When a person hyperventilates, they expel CO₂ faster than their body produces it — blood pH rises above 7.45 within minutes, causing dizziness, tingling, and muscle cramps. Every calculation in this lesson is the mathematics behind why a shift of 0.1 pH units in blood has clinical consequences.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A patient arrives at an emergency department breathing rapidly and shallowly after a panic attack. A blood gas test shows pH 7.56 — their blood is significantly more basic than the normal range of 7.35–7.45. The attending doctor explains: "Hyperventilation removes CO₂ from the blood faster than it is produced. CO₂ dissolves in blood as carbonic acid — when it is removed, [H⁺] in blood drops and pH rises."
The patient's nurse asks: "If pH 7.56 seems very close to normal, why is the patient dizzy and having muscle cramps?"
Before you read on, write down what you think. If the pH scale is logarithmic, what does a change of 0.1 pH units actually mean in terms of [H₃O⁺]? And why does blood pH need to be controlled to within such a narrow range? You will return to this at the end of the lesson.
Before any calculation is performed, the physical meaning of pH must be clear — because the logarithmic nature of the scale is not a mathematical convenience, it is a reflection of the enormous range of [H₃O⁺] values that exist in chemistry and biology, and misunderstanding it leads to systematic errors in interpreting pH differences.
pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration: pH = −log₁₀[H₃O⁺]. The negative sign is applied because [H₃O⁺] in aqueous solutions is always a very small number (between ~10⁻¹⁴ and ~10⁰ mol/L) — the negative sign converts these small decimals into positive, manageable numbers.
The logarithmic nature means each unit change in pH corresponds to a tenfold change in [H₃O⁺]. A solution at pH 3 has [H₃O⁺] = 10⁻³ mol/L; a solution at pH 5 has [H₃O⁺] = 10⁻⁵ mol/L — the pH 3 solution has 100 times more H₃O⁺ than the pH 5 solution despite appearing "only 2 units" different on the scale.
This is the answer to the Think First: a blood pH change from 7.45 (normal) to 7.56 (hyperventilation) represents a [H₃O⁺] decrease by a factor of 10⁰·¹¹ ≈ 1.3 — blood [H₃O⁺] has dropped by 23% in a system calibrated to a narrow window. Enzymes, ion channels, and nerve conduction are all sensitive to [H₃O⁺] at the 10⁻⁸ mol/L scale — a 23% change is enormous in physiological terms even though 0.11 pH units appears trivial on paper.
| pH | [H₃O⁺] (mol/L) | Example | Relative [H₃O⁺] vs pH 7 |
|---|---|---|---|
| 0 | 1.0 | Battery acid (conc. H₂SO₄) | 10,000,000× |
| 1 | 0.1 | Stomach acid (HCl) | 1,000,000× |
| 3 | 0.001 | Vinegar (CH₃COOH ~5%) | 10,000× |
| 5 | 0.00001 | Black coffee | 100× |
| 7 | 0.0000001 | Pure water at 25°C | 1× (reference) |
| 9 | 0.000000001 | Baking soda solution | 0.01× |
| 11 | 0.00000000001 | Household ammonia | 0.0001× |
| 13 | 0.0000000000001 | Oven cleaner (NaOH) | 0.000001× |
For a strong acid, the [H₃O⁺] calculation is the simplest possible — because 100% ionisation means the concentration of H₃O⁺ is directly equal to the concentration of the acid (adjusted for the number of protons donated per formula unit), with no equilibrium to solve.
For a monoprotic strong acid (HCl, HNO₃, HBr, HI, HClO₄): complete ionisation means [H₃O⁺] = c(acid) exactly. Then pH = −log₁₀(c(acid)).
For diprotic H₂SO₄ in dilute solution, both protons are assumed to be fully donated: [H₃O⁺] = 2 × c(H₂SO₄). This is the standard HSC treatment for dilute H₂SO₄.
Strong base pH calculations require one additional step compared to strong acid calculations — because the starting information is [OH⁻] rather than [H₃O⁺], and the route to pH passes through pOH and the relationship pH + pOH = 14.
For a strong base, complete dissociation means [OH⁻] = c(base) × n(OH⁻ per formula unit). For NaOH and KOH (monoprotic): [OH⁻] = c(base). For Ca(OH)₂ and Ba(OH)₂ (diprotic): [OH⁻] = 2 × c(base).
Diluting an acid or base with water decreases the concentration of all ions — but because pH is logarithmic, the effect on pH is not proportional to the dilution factor, and there is a physical limit beyond which further dilution cannot push pH.
When a strong acid solution is diluted, [H₃O⁺] decreases because the same number of H⁺ ions are distributed through a larger volume. Use c₁V₁ = c₂V₂ (moles conserved — adding water does not add or remove H⁺) to find the new concentration, then calculate pH from the new [H₃O⁺].
A critical physical limit applies: no matter how much a strong acid is diluted, pH cannot exceed 7 at 25°C — because water itself contributes [H₃O⁺] = 1.0 × 10⁻⁷ mol/L through autoionisation. When the acid concentration drops to approximately 10⁻⁶ mol/L or below, water's autoionisation contribution becomes significant. In HSC calculations, you will not be asked to calculate pH for solutions where c(acid) < 10⁻⁶ mol/L, but you must be able to state that pH approaches (but never reaches or exceeds) 7 upon extreme dilution of an acid.
When a strong acid and a strong base are mixed, the fundamental question is always which one is in excess — because the excess species determines whether the final solution is acidic, basic, or neutral, and calculating its concentration after mixing gives everything needed for the pH calculation.
When H⁺ and OH⁻ are mixed, the neutralisation reaction H⁺ + OH⁻ → H₂O occurs. The excess species remains unreacted.
"A pH difference of 0.3 is insignificant." — A change of 0.3 pH units corresponds to a factor of 10⁰·³ ≈ 2 in [H₃O⁺]. Blood pH dropping from 7.4 to 7.1 doubles [H₃O⁺] — enough to denature enzymes and alter haemoglobin's oxygen affinity.
"0.10 mol/L H₂SO₄ has pH = 1.0." — H₂SO₄ is diprotic. [H₃O⁺] = 2 × 0.10 = 0.20 mol/L. pH = −log(0.20) = 0.70, not 1.0. Always multiply by 2 for dilute H₂SO₄.
"pH approaches 0 when you add more and more acid." — pH approaches the limit set by the acid's concentration. With a very concentrated strong acid (e.g. 10 mol/L HCl), pH = −log(10) = −1. Negative pH values are physically possible (just rare in everyday contexts).
"I can use V(acid) as the total volume when calculating pH after mixing." — No. V(total) = V(acid) + V(base). The excess species is diluted into the combined volume. Using only V(acid) overestimates [excess] and gives a pH that is too extreme.
GIVEN: (a) 0.025 mol/L HNO₃; (b) 0.0040 mol/L Ca(OH)₂; (c) 0.80 g NaOH in 500 mL.
FIND: pH of each solution.
METHOD (a — HNO₃): HNO₃ is a strong monoprotic acid → [H₃O⁺] = c(HNO₃) = 0.025 mol/L. pH = −log(0.025) = −log(2.5 × 10⁻²) = 2 − log(2.5) = 2 − 0.398 = 1.60. Sanity check: pH 1.60 < 7 ✓
METHOD (b — Ca(OH)₂): Ca(OH)₂ is a strong diprotic base → [OH⁻] = 2 × c(Ca(OH)₂) = 2 × 0.0040 = 0.0080 mol/L. pOH = −log(0.0080) = −log(8.0 × 10⁻³) = 3 − log(8.0) = 3 − 0.903 = 2.10. pH = 14.00 − 2.10 = 11.90. Sanity check: pH 11.90 > 7 ✓
METHOD (c — NaOH dissolved): n(NaOH) = mass/M = 0.80/40.0 = 0.020 mol. V = 500 mL = 0.500 L. c(NaOH) = n/V = 0.020/0.500 = 0.040 mol/L. NaOH is strong monoprotic → [OH⁻] = 0.040 mol/L. pOH = −log(0.040) = −log(4.0 × 10⁻²) = 2 − log(4.0) = 2 − 0.602 = 1.40. pH = 14.00 − 1.40 = 12.60. Sanity check: pH 12.60 > 7 ✓
ANSWER: (a) pH = 1.60. (b) pH = 11.90. (c) pH = 12.60.
GIVEN: (a) 25.0 mL of 0.200 mol/L HCl diluted to 500 mL; (b) pH = 3.40; (c) further dilution to 5000 mL.
METHOD (a — dilution): n(HCl) = c × V = 0.200 × 0.0250 = 5.00 × 10⁻³ mol (moles conserved). V(new) = 500 mL = 0.500 L. c(new) = n/V = (5.00 × 10⁻³)/0.500 = 0.0100 mol/L. [H₃O⁺] = 0.0100 mol/L. pH = −log(0.0100) = 2.00.
METHOD (b — [H₃O⁺] and [OH⁻] from pH): [H₃O⁺] = 10⁻ᵖᴴ = 10⁻³·⁴⁰. Calculate: 10⁻³·⁴⁰ = 10⁻⁴ × 10⁰·⁶⁰ = 10⁻⁴ × 3.981 = 3.98 × 10⁻⁴ mol/L. [OH⁻] = Kw/[H₃O⁺] = (1.0 × 10⁻¹⁴)/(3.98 × 10⁻⁴) = 2.51 × 10⁻¹¹ mol/L. Check: pH + pOH = 3.40 + 10.60 = 14.00 ✓
METHOD (c — further dilution to 5000 mL): Same n = 5.00 × 10⁻³ mol HCl in 5000 mL = 5.000 L. c(new) = (5.00 × 10⁻³)/5.000 = 1.00 × 10⁻³ mol/L. pH = −log(1.00 × 10⁻³) = 3.00. pH cannot rise indefinitely — water's autoionisation sets a minimum [H₃O⁺] of 1.0 × 10⁻⁷ mol/L at 25°C. As [H⁺] from HCl approaches this value, pH asymptotically approaches 7 but can never reach or exceed 7 for an acid at any finite dilution.
ANSWER: (a) pH = 2.00. (b) [H₃O⁺] = 3.98 × 10⁻⁴ mol/L; [OH⁻] = 2.51 × 10⁻¹¹ mol/L. (c) pH = 3.00; pH cannot exceed 7 — water autoionisation sets a minimum [H₃O⁺] of 1.0 × 10⁻⁷ mol/L, so pH asymptotically approaches 7 upon extreme dilution.
GIVEN: 40.0 mL of 0.150 mol/L HCl + 60.0 mL of 0.0800 mol/L Ba(OH)₂; blood pH 7.40 → 7.10.
FIND: pH of mixture; error analysis; [H₃O⁺] ratio and clinical significance.
METHOD (a — moles): n(H⁺) from HCl = 0.150 × 0.0400 = 6.00 × 10⁻³ mol. n(OH⁻) from Ba(OH)₂ = 0.0800 × 0.0600 × 2 = 9.60 × 10⁻³ mol. OH⁻ is in excess: n(excess OH⁻) = 9.60 × 10⁻³ − 6.00 × 10⁻³ = 3.60 × 10⁻³ mol.
METHOD (a — concentration and pH): V(total) = 40.0 + 60.0 = 100.0 mL = 0.1000 L. c(OH⁻) = (3.60 × 10⁻³)/0.1000 = 0.0360 mol/L. pOH = −log(0.0360) = −log(3.60 × 10⁻²) = 2 − log(3.60) = 2 − 0.556 = 1.444. pH = 14.00 − 1.444 = 12.56. Sanity: excess OH⁻ → pH > 7 ✓
METHOD (b — error analysis): Second student uses V = 40.0 mL = 0.0400 L (acid volume only). c(OH⁻)(wrong) = (3.60 × 10⁻³)/0.0400 = 0.0900 mol/L. pOH(wrong) = −log(0.0900) = 1.046. pH(wrong) = 14.00 − 1.046 = 12.95. Error = 12.95 − 12.56 = 0.39 pH units. The volume used is 2.5× too small → [OH⁻] is 2.5× too large → pH is 0.39 units too high.
METHOD (c — clinical ratio): [H₃O⁺] at pH 7.10 = 10⁻⁷·¹⁰ = 7.94 × 10⁻⁸ mol/L. [H₃O⁺] at pH 7.40 = 10⁻⁷·⁴⁰ = 3.98 × 10⁻⁸ mol/L. Ratio = 7.94/3.98 = 2.0. A drop of 0.30 pH units corresponds to a doubling of [H₃O⁺] in blood. Clinically significant because: (1) enzyme active site residues are sensitive to [H₃O⁺] at 10⁻⁸ mol/L — a 100% increase alters their ionisation state; (2) haemoglobin's oxygen affinity decreases at lower pH (Bohr effect) — tissues receive less oxygen; (3) cardiac ion channel gating is pH-dependent — risk of arrhythmia increases.
ANSWER: (a) n(excess OH⁻) = 3.60 × 10⁻³ mol; V(total) = 0.100 L; [OH⁻] = 0.0360 mol/L; pOH = 1.44; pH = 12.56. (b) Error: used V = 0.040 L instead of 0.100 L → [OH⁻] overestimated 2.5×; pH(wrong) = 12.95; error = +0.39 pH units. (c) [H₃O⁺] doubles from pH 7.40 to 7.10 (ratio = 2.0); clinically significant because enzyme activity, haemoglobin oxygen affinity, and cardiac ion channels are all sensitive to [H₃O⁺] at this scale — a 100% increase in [H₃O⁺] is physiologically substantial despite appearing small on the pH scale.
These are the minimum formulas and procedures to have in your notes for Lesson 8.
Calculate the pH of each solution. Show all working including the sanity check.
| # | Solution | [H₃O⁺] or [OH⁻] calculation | pH | Sanity check |
|---|---|---|---|---|
| 1 | 0.050 mol/L HCl | |||
| 2 | 0.020 mol/L H₂SO₄ (dilute) | |||
| 3 | 0.0015 mol/L KOH | |||
| 4 | 0.025 mol/L Ba(OH)₂ | |||
| 5 | 3.65 g HCl (M = 36.5) in 2.00 L |
Each student response below contains at least one error. Identify each error precisely and provide the correct calculation.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
1. A student calculates the pH of 0.020 mol/L Ba(OH)₂. Which of the following is the correct pH?
2. 30.0 mL of 0.100 mol/L NaOH is mixed with 20.0 mL of 0.100 mol/L HCl. Which of the following correctly identifies the excess species and calculates the pH?
3. A solution of HCl is diluted from 0.50 mol/L to 5.0 × 10⁻⁴ mol/L by successive additions of water. Which statement correctly describes the effect on pH and acid strength?
4. A solution has [OH⁻] = 4.0 × 10⁻⁵ mol/L at 25°C. What is the pH of this solution?
5. A student mixes 25.0 mL of 0.200 mol/L H₂SO₄ with 80.0 mL of 0.200 mol/L NaOH. Which of the following correctly identifies the excess and gives the pH?
Question 6. Calculate the pH of each of the following solutions at 25°C. Show all working including the sanity check.
(a) 0.0050 mol/L H₂SO₄ (dilute) (b) 0.0020 mol/L Ca(OH)₂ (c) a solution prepared by dissolving 1.12 g of KOH (M = 56.0 g/mol) in enough water to make 250 mL of solution
Question 7. 50.0 mL of 0.150 mol/L HNO₃ is mixed with 30.0 mL of 0.200 mol/L KOH.
(a) Calculate the moles of H⁺ and OH⁻ present. (b) Identify the excess species and calculate its moles. (c) Calculate the pH of the resulting mixture at 25°C. (d) What volume of 0.200 mol/L KOH would be required to exactly neutralise 50.0 mL of 0.150 mol/L HNO₃?
Question 8. A laboratory technician prepares a solution by mixing 100 mL of 0.250 mol/L HCl with 150 mL of 0.100 mol/L Ba(OH)₂.
(a) Calculate the pH of the resulting mixture. (3 marks) (b) A colleague argues: "Since both HCl and Ba(OH)₂ are strong, and they neutralise each other, the pH must always be 7 when you mix them." Evaluate this claim. (2 marks) (c) Calculate the volume of 0.250 mol/L HCl that would need to be added to the mixture calculated in (a) to reduce the pH to exactly 7.00. Show all steps. (2 marks)
Return to your Think First response about the hyperventilating patient. You can now answer this precisely:
Q1: B — pH = 12.60
Ba(OH)₂ is a strong diprotic base → [OH⁻] = 2 × 0.020 = 0.040 mol/L. pOH = −log(0.040) = 1.40. pH = 14.00 − 1.40 = 12.60. Option A (12.30) is the error of forgetting to multiply [OH⁻] by 2 for a diprotic base — this uses [OH⁻] = 0.020 mol/L (pOH = 1.70; pH = 12.30). Option C forgets to subtract pOH from 14. Option D uses an incorrect concentration.
Q2: B
n(NaOH) = 0.100 × 0.0300 = 3.00 × 10⁻³ mol OH⁻. n(HCl) = 0.100 × 0.0200 = 2.00 × 10⁻³ mol H⁺. Excess OH⁻ = 1.00 × 10⁻³ mol. V(total) = 50.0 mL = 0.0500 L. c(OH⁻) = (1.00 × 10⁻³)/0.0500 = 0.020 mol/L. pOH = −log(0.020) = 1.70. pH = 14.00 − 1.70 = 12.30. Option A uses wrong excess species (HCl). Option C is wrong — n(NaOH) ≠ n(HCl). Option D uses V = 0.030 L (base volume only — wrong).
Q3: B
pH(initial) = −log(0.50) = 0.30. pH(final) = −log(5.0 × 10⁻⁴) = 3.30. HCl remains strong — dilution changes c, not Ka. Option A correctly states the pH but incorrectly claims HCl becomes weak. Option C has wrong pH direction. Option D is wrong — pH asymptotically approaches 7 but never reaches it.
Q4: C
pOH = −log(4.0 × 10⁻⁵) = −log(4.0) + 5 = 5 − 0.602 = 4.40. pH = 14.00 − 4.40 = 9.60. The solution is basic (pH > 7), consistent with [OH⁻] > 1.0 × 10⁻⁷. Option B states the correct pH but describes an incorrect method (you cannot take −log[OH⁻] to get pH directly). Options A and D are incorrect values.
Q5: D
n(H⁺) from H₂SO₄ = 2 × 0.200 × 0.0250 = 0.0100 mol (×2 for diprotic). n(OH⁻) from NaOH = 0.200 × 0.0800 = 0.0160 mol. Excess OH⁻ = 0.0160 − 0.0100 = 0.0060 mol. V(total) = 25.0 + 80.0 = 105.0 mL = 0.105 L. c(OH⁻) = 0.0060/0.105 = 0.0571 mol/L. pOH = −log(0.0571) = 1.243. pH = 14 − 1.243 = 12.76. Options A and B are wrong (acid is in excess in A — incorrect; B ignores that H₂SO₄ is diprotic). Option C uses wrong V(total).
(a) 0.0050 mol/L H₂SO₄ (dilute):
[H₃O⁺] = 2 × c(H₂SO₄) = 2 × 0.0050 = 0.0100 mol/L (×2 for diprotic strong acid)
pH = −log(0.0100) = 2.00. Sanity: pH < 7 ✓ (acid)
(b) 0.0020 mol/L Ca(OH)₂:
[OH⁻] = 2 × c(Ca(OH)₂) = 2 × 0.0020 = 0.0040 mol/L
pOH = −log(0.0040) = −log(4.0 × 10⁻³) = 3 − log(4.0) = 3 − 0.602 = 2.40
pH = 14.00 − 2.40 = 11.60. Sanity: pH > 7 ✓ (base)
(c) 1.12 g of KOH in 250 mL:
n(KOH) = 1.12/56.0 = 0.0200 mol
V = 250 mL = 0.250 L; c(KOH) = 0.0200/0.250 = 0.0800 mol/L
[OH⁻] = c(KOH) = 0.0800 mol/L (KOH is strong monoprotic base)
pOH = −log(0.0800) = −log(8.0 × 10⁻²) = 2 − log(8.0) = 2 − 0.903 = 1.10
pH = 14.00 − 1.10 = 12.90. Sanity: pH > 7 ✓ (base)
(a) Moles:
n(H⁺) = c(HNO₃) × V = 0.150 × 0.0500 = 7.50 × 10⁻³ mol
n(OH⁻) = c(KOH) × V = 0.200 × 0.0300 = 6.00 × 10⁻³ mol
(b) Excess species:
HNO₃ is in excess: n(excess H⁺) = 7.50 × 10⁻³ − 6.00 × 10⁻³ = 1.50 × 10⁻³ mol
(c) pH of mixture:
V(total) = 50.0 + 30.0 = 80.0 mL = 0.0800 L
c(H⁺) = (1.50 × 10⁻³)/0.0800 = 0.01875 mol/L
pH = −log(0.01875) = −log(1.875 × 10⁻²) = 2 − log(1.875) = 2 − 0.273 = 1.73
Sanity: pH < 7 ✓ (excess acid)
(d) Volume to neutralise:
n(H⁺) = 0.150 × 0.0500 = 7.50 × 10⁻³ mol. At equivalence: n(KOH) = n(H⁺) = 7.50 × 10⁻³ mol
V(KOH) = n/c = (7.50 × 10⁻³)/0.200 = 0.0375 L = 37.5 mL
(a) pH of mixture:
n(H⁺) from HCl = 0.250 × 0.100 = 0.0250 mol
n(OH⁻) from Ba(OH)₂ = 2 × 0.100 × 0.150 = 0.0300 mol (×2 for diprotic base)
Excess OH⁻: n = 0.0300 − 0.0250 = 0.0050 mol
V(total) = 100 + 150 = 250 mL = 0.250 L
c(OH⁻) = 0.0050/0.250 = 0.020 mol/L
pOH = −log(0.020) = 1.70; pH = 14.00 − 1.70 = 12.30
(b) Evaluate the claim:
The claim is incorrect. pH = 7.00 occurs only at the equivalence point — when n(H⁺) = n(OH⁻) exactly (equal moles, not just equal concentrations). In this case, the moles are not equal (n(H⁺) = 0.025 mol; n(OH⁻) = 0.030 mol), so there is an excess of OH⁻ and the solution is basic (pH 12.30). Mixing a strong acid and strong base only gives pH 7 when moles of H⁺ and OH⁻ are exactly equal. Concentration and volume must both be considered.
(c) Volume of HCl to reach pH 7.00:
At pH 7.00, the mixture is neutral — all OH⁻ must be neutralised. From (a), the mixture contains excess n(OH⁻) = 0.0050 mol in 250 mL.
To reach neutrality, must add n(H⁺) = 0.0050 mol HCl (to react with all excess OH⁻).
V(HCl) = n/c = 0.0050/0.250 = 0.0200 L = 20.0 mL
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