in excess: n(excess H⁺) = 3.60 × 10−3 mol. c(H⁺) = 3.60 × 10−3 / 0.1400 = 0.02571 mol/L. pH = −log(0.02571) = 1.59. Conceptual mistake: Ba(OH)₂ releases 2 OH⁻ ions per formula unit; treating it as monoprotic halves the calculated [OH⁻], incorrectly reversing which species is in excess.
Evaluation of errors:
Error (i): |pH deviation| = |12.48 − 12.23| ≈ 0.25 pH units.
Error (ii): |pH deviation| = |1.59 − 12.23| = 10.64 pH units (reversed excess species).
Error (ii) produces by far the larger deviation — it changes the nature of the excess species entirely, predicting an acidic solution when the true solution is strongly basic. This is a category error, not just a numerical imprecision.
Why pH ≠ 7.00 and equivalence volume:
The two solutions are not at equivalence: n(OH⁻) > n(H⁺). To reach pH 7.00, the moles of H⁺ must equal the moles of OH⁻: n(H⁺) = n(OH⁻). 0.120 × V(HCl) = 0.100 × 2 × V(Ba(OH)₂). At V(Ba(OH)₂) = 60.0 mL: V(HCl) needed = 0.200 × 0.0600 / 0.120 = 0.100 L = 100.0 mL. Using 100.0 mL HCl + 60.0 mL Ba(OH)₂ would give n(H⁺) = n(OH⁻) and pH = 7.00.
Marking notes. 1 mark: correct n(H⁺) and n(OH⁻) with diprotic factor. 1 mark: correct identification of excess and n(excess). 1 mark: correct V(total) and c(OH⁻). 1 mark: correct pH via pOH route. 1 mark: correct error (i) pH and explanation. 1 mark: correct error (ii) identification and explanation. 1 mark: correct comparison of error magnitudes with justification. 1 mark: correct equivalence volume calculation or reasoning.
Q2 — Source critique: ocean acidification
Calculations:
[H⁺] at pH 8.2 = 10−8.2 = 6.31 × 10−9 mol/L.
[H⁺] at pH 8.1 = 10−8.1 = 7.94 × 10−9 mol/L.
Factor = 7.94 / 6.31 = 1.26. [H⁺] has increased by 26%.
Scientific flaw: The commenter treats pH as a linear scale. The scale is logarithmic, so a 0.1 unit decrease in pH corresponds to a 100.1 ≈ 26% increase in [H⁺] — not a trivial change. The claim that the change is “barely measurable” is false: a 26% increase in [H⁺] is readily detectable with modern pH electrodes (precision ± 0.01 pH units) and by biological response.
Chemical significance: A 26% increase in [H⁺] shifts the carbonate equilibrium system: CO₂ + H₂O ↔ H₂CO₃ ↔ H⁺ + HCO₃⁻ ↔ 2H⁺ + CO₃²−. Higher [H⁺] drives the equilibrium left, reducing [CO₃²−], which is needed by marine organisms to precipitate CaCO₃ (shells, skeletons). This is a direct chemical consequence of the pH shift.
Experimental test: Place known masses of coral or mussel shells in seawater adjusted to pH 8.2 vs pH 8.1 for 30 days. Measure mass loss by dissolution. The pH 8.1 solution will dissolve significantly more CaCO₃, demonstrating that the 0.1 unit change is biologically significant.
Exaggeration or misunderstanding: This is a misunderstanding of scale, not exaggeration by scientists. The scientific consensus uses precise quantitative language (a 26% increase in [H⁺] since the Industrial Revolution). The commenter’s error is applying linear intuition to a logarithmic quantity, which is a mathematical misinterpretation, not a bias in the scientific reporting.
Named organism and mechanism: Coral (Acropora spp., Great Barrier Reef) — elevated [H⁺] reduces [CO₃²−], decreasing the saturation state of aragonite (a CaCO₃ polymorph). Below aragonite saturation, existing skeletons dissolve and new skeletal deposition is energetically costlier, slowing growth and weakening reef structure. Studies on the GBR show reduced calcification rates of 10–25% at pH projected for 2100.
Marking notes. 1 mark: correct [H⁺] at both pH values. 1 mark: correct factor (1.26 ± 0.02). 1 mark: identifying the flaw (linear vs logarithmic scale). 1 mark: explaining carbonate chemistry link. 1 mark: valid experimental test described. 1 mark: assessment (misunderstanding, not exaggeration) with justification. 1 mark: named organism + mechanism with specificity.