9701 Chemistry · Topics 7 & 25 · AS + A Level
Equilibria and Acid–Base Cheat Sheet — A Level Chemistry 9701
Equilibrium questions are usually two marks of theory and four marks of careful arithmetic. These sheets set out Le Chatelier’s principle and the effect of a catalyst, how to build a Kc or Kp expression with the right units, and then the whole A2 acid–base toolkit — pH, Ka and pKa, buffers, solubility product and reading a titration curve to pick the right indicator.
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What’s on this cheat sheet
Equilibria — AS
01 · Dynamic equilibrium
Reached in a closed system when the forward and reverse rates are equal, so the concentrations of reactants and products stay constant. Both reactions continue — it is dynamic, not stopped.
Equilibrium can be approached from either direction and gives the same position under the same conditions.
02 · Le Chatelier’s principle
If a change is made to a system at equilibrium, the position shifts so as to oppose that change.
| Change | Shift |
|---|---|
| [reactant] ↑ | to the right |
| pressure ↑ | to the side with fewer gas moles |
| temperature ↑ | in the endothermic direction |
| catalyst | no shift — equilibrium reached sooner |
Only temperature changes Kc or Kp. Concentration and pressure move the position but leave K unchanged.
03 · Kc and Kp
Kc = [C]c[D]d ÷ [A]a[B]b
Kp = p(C)cp(D)d ÷ p(A)ap(B)b
Products on top. Solids and pure liquids are left out. Work the units out each time from the powers — they may cancel entirely.
Partial pressure = mole fraction × total pressure. Mole fractions sum to 1.
04 · Worked example — Kc by ICE
I 1.00 / 1.00 / 0 / 0
C −0.667 / −0.667 / +0.667 / +0.667
E 0.333 / 0.333 / 0.667 / 0.667
Kc = (0.667 × 0.667) ÷ (0.333 × 0.333)
Kc = 4.01, no units
05 · Industrial compromise
Haber N₂ + 3H₂ ⇌ 2NH₃, ΔH negative. High pressure favours NH₃, low temperature favours yield but is too slow — so ~450 °C, ~200 atm, Fe catalyst.
Contact 2SO₂ + O₂ ⇌ 2SO₃, ΔH negative. ~450 °C, ~2 atm (yield is already high, so high pressure is not worth the cost), V₂O₅ catalyst.
06 · Brønsted–Lowry
Acid — proton donor. Base — proton acceptor. Every acid has a conjugate base formed by losing H⁺; the pairs differ by one proton.
HCl + H₂O → H₃O⁺ + Cl⁻ — the pairs are HCl/Cl⁻ and H₂O/H₃O⁺.
Strong acid — fully dissociated (HCl, HNO₃, H₂SO₄). Weak — partially dissociated (CH₃COOH), an equilibrium. Strength is not the same as concentration.
07 · Homogeneous and heterogeneous
Homogeneous — all species in the same phase, so every one appears in the K expression.
Heterogeneous — more than one phase; solids and pure liquids have constant concentration and are omitted. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = p(CO₂) alone.
08 · Worked example — Kp
p(N₂) = 0.20 × 200 = 40 atm
p(H₂) = 0.60 × 200 = 120 atm
p(NH₃) = 0.20 × 200 = 40 atm
Kp = 40² ÷ (40 × 120³)
Kp = 1600 ÷ 6.91 × 10⁷ = 2.3 × 10⁻⁵ atm⁻²
09 · What the size of K tells you
| Value | Position |
|---|---|
| K ≫ 1 | products dominate |
| K ≈ 1 | comparable amounts |
| K ≪ 1 | reactants dominate |
K says nothing about rate. For an exothermic forward reaction K falls as temperature rises.
10 · Indicators
An indicator is a weak acid whose acid and conjugate‑base forms have different colours: HIn ⇌ H⁺ + In⁻.
The colour changes over roughly pKIn ± 1. Methyl orange turns red → yellow over 3.1–4.4; phenolphthalein colourless → pink over 8.3–10.0.
11 · Kc or Kp?
Use Kc when the data are concentrations or moles in a stated volume. Use Kp when the data are partial pressures, mole fractions or a total pressure — always a gas‑phase system.
For a reaction with equal numbers of gas moles on both sides the units cancel and the value is dimensionless; pressure then has no effect on the position of equilibrium either.
Both are constant at a fixed temperature no matter what starting amounts are used — that is what makes them useful.
12 · Industrial conditions at a glance
| Process | T | p | Catalyst |
|---|---|---|---|
| Haber | 450 °C | 200 atm | Fe |
| Contact | 450 °C | 2 atm | V₂O₅ |
| Methanol | 250 °C | 50–100 atm | Cu/ZnO |
Every one of these is a compromise: the temperature chosen is higher than the yield alone would justify, because an acceptable rate matters more than a theoretical maximum. Unreacted gases are recycled to raise the overall conversion.
13 · Worked example — shifting the position
Raise the pressure → shifts right, 3 moles of gas become 2, yield rises
Raise the temperature → shifts left, the endothermic direction, yield falls and Kp falls
Remove SO₃ → shifts right
Add V₂O₅ → no shift, equilibrium reached sooner
Marks lost here
— Saying the reaction “stops” at equilibrium.
— Claiming a catalyst increases yield or changes K.
— Putting solids or pure liquids into a K expression.
— Using moles instead of concentrations when the volume is not 1 dm³.
Acid–Base Equilibria — A2
14 · pH and Kw
Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 298 K
For a strong base find [OH⁻], then [H⁺] = Kw ÷ [OH⁻]. Kw rises with temperature (ionisation is endothermic), so neutral water has pH below 7 above 298 K — still neutral, since [H⁺] = [OH⁻].
15 · Ka and weak acids
[H⁺] = √(Ka × [HA])
Assumptions: [H⁺] = [A⁻], and [HA] at equilibrium ≈ the initial value. Larger Ka (smaller pKa) = stronger acid.
16 · Worked example — weak acid pH
[H⁺] = √(1.74 × 10⁻⁵ × 0.100)
[H⁺] = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³
pH = −log(1.32 × 10⁻³) = 2.88
Compare 0.100 mol dm⁻³ HCl: pH = 1.00 — same concentration, far lower pH, because HCl is fully dissociated.
17 · Buffers
A solution that resists pH change on adding small amounts of acid or alkali. Made from a weak acid + its salt (CH₃COOH / CH₃COONa) or a weak base + its salt (NH₃ / NH₄Cl), or by part‑neutralising a weak acid.
Add H⁺: A⁻ + H⁺ → HA. Add OH⁻: HA + OH⁻ → A⁻ + H₂O. The large reservoirs of both species absorb the change. Buffering is best when [salt] = [acid], where the pH equals pKa.
18 · Titration curves
| Acid + base | pH at equiv. | Indicator |
|---|---|---|
| strong + strong | 7 | either |
| weak + strong base | > 7 | phenolphthalein |
| strong + weak base | < 7 | methyl orange |
| weak + weak | ~7 | none suitable |
The indicator’s pKIn must lie inside the vertical section. Half‑way to the equivalence point, the pH equals pKa — that is the buffer region.
19 · Solubility product
Applies to a saturated solution of a sparingly soluble salt. For MX, Ksp = s²; for MX₂, Ksp = 4s³. A precipitate forms when the ionic product exceeds Ksp.
Common‑ion effect: adding an ion already present shifts the equilibrium left, so solubility falls — but Ksp is unchanged.
20 · Worked example — buffer
[H⁺] = Ka × [acid] ÷ [salt]
[H⁺] = 1.74 × 10⁻⁵ × (0.200 ÷ 0.100)
[H⁺] = 3.48 × 10⁻⁵
pH = 4.46
Diluting the buffer leaves the ratio unchanged, so the pH barely moves.
21 · Titration curve — weak acid + strong base
The flat buffer region sits before the equivalence point; the vertical section is shorter than for a strong acid, so only phenolphthalein fits inside it.
22 · Worked example — Ksp
Mg(OH)₂ ⇌ Mg²⁺ + 2OH⁻
[Mg²⁺] = s, [OH⁻] = 2s
Ksp = s × (2s)² = 4s³
Ksp = 4 × (1.44 × 10⁻⁴)³
Ksp = 1.19 × 10⁻¹¹ mol³ dm⁻⁹
23 · Worked example — strong base
NaOH is fully dissociated → [OH⁻] = 0.0500
[H⁺] = Kw ÷ [OH⁻]
[H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.0500 = 2.00 × 10⁻¹³
pH = −log(2.00 × 10⁻¹³) = 12.70
For Ba(OH)₂ remember [OH⁻] is twice the salt concentration.
24 · Worked example — common ion
[Cl⁻] ≈ 0.100 from the NaCl
[Ag⁺] = Ksp ÷ [Cl⁻]
[Ag⁺] = 1.8 × 10⁻¹⁰ ÷ 0.100 = 1.8 × 10⁻⁹ mol dm⁻³
In pure water s = √Ksp = 1.3 × 10⁻⁵ — about ten thousand times more soluble.
Marks lost here
— Confusing strength with concentration when comparing pH.
— Using [H⁺] = c directly for a weak acid.
— Forgetting Kw when going from [OH⁻] to pH.
— Explaining a buffer without writing the two equations for added H⁺ and added OH⁻.
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Equilibria and Acid–Base — Frequently Asked Questions
Does a catalyst change the position of equilibrium?
No. A catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached faster but Kc and the position of equilibrium are unchanged. Only temperature changes the value of Kc.
How do you choose an indicator for a titration?
The indicator’s pH range must fall entirely within the near-vertical section of the titration curve. Methyl orange suits strong acid–weak base, phenolphthalein suits weak acid–strong base, and a weak acid–weak base titration has no vertical section so no indicator works.
How does a buffer resist a change in pH?
A buffer holds a reservoir of a weak acid and its conjugate base. Added H⁺ is removed by the conjugate base and added OH⁻ is removed by the weak acid, so the ratio [acid]/[base] — and therefore the pH — barely changes.
