Chemistry • Year 12 • Module 6 • Lesson 8
pH and pOH: Calculations for Strong Acids & Bases
Apply pH and pOH methods to real data, graph interpretation, case studies, and compare-and-contrast tasks — moving from recall to reasoning.
1. Interpret a pH data table — common Australian solutions
The table below shows [H⁺] values for five solutions relevant to Australian science and industry. Use the data to answer the questions below. 8 marks
| # | Solution | [H⁺] (mol/L) | pH (calculate) | Nature (acid/neutral/base) |
|---|---|---|---|---|
| A | Sydney Olympic Pool (target) | 3.0 × 10−8 | ||
| B | Stomach acid (gastric HCl) | 1.0 × 10−2 | ||
| C | Battery acid (conc. H₂SO₄) | 1.0 | ||
| D | Seawater (NSW coast) | 7.9 × 10−9 | ||
| E | Household bleach (NaOCl solution) | 2.0 × 10−11 |
1.1 Calculate the pH of each solution (A–E). Show working for solutions A and D only; write the final pH for B, C, and E. 5 marks
1.2 How many times more acidic is stomach acid (B) than the NSW pool target (A)? Express your answer as a ratio and as a number of pH units. 2 marks
1.3 Ocean acidification due to rising CO₂ is shifting seawater pH from 8.2 toward 8.1. By what factor does [H⁺] change when pH drops from 8.2 to 8.1? Is this significant? 1 mark
2. Interpret a log-scale graph — dilution of HCl
The graph below shows how the pH of a 1.00 mol/L HCl stock solution changes as it is progressively diluted with water at 25°C. Each step on the x-axis represents a 10-fold dilution from the previous concentration. 9 marks
2.1 Describe the relationship between pH and log₁₀[HCl] in the linear region of the graph. State the gradient and its chemical significance. 2 marks
2.2 Using the graph, estimate the pH of a 5.0 × 10−4 mol/L HCl solution. Show how you could verify this by calculation. 2 marks
2.3 Explain why the graph becomes curved and approaches a horizontal asymptote at pH 7 rather than continuing linearly. 3 marks
2.4 Sketch on the graph above (or describe in words) what the equivalent curve for 1.00 mol/L NaOH progressively diluted would look like. At what pH would its asymptote sit? 2 marks
3. Case study — swimming pool pH management in NSW
Public swimming pools in New South Wales are required by health regulations to maintain pH in the range 7.2–7.8, with a target near pH 7.5. Pool operators add HCl (“muriatic acid”) to lower pH when it rises, or sodium carbonate (soda ash) to raise pH when it drops. 7 marks
Scenario: A 500,000 L Olympic-size pool has its pH measured at 7.8 on a Monday morning. By Friday afternoon, heavy rain has diluted the pool water and the pH has shifted to 8.2. The pool operator adds concentrated HCl (36%, density 1.18 g/mL) to bring the pH back to 7.5. The target [H⁺] at pH 7.5 is 3.16 × 10−8 mol/L.
(a) Calculate [H⁺] in the pool at pH 7.8 and at pH 8.2. Express both in mol/L. 2 marks
(b) By what factor has [H⁺] changed between pH 7.8 (Monday) and pH 8.2 (Friday)? Express as a ratio and as a percentage decrease. 2 marks
(c) A student says: “A change from pH 7.8 to 8.2 is only 0.4 units — that's barely any change at all.” Using your calculations from (b), evaluate this statement. 2 marks
(d) Identify one potential health risk of swimming in a pool with pH 8.2 rather than 7.5. 1 mark
4. Compare and contrast — strong acid vs strong base pH calculations
Complete the two-column table to compare the method for calculating pH of a strong acid with the method for a strong base. 8 marks (1 per row)
| Feature | Strong acid (e.g. HCl) | Strong base (e.g. NaOH) |
|---|---|---|
| Starting ion calculated | ||
| Formula used to find [H⁺] or [OH⁻] | ||
| Factor applied for diprotic species | ||
| Intermediate step before pH | ||
| Final formula for pH | ||
| Expected pH range at 25°C | ||
| Named Australian example | ||
| Sanity check |
Q1 — Data table
A pH = −log(3.0 × 10−8) = 8 − log(3.0) = 8 − 0.477 = 7.52; basic (base).
B pH = −log(1.0 × 10−2) = 2.00; acid.
C pH = −log(1.0) = 0.00; acid.
D pH = −log(7.9 × 10−9) = 9 − log(7.9) = 9 − 0.898 = 8.10; basic (base).
E pH = −log(2.0 × 10−11) = 11 − log(2.0) = 11 − 0.301 = 10.70; basic (base).
1.2 [H⁺] ratio = 1.0 × 10−2 / 3.0 × 10−8 = 3.3 × 105 (330,000 times more acidic). pH difference = 7.52 − 2.00 = 5.52 units (so 105.52 ≈ 3.3 × 105 times).
1.3 [H⁺] at pH 8.2 / [H⁺] at pH 8.1 = 108.2−8.1 = 100.1 ≈ 1.26. [H⁺] increases by a factor of ~1.26 (26% increase). Yes, this is significant: a 26% increase in [H⁺] affects carbonate chemistry and marine organisms that build CaCO₃ shells.
Q2 — Graph interpretation
2.1 In the linear region (log[HCl] from 0 to −6), pH increases by exactly 1 unit for each 1-unit decrease in log₁₀[HCl] (i.e. each 10-fold dilution). Gradient = −1 pH unit per log unit dilution, reflecting the definition pH = −log[H⁺] for a strong acid where [H⁺] = [HCl].
2.2 5.0 × 10−4 mol/L: log₁₀[HCl] = −3.30. From the graph, estimated pH ≈ 3.30. Verification: pH = −log(5.0 × 10−4) = 4 − log(5.0) = 4 − 0.699 = 3.30.
2.3 At very low [HCl], the H⁺ ions contributed by the acid become comparable to those from water autoionisation (K₩ = 10−14; water alone gives [H⁺] = 10−7 mol/L). The total [H⁺] is the sum of contributions from HCl and from water. At extreme dilution, the water contribution dominates and [H⁺] → 10−7 mol/L, so pH → 7. The graph asymptotically approaches pH 7 because the acid cannot push [H⁺] below the water autoionisation value.
2.4 The NaOH curve would be a mirror image: starting at pH 14 at 1.00 mol/L, decreasing linearly (1 pH unit per 10× dilution) down to pH ≈ 8–9, then curving asymptotically toward pH 7 from above. Its asymptote would sit at pH 7 (from below, i.e. the base curve approaches 7 from above).
Q3 — Pool case study
(a) [H⁺] at pH 7.8 = 10−7.8 = 1.58 × 10−8 mol/L. [H⁺] at pH 8.2 = 10−8.2 = 6.31 × 10−9 mol/L.
(b) Ratio = 1.58 × 10−8 / 6.31 × 10−9 = 2.50. [H⁺] is 2.5 times higher at pH 7.8. % decrease going from pH 7.8 to 8.2: change = (1.58 − 0.631) × 10−8 = 9.5 × 10−9. % decrease = 9.5/15.8 × 100 = 60% decrease in [H⁺].
(c) The student's statement is incorrect. A shift of only 0.4 pH units corresponds to a 2.5-fold change in [H⁺] (60% decrease). The logarithmic nature of the pH scale means that small numerical differences represent large actual changes in acid concentration. This is chemically and biologically significant for pool disinfection efficiency and swimmer comfort.
(d) At pH 8.2, chlorine disinfectant (hypochlorite) exists more as the less-active OCl⁻ ion rather than HOCl, reducing disinfection efficiency and increasing the risk of bacterial contamination / swimmer infection. Alternatively: high pH irritates eyes (teardrop pH is ~7.5; pH 8.2 causes chemical irritation).
Q4 — Compare and contrast table
| Feature | Strong acid (e.g. HCl) | Strong base (e.g. NaOH) |
|---|---|---|
| Starting ion | [H⁺] | [OH⁻] |
| Formula | [H⁺] = c(acid) × n(H⁺) | [OH⁻] = c(base) × n(OH⁻) |
| Diprotic factor | ×2 for H₂SO₄ (dilute) | ×2 for Ca(OH)₂, Ba(OH)₂ |
| Intermediate step | None — go directly to pH | Calculate pOH = −log[OH⁻] |
| Final formula | pH = −log[H⁺] | pH = 14 − pOH |
| Expected pH | < 7 at 25°C | > 7 at 25°C |
| Australian example | HCl in stomach acid; H₂SO₄ in car battery | NaOH in oven cleaner; Ca(OH)₂ in lime water |
| Sanity check | pH < 7; if > 7, check calculation | pH > 7; if < 7, check for pH = pOH error |