Week 15 · Term 1 review (Chapters 1–3) and Darslab’s own review tasks · 3 lessons
3
Review: equilibrium, ionisation and pH
Chemistry Sec 3 Term 1, Chapters 1–3: Review
GoalRevise weeks 10–14 with new tasks: rate, Kc and Kp, Le Chatelier’s principle, the degree of ionisation, pH and Ksp.
New words
rate of reaction · the change in concentration divided by the time taken, in mol/(L·s)degree of ionisation α · the fraction of the dissolved molecules that have ionised; α = √(Ka ÷ C) for a weak acidequilibrium shift · the change in the position of an equilibrium that opposes a disturbance, by Le Chatelier’s principle
Explanation
Match the question to the tool. Rate = change in concentration ÷ time. Kc = [products] ÷ [reactants], each raised to its coefficient, with solids left out; Kp is the same with partial pressures. A big Kc means mostly products. Le Chatelier: a stress makes the system shift to oppose it; a catalyst changes only the speed, never Kc or the position. Weak acid: [H₃O⁺] = √(Ka × C) and α = √(Ka ÷ C); diluting by 100 makes α ten times larger, although [H⁺] itself falls. Water: Kw = [H⁺][OH⁻] = 10⁻¹⁴, pH = −log[H⁺], pH + pOH = 14. Use only exact powers of ten or the given log 2 = 0.30, log 3 = 0.48 and log 5 = 0.70. Sparingly soluble salt: Ksp is the product of the ion concentrations raised to their coefficients, so AB gives s² and AB₂ gives 4s³; the solid is left out. Always keep mol/L and g/L apart.
Question
Tool
Key relation
How fast?
rate
Δc ÷ t
How far?
Kc, Kp
products ÷ reactants
Which way?
Le Chatelier
opposes the stress
How much ionised?
α
√(Ka ÷ C)
How acidic?
pH
−log[H⁺]
How soluble?
Ksp
s² , 4s³
Worked examples
Rate and Kc. The concentration of a reactant falls from 0.50 to 0.20 mol/L in 25 s, so the rate is (0.50 − 0.20) ÷ 25 = 0.012 mol/(L·s). For PCl₅ ⇌ PCl₃ + Cl₂ at equilibrium, [PCl₅] = 0.40, [PCl₃] = 0.20 and [Cl₂] = 0.20 mol/L, so Kc = 0.20 × 0.20 ÷ 0.40 = 0.10. Taking in heat and making two gas molecules from one, this equilibrium moves to the right when the temperature rises or the pressure falls. A catalyst reaches the same Kc faster.
α and pH. An invented weak acid, Ka = 9.0 × 10⁻⁸, at C = 0.010 mol/L: α = √(Ka ÷ C) = √(9.0 × 10⁻⁶) = 3.0 × 10⁻³ (0.3 %), [H⁺] = α × C = 3.0 × 10⁻⁵, and pH = 5 − log 3 = 5 − 0.48 = 4.52. Diluted 100 times to C = 1.0 × 10⁻⁴: α = √(9.0 × 10⁻⁴) = 3.0 × 10⁻², ten times larger, but [H⁺] = 3.0 × 10⁻⁶ is smaller, and pH = 6 − 0.48 = 5.52, which is 1 unit higher, not 2.
Ksp both ways. An invented salt AB with s = 4.0 × 10⁻⁵ mol/L: Ksp = s² = 1.6 × 10⁻⁹. An invented AB₂ with Ksp = 4.0 × 10⁻¹⁵: 4s³ = 4.0 × 10⁻¹⁵, so s³ = 1.0 × 10⁻¹⁵ and s = 1.0 × 10⁻⁵ mol/L; then [A²⁺] = 1.0 × 10⁻⁵ and [B⁻] = 2.0 × 10⁻⁵ mol/L. Check: 1.0 × 10⁻⁵ × (2.0 × 10⁻⁵)² = 4.0 × 10⁻¹⁵.
Class activity
One-page summary, paper only. Fold a sheet into six boxes titled rate, Kc and Kp, Le Chatelier, α, pH and Ksp. In each box write the key relation and one example of your own with fresh numbers, solved without a calculator. Add a small drawing for Le Chatelier. Trade sheets with a friend and check three answers each.
Practice
1
A reactant falls from 1.20 to 0.60 mol/L in 40 s. Find the average rate.
(1.20 − 0.60) ÷ 40 = 0.015 mol/(L·s).
2
For PCl₅ ⇌ PCl₃ + Cl₂ the partial pressures at equilibrium are PCl₃ 0.30 atm, Cl₂ 0.60 atm and PCl₅ 0.90 atm. Find Kp.
Kp = (0.30 × 0.60) ÷ 0.90 = 0.20.
3
The forward reaction PCl₅ ⇌ PCl₃ + Cl₂ takes in heat. Which changes raise the yield of PCl₃: lower pressure, a catalyst, more heat, extra Cl₂?
Lower pressure (fewer gas molecules are on the left, so it shifts right) and more heat (it shifts to take in heat). A catalyst changes nothing about the yield, and extra Cl₂ shifts the equilibrium to the left.
4
An invented acid has Ka = 1.0 × 10⁻⁷ and C = 0.10 mol/L. Find α, [H⁺] and pH.
