IB Chemistry · Unit 10 · Reactivity

Protons,
in transit.

Sour, bitter, conductive, corrosive. Proton transfer reactions explain pH, buffers, indicators and titrations.

Lessons HL extension Key terms
11Lessons
4HL extensions
42Key terms
SL+HLLevel
1 2 3 4 5 6 7 8 9 10 pH scale ACID BASE
Unit 10 · Standard Level

7 lessons to work through.

The required syllabus content for Unit 10, in order. Each card is one lesson-sized checkpoint.

Lesson 1-2

Acid-Base theories

4. Which of the following is an example of an alkali? a. Hydrochloric acid b. Sodium hydroxide c. Water d. Ammonium chloride

Lesson 3

The pH Scale

The pH scale was introduced to simplify big numbers since the concentration of H+ ions in most substances is very low (e.g. The H+ concentration in blood is 4.6 x 10-8 mol/dm3

Lesson 4

The Ionic Product of Water

If the concentration of either H+ or OH- is known, the other can be calculated from the value of Kw.

Lesson 5

Reaction of acids

Lesson 5 of Unit 10.

Lesson 6

Strong and weak acids

Compared with strong acids of the same concentration, weak acids:

Lesson 7

Titrations

The two reactants are slowly combined until they have reached their exact stoichiometric proportions – they have then reached their equivalence point.

Lesson 13

Indicators

Disadvantage: Subjectivity, as the user’s eyes need to interpret the colour.

Lessons in detail

The unit, lesson by lesson.

Each lesson card below mirrors the original teacher deck — syllabus refs, content, worked examples and practice questions in order.

Lesson 1-2

Acid-base theories

Reactivity 3.1.1

Lesson outcomes

Arrhenius (1884)

Acid: produces H⁺ in aqueous solution. Base: produces OH⁻. Limited to water as the solvent.

Brønsted-Lowry (1923)

Acid = proton donor. Base = proton acceptor. Works in any solvent and with non-aqueous reactions. Gives the powerful concept of conjugate pairs:

HCl + H₂O → Cl⁻ + H₃O⁺
acid   base   conj. base   conj. acid

The two species of a conjugate pair differ by exactly one H⁺.

Amphoteric / amphiprotic

A species that can act as either an acid or a base depending on the partner. Examples: H₂O (acts as base toward HCl; acts as acid toward NH₃), HCO₃⁻, amino acids (the –COOH and –NH₂ together).

Worked example

Identifying conjugate pairs

Problem. For NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, identify the acid, base, conjugate acid and conjugate base.
Solution. NH₃ gains H⁺ → NH₄⁺: NH₃ is the base, NH₄⁺ is its conjugate acid.
H₂O loses H⁺ → OH⁻: H₂O is the acid, OH⁻ is its conjugate base.
Pairs: (NH₃ / NH₄⁺) and (H₂O / OH⁻).

Try these

  1. State the Arrhenius and Brønsted-Lowry definitions of acids and bases.
    Show answer
    Arrhenius: acid produces H⁺ in water; base produces OH⁻. Brønsted-Lowry: acid is a proton (H⁺) donor; base is a proton acceptor. B-L works in any solvent, not just water.
  2. Why is the Brønsted-Lowry definition more useful than Arrhenius?
    Show answer
    Works in any solvent (not just water). Captures reactions like NH₃(g) + HCl(g) → NH₄Cl(s) where Arrhenius doesn't apply. Introduces the powerful concept of conjugate pairs.
  3. For each acid below, write its conjugate base: (a) HCl, (b) H₂SO₄, (c) NH₄⁺, (d) HCO₃⁻.
    Show answer
    (a) Cl⁻. (b) HSO₄⁻ (then SO₄²⁻ on losing the second H). (c) NH₃. (d) CO₃²⁻.
  4. For each base below, write its conjugate acid: (a) OH⁻, (b) NH₃, (c) HCO₃⁻.
    Show answer
    (a) H₂O. (b) NH₄⁺. (c) H₂CO₃.
  5. Identify the acid, base, conjugate acid, and conjugate base in: HF(aq) + H₂O(l) ⇌ F⁻(aq) + H₃O⁺(aq).
    Show answer
    HF = acid (loses H⁺); H₂O = base (gains H⁺). F⁻ = conjugate base of HF; H₃O⁺ = conjugate acid of H₂O.
  6. Show that HCO₃⁻ is amphoteric by writing two equations — one where it acts as an acid and one where it acts as a base.
    Show answer
    As acid: HCO₃⁻ + H₂O ⇌ CO₃²⁻ + H₃O⁺. As base: HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻.
Lesson 3

The pH scale

Reactivity 3.1.2

Lesson outcomes

The pH scale compresses a huge range of [H⁺] into manageable numbers:

pH = −log₁₀[H⁺]    [H⁺] = 10⁻ᵖᴴ

Each unit corresponds to a 10× change in [H⁺]. pH 1 = 0.1 M H⁺; pH 7 = 10⁻⁷ M H⁺ (neutral water at 25 °C); pH 14 = 10⁻¹⁴ M H⁺.

Worked example

pH from [H⁺]

Problem. Calculate the pH of (a) 0.025 M HCl, (b) 3.2 × 10⁻⁴ M H₃O⁺.
Solution. (a) pH = −log(0.025) = 1.60. (b) pH = −log(3.2 × 10⁻⁴) = 3.49.
Worked example

[H⁺] from pH

Problem. Find [H⁺] for blood at pH 7.40.
Solution. [H⁺] = 10⁻⁷·⁴⁰ = 4.0 × 10⁻⁸ mol dm⁻³.

Try these

  1. Define pH.
    Show answer
    pH = −log₁₀([H⁺]/mol dm⁻³). Equivalently, [H⁺] = 10⁻ᵖᴴ. Logarithmic scale: each pH unit represents a factor of 10 in [H⁺].
  2. Calculate the pH of: (a) 0.10 mol dm⁻³ HCl, (b) 1.0 × 10⁻⁴ mol dm⁻³ HCl.
    Show answer
    (a) pH = −log(0.10) = 1.00. (b) pH = −log(1.0 × 10⁻⁴) = 4.00.
  3. Calculate [H⁺] for solutions of pH (a) 3.00, (b) 8.50.
    Show answer
    (a) [H⁺] = 10⁻³ = 1.0 × 10⁻³ mol dm⁻³. (b) [H⁺] = 10⁻⁸·⁵⁰ = 3.16 × 10⁻⁹ mol dm⁻³.
  4. If pH drops from 5 to 2, by what factor has [H⁺] increased?
    Show answer
    Each pH unit is a factor of 10. Three units → factor of 1000 increase.
  5. Why does the pH scale typically span 0–14 even though it could be extended further?
    Show answer
    pH 0 = 1 M H⁺ (concentrated strong acid). pH 14 = 10⁻¹⁴ M H⁺ (so [OH⁻] = 1 M, concentrated strong base). At room T, this covers the range from concentrated acid to concentrated base. Specialty applications can go negative (super-concentrated HCl) or above 14 (concentrated NaOH).
Lesson 4

The ionic product of water (K_w)

Reactivity 3.1.2

Lesson outcomes

Pure water self-ionises slightly:

2 H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

The equilibrium constant (excluding water) is the ionic product:

Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 25 °C

Take −log of both sides: pH + pOH = 14 (at 25 °C only).

Temperature dependence

The autoionisation is endothermic (ΔH = +57 kJ mol⁻¹). Heating water shifts the equilibrium right → more H⁺ and OH⁻ → Kw increases → neutral pH drops below 7 (but the water is still neutral because [H⁺] = [OH⁻]).

Worked example

[OH⁻] from pH

Problem. Calculate [OH⁻] in a solution of pH 4.0 at 25 °C.
Solution. [H⁺] = 10⁻⁴. [OH⁻] = Kw / [H⁺] = 10⁻¹⁴ / 10⁻⁴ = 10⁻¹⁰ mol dm⁻³.

Try these

  1. Write the autoionisation equation for water and the Kw expression.
    Show answer
    2 H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq). Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 25 °C.
  2. Calculate [OH⁻] in 0.025 M HCl at 25 °C.
    Show answer
    Strong acid → [H⁺] = 0.025 M. [OH⁻] = Kw/[H⁺] = 10⁻¹⁴/0.025 = 4.0 × 10⁻¹³ mol dm⁻³. Tiny — most OH⁻ has been consumed by added H⁺.
  3. Calculate pH and pOH for a 0.005 M Ca(OH)₂ solution. (Strong base, dissociates completely.)
    Show answer
    [OH⁻] = 2 × 0.005 = 0.010 M (two OH⁻ per formula unit). pOH = −log(0.010) = 2.00. pH = 14 − 2.00 = 12.00.
  4. At 50 °C, Kw = 5.5 × 10⁻¹⁴. What is the pH of pure water at this temperature?
    Show answer
    [H⁺] = √Kw = √(5.5 × 10⁻¹⁴) = 2.35 × 10⁻⁷. pH = −log(2.35 × 10⁻⁷) = 6.63. Pure water is still neutral; the value '7' is only true at 25 °C.
  5. Why does Kw increase with temperature? What does this say about the autoionisation of water?
    Show answer
    Heating water shifts the autoionisation equilibrium right (more H⁺ and OH⁻ form). By Le Chatelier, the forward reaction must be endothermic — energy is absorbed when water ionises. ΔH(autoion) ≈ +57 kJ mol⁻¹.
Lesson 5

Reactions of acids

Reactivity 3.1.3

Lesson outcomes

Four classic acid reactions

ReactionProductsTest
Acid + reactive metalSalt + H₂(g)"Squeaky pop" test for H₂
Acid + metal oxideSalt + H₂OSolid dissolves
Acid + metal carbonateSalt + H₂O + CO₂(g)CO₂ turns limewater milky
Acid + base (neutralisation)Salt + H₂OpH approaches 7 (strong+strong)

Net ionic equation for neutralisation

Stripping out spectators: H⁺(aq) + OH⁻(aq) → H₂O(l). This is the universal neutralisation chemistry.

Try these

  1. Write the general word equations for acid + base, acid + carbonate, and acid + reactive metal.
    Show answer
    Acid + base → salt + water (neutralisation). Acid + carbonate → salt + water + CO₂. Acid + metal (above H in reactivity series) → salt + H₂.
  2. Write the balanced equation for: (a) 2 HCl + CaCO₃, (b) Mg + 2 HCl, (c) CuO + 2 HCl.
    Show answer
    (a) 2 HCl(aq) + CaCO₃(s) → CaCl₂(aq) + H₂O(l) + CO₂(g). (b) Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g). (c) CuO(s) + 2 HCl(aq) → CuCl₂(aq) + H₂O(l).
  3. Write the net ionic equation for neutralisation of any strong acid by any strong base.
    Show answer
    H⁺(aq) + OH⁻(aq) → H₂O(l). All the spectator ions (e.g. Na⁺, Cl⁻) are unchanged.
  4. What gas is produced when an acid reacts with a carbonate? How would you test for it?
    Show answer
    CO₂. Test: bubble through limewater (saturated Ca(OH)₂(aq)). If CO₂ is present, the limewater turns milky/cloudy (forms CaCO₃ precipitate).
  5. How would you test for the gas produced when an acid reacts with a reactive metal?
    Show answer
    H₂. 'Squeaky pop' test: a lit splint at the mouth of a test tube of H₂ produces a sharp 'pop' as the H₂ ignites with the air.
Lesson 6

Strong vs weak acids

Reactivity 3.1.4

Lesson outcomes

A strong acid is one that is essentially fully dissociated in water: HA + H₂O → A⁻ + H₃O⁺ (one-way arrow). A weak acid partly dissociates and sets up an equilibrium.

Common strong species

Common weak species

Distinguishing experimentally

For equal concentration solutions, the strong species has:

But: the same volume of strong and weak acid of the same concentration require the same volume of NaOH to neutralise — total moles of H⁺ available is identical. Strength affects rate and pH, not stoichiometry.

Worked example

pH of strong acid

Problem. Calculate the pH of 0.0500 mol dm⁻³ HNO₃.
Solution. HNO₃ is strong → fully dissociated. [H⁺] = 0.0500 M. pH = −log(0.0500) = 1.30.
Worked example

pH of strong base

Problem. Calculate the pH of 0.025 mol dm⁻³ NaOH at 25 °C.
Solution. [OH⁻] = 0.025 M. pOH = −log(0.025) = 1.60. pH = 14 − 1.60 = 12.40.

Try these

  1. Define strong acid and weak acid in terms of dissociation.
    Show answer
    Strong acid: essentially fully dissociated in water (one-way arrow: HA → H⁺ + A⁻). [H⁺] = [HA]initial. Weak acid: partially dissociated (equilibrium: HA ⇌ H⁺ + A⁻). [H⁺] < [HA]initial.
  2. Name three common strong acids and three common weak acids.
    Show answer
    Strong: HCl, HNO₃, H₂SO₄ (also HBr, HI, HClO₄). Weak: ethanoic acid CH₃COOH, methanoic acid HCOOH, hydrofluoric acid HF, carbonic acid H₂CO₃.
  3. Equal volumes and concentrations of HCl and CH₃COOH are tested with universal indicator. Which is more acidic? Why?
    Show answer
    HCl is more acidic (lower pH) because it's fully dissociated, giving higher [H⁺]. Ethanoic acid is only partly dissociated.
  4. Compare the conductivities of 0.10 M HCl and 0.10 M CH₃COOH. Which is higher? Why?
    Show answer
    HCl has higher conductivity. Full dissociation → many ions in solution. CH₃COOH has fewer ions (most stays as undissociated molecules) → lower conductivity.
  5. Equal volumes of equal-concentration HCl and CH₃COOH require how much NaOH to be neutralised? Compare and explain.
    Show answer
    The same volume of NaOH. Both have the same total amount of acid (same n moles per volume). The number of moles of H⁺ that can ever be released is identical — strong vs weak only affects how fast and at what pH it's released, not the total.
  6. Which would react faster with magnesium: 0.1 M HCl or 0.1 M CH₃COOH? Why?
    Show answer
    HCl. Both have the same total moles of H⁺ available, but HCl releases them immediately ([H⁺] = 0.1 M), while CH₃COOH releases them slowly as the equilibrium shifts. Higher instantaneous [H⁺] → more collisions per second with Mg → faster reaction.
Lesson 7

Acid-base titrations

Reactivity 3.1.5

Lesson outcomes

Recall the titration procedure from Unit 2: pipette a known volume of one solution (analyte) into a flask with indicator; burette the other (titrant) until endpoint colour change; record titre.

Calculation

For a balanced reaction aA + bB → products:

cAVA / a = cBVB / b

Indicator choice

IndicatorRangeSuitable for
Methyl orange3.1–4.4Strong acid + weak base
Bromothymol blue6.0–7.6Strong acid + strong base
Phenolphthalein8.3–10.0Weak acid + strong base
Worked example

Titration calculation · H₂SO₄ + NaOH

Problem. 25.0 cm³ of 0.150 mol dm⁻³ NaOH requires 18.75 cm³ H₂SO₄ to reach the endpoint. Calculate [H₂SO₄].
Solution. 2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O. So n(H₂SO₄) = ½ × n(NaOH).
n(NaOH) = 0.150 × 25.0/1000 = 3.75 × 10⁻³ mol.
n(H₂SO₄) = 0.5 × 3.75 × 10⁻³ = 1.875 × 10⁻³ mol.
[H₂SO₄] = 1.875 × 10⁻³ / (18.75/1000) = 0.100 mol dm⁻³.

Try these

  1. What is the difference between an endpoint and an equivalence point?
    Show answer
    Equivalence point: where stoichiometrically equal moles of acid and base have been mixed. Endpoint: where the indicator changes colour, signalling that we've reached equivalence (approximately). A well-chosen indicator makes endpoint ≈ equivalence point.
  2. Why must the burette be rinsed with the titrant solution before use, but the conical flask with deionised water only?
    Show answer
    Burette: any water on the walls would dilute the titrant, lowering its effective concentration → larger titre → overestimated [analyte]. Conical flask: contains the analyte; water doesn't change the amount of analyte (just dilutes it harmlessly), so deionised water is fine to rinse residues.
  3. 23.50 cm³ of 0.100 mol dm⁻³ HCl is titrated with NaOH; the titre is 25.00 cm³. Calculate [NaOH].
    Show answer
    n(HCl) = 0.100 × 23.50/1000 = 2.35 × 10⁻³ mol. n(NaOH) = n(HCl) = 2.35 × 10⁻³ mol (1:1 ratio). [NaOH] = 2.35 × 10⁻³ / 25.00/1000 = 0.0940 mol dm⁻³.
Lesson 13

Indicators

Reactivity 3.1.6

Lesson outcomes

An acid-base indicator HIn is a weak acid (or base) whose dissociated and protonated forms have distinctly different colours:

HIn (colour A) ⇌ H⁺ + In⁻ (colour B)

Adding acid pushes the equilibrium left → colour A dominates. Adding base pushes right → colour B.

The colour change is most sensitive when [HIn] = [In⁻], which happens when pH = pKIn. The visible colour change spans approximately pKIn ± 1 unit.

Choosing an indicator

Pick one whose pKIn is within the steep section of the titration curve — close to the equivalence pH.

Try these

  1. Why does an acid-base indicator change colour?
    Show answer
    Indicators are weak acids (or bases): HIn ⇌ H⁺ + In⁻ where HIn and In⁻ have different colours. Adding acid shifts equilibrium left (more HIn, one colour). Adding base removes H⁺, shifts right (more In⁻, other colour). pH change → different ratio → different colour.
  2. Why does a colour change occur over a pH range of about pKIn ± 1?
    Show answer
    The colour we see depends on the [HIn]/[In⁻] ratio. When this ratio is < 1/10 or > 10/1 (i.e. about 1 pH unit either side of pKIn), one form dominates. In between, we see a transition (mixture of colours).
  3. What is the colour change for: methyl orange, bromothymol blue, phenolphthalein?
    Show answer
    Methyl orange (pKIn ~3.7): red (low pH) → yellow (high pH). Bromothymol blue (pKIn ~7): yellow → blue. Phenolphthalein (pKIn ~9): colourless → pink.
  4. Litmus has a wide pH range of 5-8. Why is litmus not suitable for accurate titrations?
    Show answer
    The wide colour-change range (3 pH units) means the endpoint is not sharp — you can't tell precisely when neutralisation occurred. Indicators with narrow ranges (~2 units) give sharper endpoints.
HL extension

Buffers
and titration
curves.

HL: Ka, Kb, pKa relationships, buffer chemistry, and the analytical power of pH titration curves.

HL only

Ka/pKa and Kb/pKb HL Only

Lesson 8 of Unit 10.

HL only

Weak acid/base calculations HL Only

Lesson 9 of Unit 10.

HL only

pH Curves and Salt Hydrolysis HL Only

Lesson 11 of Unit 10.

HL only

Buffer Solutions HL Only

Lesson 12 of Unit 10.

Lesson 8 HL only

K_a, pK_a, K_b, pK_b

Reactivity 3.1.7 (HL)

Lesson outcomes

  • Write the equilibrium expression for the dissociation of a weak acid (Ka) and weak base (Kb).
  • Define pKa = −log Ka; smaller pKa = stronger acid.
  • Use Ka × Kb = Kw for conjugate pairs.

For a weak acid HA: HA + H₂O ⇌ H₃O⁺ + A⁻. The equilibrium constant (with water excluded as solvent) is the acid dissociation constant:

Ka = [H⁺][A⁻] / [HA]    pKa = −log Ka

Larger Ka = stronger acid. Smaller pKa = stronger acid. Most weak acids have pKa between 3 and 10.

Base equivalents

For weak base B + H₂O ⇌ BH⁺ + OH⁻: Kb = [BH⁺][OH⁻] / [B].

Conjugate pair relationship

Ka × Kb = Kw    (or pKa + pKb = 14 at 25 °C)

Stronger acid → weaker conjugate base. Powerful tool: knowing Ka of an acid gives you Kb of its conjugate base instantly.

Worked example

K_b of acetate from K_a of acetic acid

Problem. Ka(CH₃COOH) = 1.8 × 10⁻⁵. Calculate Kb(CH₃COO⁻).
Solution. Kb = Kw / Ka = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰. Acetate is a very weak base (consistent with acetic acid being a moderately strong weak acid).

Try these

  1. What does Ka tell you about an acid? What about pKa?
    Show answer
    Ka = [H⁺][A⁻]/[HA]. Larger Ka = stronger acid (more dissociation). pKa = −log Ka. Smaller pKa = stronger acid (more H⁺ released).
  2. Rank in order of acid strength: HF (pKa 3.17), HClO (pKa 7.5), CH₃COOH (pKa 4.76), H₂SO₃ (pKa 1.85).
    Show answer
    Smaller pKa = stronger. H₂SO₃ > HF > CH₃COOH > HClO.
  3. If Ka(HCN) = 6.2 × 10⁻¹⁰, calculate Kb(CN⁻).
    Show answer
    Kb = Kw/Ka = 10⁻¹⁴/6.2 × 10⁻¹⁰ = 1.6 × 10⁻⁵. Note: HCN is very weak acid → CN⁻ is moderately strong base (relatively).
Lesson 9 HL only

Weak acid / base calculations

Reactivity 3.1.7 (HL)Reactivity 3.1.8 (HL)

Lesson outcomes

  • Calculate the pH of a weak acid solution using [H⁺] ≈ √(Ka × [HA]).
  • Calculate Ka from a measured pH and concentration.
  • Recognise when the approximation [HA]eq ≈ [HA]initial breaks down.

For a weak monoprotic acid HA at initial concentration c, with Ka:

HA ⇌ H⁺ + A⁻. If small amount x dissociates: [H⁺] = [A⁻] = x; [HA] = c − x.

Ka = x² / (c − x)

Approximation (valid when x ≪ c, typically c > ~10×Ka): x ≈ √(Ka × c). Then pH ≈ ½ (pKa − log c).

Worked example

pH of weak acid

Problem. Calculate the pH of 0.100 mol dm⁻³ ethanoic acid. Ka = 1.8 × 10⁻⁵.
Solution. [H⁺] = √(Ka × c) = √(1.8 × 10⁻⁵ × 0.100) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³.
pH = −log(1.34 × 10⁻³) = 2.87.
Check approximation: [HA]eq ≈ 0.100 − 0.00134 = 0.0987 — close to 0.100, so the approximation is valid.
Worked example

K_a from pH

Problem. 0.250 M HF has pH = 2.05. Calculate Ka(HF).
Solution. [H⁺] = 10⁻²·⁰⁵ = 8.91 × 10⁻³. Ka = [H⁺]² / [HA] = (8.91 × 10⁻³)² / 0.250 = 3.17 × 10⁻⁴.

Try these

  1. When can you use [H⁺] ≈ √(Ka × [HA]) for a weak acid?
    Show answer
    When [HA]eq ≈ [HA]initial — i.e. very little of the acid actually dissociates. This is true when Ka is much smaller than [HA]initial (typically [HA] > 10² × Ka). For ethanoic acid (Ka = 1.8 × 10⁻⁵) at 0.1 M, approximation is excellent.
  2. Calculate the pH of 0.025 M ethanoic acid. Ka = 1.8 × 10⁻⁵.
    Show answer
    [H⁺] = √(1.8 × 10⁻⁵ × 0.025) = √(4.5 × 10⁻⁷) = 6.71 × 10⁻⁴. pH = −log(6.71 × 10⁻⁴) = 3.17.
  3. Calculate the pH of 0.10 mol dm⁻³ NH₃ (a weak base, Kb = 1.8 × 10⁻⁵).
    Show answer
    [OH⁻] = √(Kb × c) = √(1.8 × 10⁻⁵ × 0.10) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³. pOH = 2.87. pH = 14 − 2.87 = 11.13.
Lesson 11 HL only

pH curves and salt hydrolysis

Reactivity 3.1.9 (HL)

Lesson outcomes

  • Draw and interpret pH titration curves for the four types of titration.
  • Identify the equivalence point and the half-equivalence point on a curve.
  • Read pKa directly from a weak-acid titration curve (= pH at half-equivalence).
  • Predict whether the salt formed from a titration is acidic, basic, or neutral.

Four titration curve types

TypepH at equivSuitable indicator
Strong acid + strong base7.0Any (pKa 4-10): methyl orange, phenolphthalein, bromothymol blue
Weak acid + strong base>7 (salt of weak acid is alkaline)Phenolphthalein (pKa 9)
Strong acid + weak base<7 (salt of weak base is acidic)Methyl orange (pKa 3.5)
Weak + weakVariable, no sharp jumpNo good indicator → use pH meter

Reading pK_a from a weak-acid curve

At the half-equivalence point, [HA] = [A⁻]. From the Henderson-Hasselbalch equation, pH = pKa. Read the pH at half the equivalence volume — that's the pKa directly.

Salt hydrolysis

A salt of a strong acid and strong base (NaCl) is neutral. Salt of weak acid + strong base (CH₃COONa) is basic — the conjugate base hydrolyses water. Salt of strong acid + weak base (NH₄Cl) is acidic — the conjugate acid hydrolyses.

Try these

  1. Sketch the pH vs volume curve for adding strong base (NaOH) to strong acid (HCl). Mark the equivalence point.
    Show answer
    Starts at low pH (~1), rises slowly, then a near-vertical jump from pH ~3 to ~11 right at the equivalence volume, then levels off near pH ~13. Equivalence point = midpoint of the vertical jump (pH 7).
  2. What is the half-equivalence point on a weak-acid titration curve? What does it tell you?
    Show answer
    The point at which half the weak acid has been neutralised — i.e. [HA] = [A⁻]. From Henderson-Hasselbalch, pH = pKa at this point. So you can read pKa directly off a weak-acid titration curve at half the equivalence volume.
  3. Predict the pH at the equivalence point of each titration: (a) HCl + NaOH, (b) CH₃COOH + NaOH, (c) HCl + NH₃, (d) CH₃COOH + NH₃.
    Show answer
    (a) pH 7 (strong + strong → neutral salt). (b) pH > 7 (weak acid + strong base → CH₃COO⁻ is basic). (c) pH < 7 (strong acid + weak base → NH₄⁺ is acidic). (d) variable (no sharp jump, depends on Ka vs Kb).
  4. Why is phenolphthalein chosen for the titration of CH₃COOH with NaOH but not for HCl with NH₃?
    Show answer
    CH₃COOH + NaOH gives CH₃COO⁻ (basic salt) → equivalence pH ~8.7. Phenolphthalein (pKa 9) changes in this range. HCl + NH₃ gives NH₄⁺ (acidic salt) → equivalence pH ~5.3. Phenolphthalein would miss the endpoint completely; methyl orange (pKa 3.5) is the right choice.
Lesson 12 HL only

Buffer solutions

Reactivity 3.1.10 (HL)

Lesson outcomes

  • Define a buffer and explain its action with reference to Le Chatelier's principle.
  • Identify a buffer system (weak acid + conjugate base, or weak base + conjugate acid).
  • Use the Henderson-Hasselbalch equation to design or analyse a buffer.
  • Recognise the importance of biological buffers (carbonate, phosphate).

A buffer resists pH change when small amounts of acid or base are added. Made from a weak acid + its conjugate base in similar concentrations (or a weak base + its conjugate acid).

How it works (acid buffer)

The acid (HA) absorbs added OH⁻: HA + OH⁻ → A⁻ + H₂O. The conjugate base (A⁻) absorbs added H⁺: A⁻ + H⁺ → HA. Both reservoirs deplete only slowly, so pH stays nearly constant.

Henderson-Hasselbalch equation

pH = pKa + log ( [A⁻] / [HA] )

When [A⁻] = [HA], pH = pKa. Choose a weak acid whose pKa is near the target buffer pH, then adjust the ratio.

Buffer capacity

The amount of acid or base a buffer can absorb before pH changes by more than ±1 unit. Higher absolute concentrations of HA and A⁻ → higher capacity.

Biological buffers

  • Blood: H₂CO₃ / HCO₃⁻. Maintains pH 7.40 ± 0.05.
  • Intracellular: H₂PO₄⁻ / HPO₄²⁻. pKa ≈ 7.2, perfect for cytoplasmic pH.
Worked example

Designing a buffer at pH 5

Problem. Design a buffer with pH 5.00 using ethanoic acid (pKa = 4.76) and sodium ethanoate. State the required ratio [CH₃COO⁻] / [CH₃COOH].
Solution. 5.00 = 4.76 + log(ratio) → log(ratio) = 0.24 → ratio = 10⁰·²⁴ = 1.74.
So make a solution containing 1.74 mol of CH₃COO⁻ for every 1 mol of CH₃COOH.

Try these

  1. What is a buffer? What components does it contain?
    Show answer
    A solution that resists pH change when small amounts of acid or base are added. Made from a weak acid + its conjugate base (or weak base + its conjugate acid) in comparable concentrations.
  2. Explain how a CH₃COOH/CH₃COO⁻ buffer resists pH change when (a) a little HCl is added, (b) a little NaOH is added.
    Show answer
    (a) Added H⁺ is mopped up by the conjugate base: CH₃COO⁻ + H⁺ → CH₃COOH. (b) Added OH⁻ is mopped up by the weak acid: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. The buffer absorbs the change instead of pH dropping/rising sharply.
  3. Design a buffer with pH 4.50 using ethanoic acid (pKa 4.76) and sodium ethanoate.
    Show answer
    pH = pKa + log([A⁻]/[HA]). 4.50 = 4.76 + log(ratio). log(ratio) = −0.26. Ratio = 10⁻⁰·²⁶ = 0.55. So [CH₃COO⁻]/[CH₃COOH] = 0.55 — i.e. more acid than salt. E.g. 0.10 M CH₃COOH + 0.055 M CH₃COONa.
  4. Adding 1 cm³ of 0.1 M HCl to (a) pure water at pH 7, and (b) a buffer at pH 7 — predict and explain the pH change in each case.
    Show answer
    (a) Water has no buffering capacity; even a tiny amount of HCl drops pH dramatically (to ~3 if no dilution). (b) The buffer's conjugate base neutralises the added H⁺ — pH change is small (perhaps ±0.05 unit).
  5. Why are blood buffers (especially the H₂CO₃/HCO₃⁻ system) essential for life?
    Show answer
    Many cellular reactions only work in a narrow pH range (~7.35–7.45 in blood). Without the carbonate buffer, normal metabolism (which produces CO₂ and acid waste) would acidify blood within minutes — fatal at pH below 7.0. The buffer maintains the safe pH window despite continuous acid/base challenges.
Vocabulary

42 terms to own.

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elementcompoundmixtureatomprotonelectronorbitalmoleconcentrationsolutesolventsolutiontitrationsignificant figuresgraphrate of reactiontemperatureequilibriumequilibrium constantkckwexothermicenthalpy changeperiodic tableperiodgrouphalogensoxideamphotericcomplex ionligandcovalent bonddouble bonddelocalised electronsdipolelone pairdipole-dipolehybridisationhydrocarbonamineadditionoxidation