KHAEMENES ACADEMY · CHEMISTRY 11 · WEEK 17 · DAY 03

Solution Stoichiometry & Equivalence

Use concentration and volume to determine reacting moles in solution.

Safety boundary

Use oxidation-number analysis, supplied datasets, and approved simulations. Any solution work should remain virtual or teacher-supervised with classroom-safe materials.

Explicit chemistry instruction

Solution stoichiometry begins by converting concentration and volume into moles using n=MV with V in litres.

Balanced-equation mole ratios then connect one dissolved reactant to another or to product.

At equivalence in a stoichiometric analysis, reacting amounts satisfy the balanced-equation ratio; equal volumes or equal concentrations are not universally required.

Vocabulary: solution stoichiometry; molarity; equivalence; aliquot; mole ratio.

Worked example

25.0 mL of 0.200 M NaOH contains 0.00500 mol NaOH. A 1:1 reaction therefore requires 0.00500 mol of the partner reactant at equivalence.

Guided question

How many moles are in 35.0 mL of 0.150 M solution?

Independent work

  1. Solve four n=MV problems.
  2. Use two balanced equations to convert solution moles.
  3. Explain why equal volumes do not necessarily mean equivalence.

Misconception check

Equivalence is a stoichiometric condition, not simply 'same volume.'

Formative check

What two pieces of information determine moles in a solution aliquot?

Reveal response

Molarity and volume in litres.

Mentor evidence conference

Ask Archaemenes to audit oxidation-number changes, mole accounting, equivalence assumptions, and uncertainty.

Teacher guidance

Foundation: Use M-V-n tables.

Core: Solve solution-stoichiometry chains.

Extended: Analyze uncertainty from burette/volume readings conceptually.