NCEA Level 3 Chemistry 91392 Aqueous Equilibria: past papers, practice and marking
- NCEA Level 3
- Chemistry
- AS91392
- 5 credits
- External exam
About 91392
AS91392, Demonstrate understanding of equilibrium principles in aqueous systems. Solubility and Ks, precipitation, species in solution, weak acid and base pH, buffers and titration curves: the most calculation-heavy chemistry external, and the most predictable.
What 91392 covers
- SolubilityEquation and Ks expression for AB, AB₂ and A₂B solids; solubility, ion concentrations and Ks, including at a specified pH
- PrecipitationDilute each ion for the mixed volume, allow for two ions per formula, then compare the ionic product (IP) with Ks
- Common ion effectEffect on solubility of a common ion, complex ion formation and added acid, explained with equations
- Weak acids and basesRank species in solution; pH of weak acids, weak bases and salts (Ka and Kw); pH and conductivity comparisons
- BuffersHow a buffer removes added acid and base, pH by the Ka method, mass of salt for a set pH, and effectiveness
- Titration curvesThe pH at the start, half-equivalence, equivalence and beyond; curve shape; choosing an indicator
91392 past papers
Open a 91392 past paper beside your own Ako page, then get it marked, part by part.
Work through a past paper All NCEA Level 3 Chemistry past papers
Practice questions
Question 1: Solubility
Limewater is a saturated solution of calcium hydroxide, Ca(OH)₂, a sparingly soluble solid. Ks(Ca(OH)₂) = 5.02 × 10⁻⁶ at 25 °C.
- Write the equation for the equilibrium occurring in a saturated solution of Ca(OH)₂.(up to Achieved)
- Write the expression for Ks(Ca(OH)₂).(up to Achieved)
Check your answer
Part a: what gets the marks
- Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq), with the equilibrium arrow, the states and the correct charges.
A full answer: Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq)
Part b: what gets the marks
- Ks = [Ca²⁺][OH⁻]², with the hydroxide concentration squared and no term for the solid.
A full answer: Ks = [Ca²⁺][OH⁻]². The solid does not appear in the expression.
What gets each grade
Question 2: Solubility
Lead(II) iodide, PbI₂, is a bright yellow, sparingly soluble solid. A student finds that 0.622 g of PbI₂ dissolves in 1.00 L of water at 25 °C. M(PbI₂) = 461.0 g mol⁻¹.
- Write the equation for the equilibrium in a saturated solution of PbI₂, and the expression for Ks(PbI₂).(up to Achieved)
- Calculate Ks(PbI₂) from the student's result.(up to Merit)
Check your answer
Part a: what gets the marks
- PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq), and Ks = [Pb²⁺][I⁻]².
A full answer: PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ks = [Pb²⁺][I⁻]²
Part b: what gets the marks
- Solubility in mol L⁻¹: s = 0.622 g ÷ 461.0 g mol⁻¹ = 1.35 × 10⁻³ mol L⁻¹
- Ks: Ks = s × (2s)² = 4s³ = 4 × (1.35 × 10⁻³)³ = 9.8 × 10⁻⁹
A full answer: s = 0.622 ÷ 461.0 = 1.35 × 10⁻³ mol L⁻¹. Ks = 4s³ = 9.8 × 10⁻⁹.
What gets each grade
Question 3: Solubility
Water passing through old lead pipes can become saturated with lead(II) iodide, PbI₂, where iodide is present. Ks(PbI₂) = 9.8 × 10⁻⁹ at 25 °C; its solubility is 1.35 × 10⁻³ mol L⁻¹. M(Pb) = 207.2 g mol⁻¹.
- Calculate [I⁻] in the saturated solution.(up to Achieved)
- Calculate the mass of lead, in mg, dissolved in a 200 mL glass of this saturated solution.(up to Excellence)
Check your answer
Part a: what gets the marks
- [I⁻] from s: [I⁻] = 2s = 2 × 1.35 × 10⁻³ = 2.70 × 10⁻³ mol L⁻¹
A full answer: [I⁻] = 2s = 2.70 × 10⁻³ mol L⁻¹
Part b: what gets the marks
- n(Pb²⁺): [Pb²⁺] = s = 1.35 × 10⁻³ mol L⁻¹; n = cV = 1.35 × 10⁻³ × 0.200 = 2.70 × 10⁻⁴ mol
- Mass of lead: m = nM = 2.70 × 10⁻⁴ × 207.2 = 0.0559 g = 55.9 mg
A full answer: [Pb²⁺] = 1.35 × 10⁻³ mol L⁻¹, so n = 1.35 × 10⁻³ × 0.200 = 2.70 × 10⁻⁴ mol. Mass of lead = 2.70 × 10⁻⁴ × 207.2 = 0.0559 g = 55.9 mg.