Year 12 Chemistry Module 6, Acid/Base Reactions ⏱ ~45 min Lesson 8 of 19 IQ2

pH and pOH, Calculations for Strong Acids & Bases

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.

Today's hook, In 1966, Poul Astrup measured blood pH of 7.62 in mountaineers hyperventilating at 7,500 m, only 0.17 pH units above normal, but a 32% drop in [H₃O⁺]. How does a "small" pH change correspond to such a large change in actual hydrogen ion concentration?

0/5QUESTS

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Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

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Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

The Hyperventilating Patient

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?

Learning Intentions

Know

pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻]
pH + pOH = 14.00 at 25°C only
Strong acids assumed 100% ionised; diprotic acids contribute 2 H⁺ per molecule

Understand

Why the pH scale is logarithmic (each unit = 10× change in [H₃O⁺])
Why dilution uses c₁V₁ = c₂V₂ before pH calculation
Why V(total) = V(acid) + V(base) is required when mixing solutions

Can Do

Calculate pH of strong monoprotic and diprotic acids
Calculate pH of strong bases including Group 2 hydroxides
Calculate pH after dilution or mixing of strong acids and bases

Key Terms

pH = −log₁₀[H⁺]; measures the acidity of a solution; decreases as [H⁺] increases.

pOH

pOH = −log₁₀[OH⁻]; measures alkalinity; at 25°C: pH + pOH = 14.

[H⁺] from strong acid

For a strong monoprotic acid: [H⁺] = concentration of acid (fully dissociates).

[OH⁻] from strong base

For NaOH: [OH⁻] = concentration of NaOH; for Ca(OH)₂: [OH⁻] = 2 × concentration.

Kw expression

[H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C; used to find [OH⁻] from [H⁺] or vice versa.

Anti-log

[H⁺] = 10⁻ᵖᴴ; used to recover concentration from a pH value.

pH, pOH, Strong Acid/Base, Dilution & Mixing

pH = −log₁₀[H₃O⁺] | [H₃O⁺] = 10⁻ᵖᴴ pOH = −log₁₀[OH⁻] | [OH⁻] = 10⁻ᵖᴼᴴ

Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C | pH + pOH = 14 at 25°C only [OH⁻] = Kw / [H₃O⁺] | pOH = 14 − pH

Strong acid: [H₃O⁺] = c(acid) × n(H⁺ per formula unit) HCl, HNO₃, HBr, HI, HClO₄: n = 1 | H₂SO₄ (dilute): n = 2 → [H₃O⁺] = 2 × c(H₂SO₄)

Strong base: [OH⁻] = c(base) × n(OH⁻ per formula unit) NaOH, KOH: n = 1 | Ca(OH)₂, Ba(OH)₂: n = 2 → [OH⁻] = 2 × c(base) | Then pOH = −log[OH⁻]; pH = 14 − pOH

Dilution: c₁V₁ = c₂V₂ (moles conserved) → find c₂ → calculate pH 10× dilution → [H₃O⁺] drops 10× → pH rises by 1 unit (for monoprotic strong acid only)

The pH scale and what it means

Core Content

Cross-lesson links: The pH and pOH calculations here (using Kw from L07) are the core numerical tool for all of L09–L17. Astrup's 1966 blood pH data reappears in the blood buffer system (L13). Dilution and mixing calculations in this lesson are foundational for titration stoichiometry (L14) and equivalence point pH (L15–L16).

The pH Scale: What It Means and Why It Is Logarithmic

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×

Must Do: When comparing two solutions, always translate the pH difference into a [H₃O⁺] ratio: "a difference of n pH units corresponds to a 10ⁿ-fold difference in [H₃O⁺]." "pH 3 is twice as acidic as pH 4" is wrong, pH 3 is ten times more acidic (10× higher [H₃O⁺]).

Common Error: Students confuse the direction of the scale: a higher pH means lower [H₃O⁺]. As [H₃O⁺] increases, pH decreases. Memorise: "lower pH = more acidic = more H₃O⁺." This must be automatic.

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?

Use the pH Calculator: a 0.010 mol/L HCl solution has pH 2, which means it is…

pH of strong acids

Calculating pH of Strong Acids: Direct Method

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.

Convention & caveat

Strictly, only the first ionisation of H₂SO₄ is complete (H₂SO₄ → H⁺ + HSO₄⁻); the second is a weak-acid equilibrium (HSO₄⁻ ⇌ H⁺ + SO₄²⁻, Ka₂ ≈ 1.2 × 10⁻²), so [H₃O⁺] is slightly less than 2 × c. At HSC level, for dilute H₂SO₄ we approximate both protons as fully donated and use [H₃O⁺] = 2 × c(H₂SO₄); just be ready to state that this is an approximation because HSO₄⁻ is itself a weak acid. The factor of 2 applies only to H₂SO₄, never to monoprotic acids such as HCl or HNO₃.

5-Step Method, Strong Acid pH

Identify the acid and confirm it is strong (on the memorised list)
Determine n(H⁺) donated per formula unit (1 for monoprotic; 2 for H₂SO₄ dilute)
Calculate [H₃O⁺] = c(acid) × n(H⁺)
Calculate pH = −log₁₀[H₃O⁺]
Sanity check: for an acid, pH must be < 7 at 25°C

HCl, 0.050 mol/L

[H₃O⁺] = 0.050
pH = −log(0.050) = 1.30

HNO₃, 0.0025 mol/L

[H₃O⁺] = 0.0025
pH = −log(0.0025) = 2.60

H₂SO₄ (dilute), 0.010 mol/L

[H₃O⁺] = 2 × 0.010 = 0.020
pH = −log(0.020) = 1.70

HClO₄, 2.5 × 10⁻⁴ mol/L

[H₃O⁺] = 2.5 × 10⁻⁴
pH = −log(2.5 × 10⁻⁴) = 3.60

Must Do: For H₂SO₄ in dilute solution, [H₃O⁺] = 2 × c(H₂SO₄) not just c(H₂SO₄). A 0.10 mol/L H₂SO₄ solution gives [H₃O⁺] = 0.20 mol/L and pH = 0.70, not pH = 1.0. Always check the number of ionisable protons before calculating [H₃O⁺].

Common Error: Students use [H₃O⁺] = Ka × c for strong acids. Ka is for weak acids only, it quantifies the fraction that ionises. For a strong acid, ionisation is 100%, [H₃O⁺] = c directly. Applying Ka to a strong acid is a category error.

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?

pH of strong bases, the pOH route

Calculating pH of Strong Bases: The pOH Route

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).

4-Step Method, Strong Base pH

Identify the base and confirm it is strong; determine n(OH⁻) per formula unit
Calculate [OH⁻] = c(base) × n(OH⁻)
Calculate pOH = −log₁₀[OH⁻]
Calculate pH = 14.00 − pOH | Sanity check: pH must be > 7 for a base

NaOH, 0.050 mol/L

[OH⁻] = 0.050
pOH = 1.30
pH = 14 − 1.30 = 12.70

KOH, 0.0025 mol/L

[OH⁻] = 0.0025
pOH = 2.60
pH = 14 − 2.60 = 11.40

Ca(OH)₂, 0.010 mol/L

[OH⁻] = 2 × 0.010 = 0.020
pOH = 1.70
pH = 14 − 1.70 = 12.30

Ba(OH)₂, 2.5 × 10⁻⁴ mol/L

[OH⁻] = 2 × 2.5 × 10⁻⁴ = 5.0 × 10⁻⁴
pOH = 3.30
pH = 14 − 3.30 = 10.70

Must Do: pH + pOH = 14 applies ONLY at 25°C. If a question specifies a different temperature and gives a different Kw, recalculate pKw = −log(Kw) and use pH + pOH = pKw at that temperature. Using 14 at non-standard temperature is a systematic error.

Common Error: Students calculate pOH correctly and then write pH = pOH instead of pH = 14 − pOH. This produces a pH below 7 for a basic solution, caught immediately by the sanity check. "For a base, pH > 7." If your answer is < 7, check whether you forgot to subtract pOH from 14.

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.

Diluting strong acids and bases

Dilution of Strong Acids and Bases

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.

What changes on dilution

[H₃O⁺] drops ×10 → pH rises by 1 unit
[H₃O⁺] drops ×100 → pH rises by 2 units
[OH⁻] drops ×10 → pOH rises by 1 unit → pH falls by 1 unit

What stays the same

n(H⁺) = c₁V₁ = c₂V₂ (moles conserved)
Ka of the acid (dilution ≠ strength change)
Identity of the acid (HCl is always strong)

Must Do: Always use c₁V₁ = c₂V₂ to find the new concentration before calculating pH. A common shortcut error is to add the dilution factor directly to pH. While this works for exactly 10× dilution of a monoprotic strong acid, it fails for any other dilution factor, diprotic acids, or base calculations.

Common Error: "Diluting an acid 10 times makes it a weak acid." Dilution changes concentration, not strength. A dilute HCl solution is still a strong acid (complete ionisation). The Ka of HCl does not change upon dilution. "Dilute strong acid" means low concentration, complete ionisation, never partial ionisation.

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⁺]?

Mixing a strong acid with a strong base

Mixing Strong Acid and Strong Base: Finding pH of the Mixture

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.

5-Step Method, Mixing Strong Acid + Strong Base

Calculate n(H⁺) = c(acid) × V(acid)
Calculate n(OH⁻) = c(base) × V(base) × n(OH⁻ per formula unit)
Determine which is in excess; n(excess) = |n(H⁺) − n(OH⁻)|
Calculate c(excess) = n(excess) / V(total), where V(total) = V(acid) + V(base)
If excess H⁺: pH = −log[H₃O⁺]. If excess OH⁻: pOH = −log[OH⁻] → pH = 14 − pOH. If exactly equal moles: pH = 7.00

More H⁺

n(H⁺) = 0.0050 mol
n(OH⁻) = 0.0030 mol
Excess: 0.0020 mol H⁺
V(total) = V(acid) + V(base)
pH = −log[H₃O⁺]

More OH⁻

n(H⁺) = 0.0030 mol
n(OH⁻) = 0.0050 mol
Excess: 0.0020 mol OH⁻
V(total) = V(acid) + V(base)
pH = 14 − (−log[OH⁻])

Equal moles

n(H⁺) = n(OH⁻) = 0.0050 mol
Excess: None
V(total) = V(acid) + V(base)
pH = 7.00

Must Do: The total volume after mixing is V(total) = V(acid) + V(base). This must be used when calculating c(excess), not V(acid) alone or V(base) alone. Using only one volume is the most common arithmetic error in mixing calculations.

Common Error: At the equivalence point, students write pH = 7 without justification. The correct reasoning: at equivalence, n(H⁺) = n(OH⁻), both completely consumed. The solution contains only water and spectator ions (e.g. Na⁺ and Cl⁻). Spectator ions are conjugates of strong acids/bases and do not hydrolyse → [H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol/L → pH = 7.00. This reasoning is required for full marks in extended response.

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?

Worked examples and misconceptions

Common Misconceptions

pH Calculations

✗ "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."

✓ 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.

Strong Acid and Strong Base pH Calculations

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.

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) = 0.020/0.500 = 0.040 mol/L. NaOH is strong monoprotic → [OH⁻] = 0.040 mol/L. pOH = −log(0.040) = 2 − log(4.0) = 2 − 0.602 = 1.40. pH = 14.00 − 1.40 = 12.60. Sanity: pH > 7 ✓

ANSWER: (a) pH = 1.60. (b) pH = 11.90. (c) pH = 12.60.

Dilution and Finding [H₃O⁺] from pH

(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.

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): n(HCl) = c × V = 0.200 × 0.0250 = 5.00 × 10⁻³ mol (moles conserved). V(new) = 500 mL = 0.500 L. c(new) = (5.00 × 10⁻³)/0.500 = 0.0100 mol/L. [H₃O⁺] = 0.0100 mol/L. pH = −log(0.0100) = 2.00.

METHOD (b): [H₃O⁺] = 10⁻ᵖᴴ = 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): 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.

Multi-Step Mixing with Clinical Context

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.

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, 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) = 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): 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) 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.

Activities

Activity

Activity A, Calculate & Interpret: Strong Acid and Base pH

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

Activity B, Spot & Fix: Common Calculation Errors

Each student response below contains at least one error. Identify each error precisely and provide the correct calculation.

Student A (0.050 mol/L Ba(OH)₂): "[OH⁻] = 0.050 mol/L; pOH = 1.30; pH = 1.30." Error: ___
Student B (50.0 mL of 0.10 mol/L NaOH + 30.0 mL of 0.10 mol/L HCl): "Excess NaOH = 0.002 mol; c(NaOH) = 0.002/0.050 = 0.040 mol/L; pOH = 1.40; pH = 12.60." Error: ___
Student C (0.10 mol/L H₂SO₄ diluted 100×): "New concentration = 0.001 mol/L; [H₃O⁺] = 0.001 mol/L; pH = 3.00." Error: ___
Student D (HCl diluted until very dilute): "If I add enough water, the HCl becomes so dilute it is essentially a weak acid with pH above 7." Error: ___

🔀 Sort the Steps +7 XP

Arrange these steps for calculating the pH of a strong base solution (e.g. 0.025 mol/L NaOH) in the correct order:

Calculate pOH = −log[OH⁻]

Calculate pH = 14 − pOH

State the final answer with correct sig figs

Write the full ionisation equation for the base

Identify [OH⁻] from the stoichiometry

Interactive · pH Stepper

Use the stepper. What is the pH of 0.010 mol/L NaOH at 25 °C?

Complete the Learn phase to unlock Practice.

✓

Practice questions

Check Your Understanding

Multiple Choice

+5 XP

A fresh set drawn from this lesson's question bank, with feedback shown immediately. +5 XP per correct · +25 XP all correct

Short Answer

Apply Band 3–4 4 marks

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

Apply Band 4 4 marks

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?

Evaluate Band 6 7 marks

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)

Calculation practice

Written calculation questions in the style and mark range of the HSC paper. Set your working out in full, because method marks are awarded for the steps as well as the final value.

ApplyBand 3(2 marks) Calculate the pH of a 0.0250 mol/L solution of hydrochloric acid at 25 °C.

ApplyBand 4(4 marks) A solution of barium hydroxide has a concentration of 5.00 × 10⁻³ mol/L at 25 °C. Calculate [OH⁻], the pOH and the pH of this solution, showing how the formula of the base affects your first step.

Answers

Q6 Sample Answer

(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 ✓

Q7 Sample Answer

(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

Q8 Sample Answer (Band 6)

(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

✓

Retrieve and reflect

Check what actually stuck

Band 6 Synthesis Challenge

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 Initial Thinking

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 does a 0.1 pH unit change mean? A change of 0.11 pH units (from 7.45 to 7.56) corresponds to a [H₃O⁺] change by a factor of 10⁰·¹¹ ≈ 1.3, a 23% decrease in [H₃O⁺]. This is calculated as 10⁻⁷·⁴⁵/10⁻⁷·⁵⁶ = 10⁰·¹¹ ≈ 1.29.
Why is this clinically significant? Blood [H₃O⁺] in the normal range is ~4 × 10⁻⁸ mol/L. Enzymes, ion channels, and haemoglobin function within a narrow [H₃O⁺] window. A 23% decrease in [H₃O⁺] alters the protonation state of histidine residues in enzyme active sites, shifts the oxygen-haemoglobin dissociation curve, and reduces calcium availability to muscles and nerves, causing the dizziness, tingling, and muscle cramps.
The logarithmic insight: The nurse was right to be surprised, pH values appear close on the scale, but the logarithmic nature masks real physiological [H₃O⁺] differences. This is the core calculation insight of Lesson 8.

Quick Recall Drill

Rapid-Fire Questions

What is [H₃O⁺] in 0.050 mol/L H₂SO₄ (dilute)?

Answer

[H₃O⁺] = 2 × 0.050 = 0.100 mol/L (×2 for diprotic)

Give the 4-step method for finding pH of a strong base.

Answer

[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?

Answer

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?

Answer

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⁺]?

Answer

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?

Answer

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.

Lesson 8 Complete, I can calculate pH and pOH for strong acids and bases (including diprotic species), perform dilution calculations using c₁V₁ = c₂V₂, and solve mixing problems using the five-step moles method with V(total).

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