In 1966, Danish physiologist Poul Astrup published data showing that mountaineers at 7,500 m have blood pH readings of up to 7.62 during a summit push — respiratory alkalosis caused by hyperventilation driving CO₂ out of the blood. A shift of just 0.17 pH units from normal (7.45) corresponds to a 32% drop in [H₃O⁺] — enough to cause tetany, loss of consciousness, and cardiac arrhythmia. Every calculation in this lesson is the mathematics that explains that 0.17.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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 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?
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.
pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration: pH = −log₁₀[H₃O⁺]. The negative sign converts small decimals ([H₃O⁺] is always between ~10⁻¹⁴ and ~10⁰ mol/L) 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%. 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× |
pH = −log₁₀[H₃O⁺]; [H₃O⁺] = 10⁻ᵖᴴ. Each 1 pH unit = 10× change in [H₃O⁺] — the scale is logarithmic because [H₃O⁺] spans 14 orders of magnitude in aqueous solutions. Lower pH = more acidic = higher [H₃O⁺]. A pH difference of 2 = 100× difference in [H₃O⁺].
Pause — copy the highlighted definition into your book before moving on.
How much more [H₃O⁺] does a pH 2 solution have compared to a pH 5 solution?
We just saw the logarithmic pH scale — each unit = 10× change in [H₃O⁺]. That raises a question: How do you actually calculate pH for the strongest, simplest case — a strong acid? This card answers it → [H₃O⁺] = c(acid) for monoprotic; [H₃O⁺] = 2 × c for dilute H₂SO₄; pH = −log[H₃O⁺].
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 acid pH: [H₃O⁺] = c(acid) for monoprotic (HCl, HNO₃, HBr, HI, HClO₄); [H₃O⁺] = 2 × c for dilute H₂SO₄; pH = −log[H₃O⁺]. Never use Ka for a strong acid — it is not an equilibrium problem. Sanity check: pH must be < 7 for an acid at 25°C.
Add the highlighted point to your notes before the check below.
What is the pH of 0.010 mol/L H₂SO₄ (dilute) at 25°C?
We just saw the direct method for strong acids — [H₃O⁺] = c, then pH = −log. That raises a question: Bases give [OH⁻] first, not [H₃O⁺] — how do you get from [OH⁻] to pH? This card answers it → use the pOH route: [OH⁻] → pOH = −log[OH⁻] → pH = 14 − pOH at 25°C; Ca(OH)₂ and Ba(OH)₂ require ×2.
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).
Strong base pH — 4-step route: (1) [OH⁻] = c × n(OH⁻); NaOH/KOH: n=1; Ca(OH)₂/Ba(OH)₂: n=2 (multiply by 2). (2) pOH = −log[OH⁻]. (3) pH = 14 − pOH at 25°C only. (4) Sanity check: pH must be > 7. If pH < 7 after this route, you wrote pH = pOH by mistake.
Pause — write the highlighted definition into your book.
True or false: 0.050 mol/L Ca(OH)₂ gives [OH⁻] = 0.050 mol/L.
We just saw the direct calculation routes for strong acids (direct) and strong bases (pOH route). That raises a question: What happens to pH when you dilute a strong acid or base — and is there a limit? This card answers it → use c₁V₁ = c₂V₂ then recalculate pH; diluting acid increases pH but never past 7; dilution changes concentration, never acid strength.
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. 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.
Dilution method: c₁V₁ = c₂V₂ (moles conserved); find new concentration, then recalculate pH. 10× dilution of monoprotic strong acid → pH rises by exactly 1 unit. Physical limit: acid pH cannot exceed 7 (water's autoionisation provides [H₃O⁺] floor at 10⁻⁷ mol/L). Dilution never changes Ka or acid strength.
Add the highlighted point to your notes before the check below.
25.0 mL of 0.200 mol/L HCl is diluted to 500 mL. What is the new [H₃O⁺]?
We just saw that dilution uses c₁V₁ = c₂V₂ and never changes acid strength. That raises a question: What if you mix a strong acid and strong base together — who wins, and how do you find the final pH? This card answers it → calculate moles of H⁺ and OH⁻, find the excess, divide by total volume, then apply the acid or base pH route.
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.
Mixing method — 5 steps: (1) n(H⁺) = c × V(acid); (2) n(OH⁻) = c × V(base) × n(OH⁻); (3) excess = |n(H⁺) − n(OH⁻)|; (4) c(excess) = n(excess) / V(total) where V(total) = V(acid) + V(base); (5) if excess H⁺: pH = −log[H₃O⁺]; if excess OH⁻: pH = 14 − pOH; if equal: pH = 7.00 (justify with spectator ion hydrolysis argument).
Pause — copy the highlighted definition into your book before moving on.
30.0 mL of 0.100 mol/L NaOH is mixed with 20.0 mL of 0.100 mol/L HCl. What is the excess species and total volume?
"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." — 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.
Calculate the pH of each of the following solutions at 25°C: (a) 0.025 mol/L HNO₃; (b) 0.0040 mol/L Ca(OH)₂; (c) a solution made by dissolving 0.80 g of NaOH (M = 40.0 g/mol) in enough water to make 500 mL of solution.
(a) 25.0 mL of 0.200 mol/L HCl is diluted to a total volume of 500 mL. Calculate the pH of the diluted solution. (b) A solution has pH 3.40 at 25°C. Calculate [H₃O⁺] and [OH⁻]. (c) The student adds more water until the total volume is 5000 mL. Calculate the new pH and explain whether pH could continue to rise indefinitely with further dilution.
40.0 mL of 0.150 mol/L HCl is mixed with 60.0 mL of 0.0800 mol/L Ba(OH)₂. (a) Calculate the pH of the resulting mixture at 25°C. (b) A second student calculates pH using only V = 40.0 mL as the total volume. Identify the error and calculate the magnitude of the pH error. (c) Blood pH drops from 7.40 to 7.10 due to metabolic acidosis. Calculate the ratio of [H₃O⁺] at pH 7.10 to [H₃O⁺] at pH 7.40, and explain why this represents a clinically significant change.
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.
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Select the best answer for each question.
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 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 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) 1.12 g of KOH (M = 56.0 g/mol) dissolved in enough water to make 250 mL
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 in (a) to reduce the pH to exactly 7.00. Show all steps. (2 marks)
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) forgets to multiply [OH⁻] by 2. 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 = 1.70. pH = 12.30. 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 D is wrong — pH approaches 7 but never reaches it.
Q4: C
pOH = −log(4.0 × 10⁻⁵) = 4.40. pH = 14.00 − 4.40 = 9.60. The solution is basic. Option B states the correct pH but uses an incorrect method (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⁻ = 0.0060 mol. V(total) = 105.0 mL = 0.105 L. c(OH⁻) = 0.0060/0.105 = 0.0571 mol/L. pOH = 1.243. pH = 12.76. Option C uses wrong V(total) (0.080 L instead of 0.105 L).
(a) 0.0050 mol/L H₂SO₄ (dilute):
[H₃O⁺] = 2 × 0.0050 = 0.0100 mol/L (×2 for diprotic strong acid)
pH = −log(0.0100) = 2.00. Sanity: pH < 7 ✓
(b) 0.0020 mol/L Ca(OH)₂:
[OH⁻] = 2 × 0.0020 = 0.0040 mol/L
pOH = −log(0.0040) = −log(4.0 × 10⁻³) = 3 − 0.602 = 2.40
pH = 14.00 − 2.40 = 11.60. Sanity: pH > 7 ✓
(c) 1.12 g of KOH in 250 mL:
n(KOH) = 1.12/56.0 = 0.0200 mol. c(KOH) = 0.0200/0.250 = 0.0800 mol/L
[OH⁻] = 0.0800 mol/L. pOH = −log(0.0800) = 2 − 0.903 = 1.10
pH = 14.00 − 1.10 = 12.90. Sanity: pH > 7 ✓
(a) Moles:
n(H⁺) = 0.150 × 0.0500 = 7.50 × 10⁻³ mol
n(OH⁻) = 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) = 2 − log(1.875) = 2 − 0.273 = 1.73. Sanity: pH < 7 ✓
(d) Volume to neutralise:
n(H⁺) = 7.50 × 10⁻³ mol; V(KOH) = (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. In this case n(H⁺) = 0.025 mol ≠ n(OH⁻) = 0.030 mol — there is excess 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:
Mixture from (a) contains excess n(OH⁻) = 0.0050 mol in 250 mL.
To reach neutrality: must add n(H⁺) = 0.0050 mol HCl.
V(HCl) = n/c = 0.0050/0.250 = 0.0200 L = 20.0 mL
A student makes the following claim: "I added equal volumes of 0.10 mol/L HCl and 0.10 mol/L NaOH to get a neutral solution. Then I added equal volumes of 0.10 mol/L H₂SO₄ and 0.10 mol/L NaOH — and I also got a neutral solution, because the concentrations are still equal."
Evaluate this claim fully. Show calculations for both mixing scenarios and explain why the second claim is incorrect. Then calculate the actual pH of the H₂SO₄ + NaOH mixture and explain what volume of NaOH would be required to reach neutrality.
Return to your Think First response about the hyperventilating patient. Recall Astrup's 1966 data: blood pH 7.62 at 7,500 m — a 0.17 unit rise from normal (7.45) — corresponds to [H₃O⁺] dropping from 3.55 × 10⁻⁸ to 2.40 × 10⁻⁸ mol/L, a 32% decrease. You can now calculate this precisely:
What is [H₃O⁺] in 0.050 mol/L H₂SO₄ (dilute)?
[H₃O⁺] = 2 × 0.050 = 0.100 mol/L (×2 for diprotic)
Give the 4-step method for finding pH of a strong base.
[OH⁻] = c(base)×n(OH⁻) → pOH = −log[OH⁻] → pH = 14 − pOH → sanity check pH > 7
Why can pH of a strong acid never exceed 7 on dilution?
Water's autoionisation provides [H₃O⁺] = 1.0 × 10⁻⁷ mol/L (pH = 7 at 25°C). This sets an asymptotic floor — the acid's [H₃O⁺] approaches but never falls below this value.
What total volume is used when 30 mL + 50 mL are mixed?
V(total) = 30 + 50 = 80 mL. NEVER use just one of the volumes alone.
A blood pH change of 0.30 corresponds to what change in [H₃O⁺]?
10⁰·³ ≈ 2× — a doubling. A drop of 0.30 pH units doubles [H₃O⁺]. This is why small pH changes in blood are clinically significant.
Why does Ka of HCl not change when you dilute it?
Ka is a thermodynamic constant at fixed temperature — it reflects the intrinsic tendency of the acid to ionise. Dilution changes concentration, not the equilibrium constant. HCl is always 100% ionised regardless of concentration.
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