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Eduqas A-level Chemistry (A410QS) · PI5 — Equilibria
Mini-Lesson

PI5 · Equilibria

Eduqas PI5 puts numbers on "how far". You will write and calculate Kc and Kp, see what really changes K, then turn to acid-base equilibria: Kw, the pH of strong and weak acids, Ka and pKa, buffers, and reading titration curves to choose an indicator.

Kc and Kp Ka, Kw and pH buffers + curves only temperature changes the value of K

Work through each screen, answer every question as you go — the multiple-choice checks, the calculations and the sorting games — and collect ⭐ stars. Press Start when you are ready.

PI5 · Kc

The equilibrium constant Kc

For aA + bB ⇌ cC + dD at equilibrium:

Kc = ([C]c[D]d) ÷ ([A]a[B]b)equilibrium concentrations only · omit solids and pure liquids · units depend on the equation

What changes Kc? Only temperature. Not concentration, not pressure, not a catalyst.

  • If the forward reaction is exothermic, raising T decreases Kc (the equilibrium shifts back to absorb the heat).
  • If it is endothermic, raising T increases Kc.

A big Kc (say > 10³) means the position of equilibrium lies well to the right. A very small Kc means barely any product forms. Kc says nothing about rate.

Calculate

Your turn — Kc for an esterification

1For CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O, the equilibrium concentrations (all mol dm⁻³) are: acid 0.33, alcohol 0.33, ester 0.67, water 0.67. Calculate Kc to 3 significant figures.
(no units)
Hint: Kc = (0.67 × 0.67) ÷ (0.33 × 0.33) = 0.4489 ÷ 0.1089.
Calculate

Your turn — Kc is not changed by volume

2For H₂(g) + I₂(g) ⇌ 2HI(g) at equilibrium a 2.00 dm³ vessel contains 0.20 mol H₂, 0.20 mol I₂ and 1.60 mol HI. Calculate Kc.
(no units)
Hint: first convert to concentrations (÷ 2.00 dm³): 0.10, 0.10 and 0.80 mol dm⁻³. Then Kc = 0.80² ÷ (0.10 × 0.10) = 0.64 ÷ 0.010.
PI5 · Kp

Kp and partial pressures

For gases we use partial pressures. The partial pressure of a gas is its mole fraction multiplied by the total pressure.

p(X) = x(X) × Ptotalmole fraction x(X) = moles of X ÷ total moles of gas · the mole fractions must sum to 1
For N₂O₄(g) ⇌ 2NO₂(g): Kp = p(NO₂)² ÷ p(N₂O₄)same layout as Kc — products over reactants, raised to the coefficients

Pressure and Kp: increasing the total pressure shifts the position of equilibrium towards the side with fewer gas moles — but Kp itself does not change. Only temperature changes Kp.

Calculate

Your turn — calculate Kp

3At equilibrium the total pressure is 100 kPa, and the mole fractions are 0.40 NO₂ and 0.60 N₂O₄. For N₂O₄(g) ⇌ 2NO₂(g), calculate Kp in kPa.
kPa
Hint: p(NO₂) = 0.40 × 100 = 40 kPa; p(N₂O₄) = 0.60 × 100 = 60 kPa. Kp = 40² ÷ 60 = 1600 ÷ 60.
Quick check

What changes K?

?Which of these changes the value of Kc?
PI5 · acids and bases

Bronsted-Lowry acids, bases and Kw

An acid is a proton donor; a base is a proton acceptor. Every acid has a conjugate base, differing by exactly one H⁺.

CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺acid₁ + base₂ ⇌ base₁ + acid₂ — two conjugate pairs

Water self-ionises, and this is governed by the ionic product of water:

Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 Kso in pure water [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ and pH = 7.00

Self-ionisation is endothermic, so heating water increases Kw and lowers the pH of pure water below 7 — yet the water is still neutral, because [H⁺] still equals [OH⁻]. Neutral means [H⁺] = [OH⁻], not "pH = 7".

Match it

Conjugate pairs

Tap an acid on the left, then the conjugate base it becomes when it donates a proton.

Acid
Conjugate base
Calculate

Your turn — pH of a strong base

4Calculate the pH of 0.100 mol dm⁻³ NaOH at 298 K, to 2 decimal places. (Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶)
pH
Hint: [OH⁻] = 0.100. So [H⁺] = Kw ÷ [OH⁻] = 1.00 × 10⁻¹⁴ ÷ 0.100 = 1.00 × 10⁻¹³. Then pH = −log₁₀[H⁺].
PI5 · weak acids

Weak acids, Ka and pKa

A weak acid is only partially dissociated, so an equilibrium is set up:

HA ⇌ H⁺ + A⁻ Ka = [H⁺][A⁻] ÷ [HA]units: mol dm⁻³ · pKa = −log₁₀Ka · a LARGER Ka (smaller pKa) means a STRONGER acid

For a solution of a weak acid alone, two approximations are made:

  • [H⁺] = [A⁻] — the dissociation is the only significant source of both (we ignore water).
  • [HA]eqm ≈ [HA]initial — so little dissociates that the loss is negligible.
Ka ≈ [H⁺]² ÷ [HA] so [H⁺] = √(Ka × [HA])the standard weak-acid pH calculation
Calculate

Your turn — pH of a weak acid

5Calculate the pH of 0.100 mol dm⁻³ ethanoic acid, to 2 decimal places. (Ka = 1.75 × 10⁻⁵ mol dm⁻³)
pH
Hint: [H⁺] = √(Ka × c) = √(1.75 × 10⁻⁵ × 0.100) = √(1.75 × 10⁻⁶) = 1.32 × 10⁻³. Then pH = −log₁₀(1.32 × 10⁻³).
Calculate

Your turn — pKa

6Calculate the pKa of ethanoic acid, to 2 decimal places. (Ka = 1.75 × 10⁻⁵ mol dm⁻³)
pKa
Hint: pKa = −log₁₀(1.75 × 10⁻⁵).
PI5 · buffers

Buffer solutions

A buffer resists a change in pH when a small amount of acid or alkali is added, or on dilution. An acidic buffer is a weak acid plus its salt (e.g. ethanoic acid + sodium ethanoate), giving a large reservoir of both HA and A⁻.

  • Add acid: the added H⁺ is mopped up by the conjugate base: A⁻ + H⁺ → HA.
  • Add alkali: the added OH⁻ is neutralised by the weak acid: HA + OH⁻ → A⁻ + H₂O.
[H⁺] = Ka × ([HA] ÷ [A⁻])or pH = pKa + log([A⁻] ÷ [HA]) — the Henderson-Hasselbalch equation

Two consequences worth knowing: (1) when [HA] = [A⁻], pH = pKa — so choose an acid whose pKa is close to the pH you want. (2) At half-neutralisation in a weak-acid titration, exactly half the acid has become salt, so the pH there reads off the pKa directly. Blood is buffered near pH 7.4 by H₂CO₃ / HCO₃⁻.

Calculate

Your turn — buffer pH

7A buffer contains 0.100 mol dm⁻³ ethanoic acid and 0.150 mol dm⁻³ sodium ethanoate. Calculate its pH, to 2 decimal places. (Ka = 1.75 × 10⁻⁵ mol dm⁻³)
pH
Hint: [H⁺] = Ka × ([HA] ÷ [A⁻]) = 1.75 × 10⁻⁵ × (0.100 ÷ 0.150) = 1.167 × 10⁻⁵. Then pH = −log₁₀ of that.
PI5 · titration curves

Titration curves and indicators

A titration curve plots pH against the volume of titrant added. The equivalence point is where the reagents have reacted in exactly the stoichiometric ratio — the centre of the near-vertical section.

  • Strong acid + strong base: a long vertical section from about pH 3 to pH 11; equivalence at pH 7. Either indicator works.
  • Weak acid + strong base: starts higher, has a buffer region, and the equivalence point is above pH 7 (about 9) — the salt is basic. Use phenolphthalein (range 8.3–10.0).
  • Strong acid + weak base: the equivalence point is below pH 7 (about 5) — the salt is acidic. Use methyl orange (range 3.1–4.4).
  • Weak acid + weak base: there is no sharp vertical section, so no indicator is suitable — use a pH meter.

The rule for choosing an indicator: its pH range must lie entirely inside the vertical section of the curve. An indicator is itself a weak acid whose conjugate base is a different colour; it changes colour when pH ≈ pKIn.

Quick check

Which indicator?

?Ethanoic acid is titrated with sodium hydroxide. Which indicator should you use?
Quick check

No good indicator at all

?Ethanoic acid is titrated with aqueous ammonia. Why can no indicator give a sharp end point?
Sort it

Read the titration curve

Tap a feature of a titration curve, then the type of titration it belongs to.

⚡ Strong acid + strong base

🧪 Weak acid + strong base

🌡️ Strong acid + weak base

Recap

The big ideas to know

Kc: products over reactants, raised to the coefficients; omit solids and pure liquids

Kp: use partial pressures; p(X) = mole fraction × total pressure

What changes K: ONLY temperature. Not pressure, not concentration, not a catalyst

Kw: [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 298 K — the route to the pH of any alkali

Weak acids: [H⁺] = √(Ka × c) · pKa = −log Ka · smaller pKa means a stronger acid

Buffers: weak acid + its salt; [H⁺] = Ka × [HA] ÷ [A⁻]; pH = pKa when they are equal

Curves: equivalence at pH 7 only for strong/strong; ~9 for weak acid; ~5 for weak base

Indicators: the range must lie inside the vertical section — none works for weak/weak

That is the whole of PI5. Press Finish to see your score.

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