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Science lesson plan · 12th grade (Year 13 in England) · 30 minutes

Titration calculations

Chemical reactions, acids and bases

CCSS NGSS.HS-PS1-7CCSS NGSS.SEP 5England: GCSE Chemistry: Chemical changes (titrations, Higher)

Objective and success criteria

Students will: Titration calculations.

  • I can answer questions on today's skill on my own, without help.
  • I can explain each step and say why it works.
  • I can check my answer and spot a mistake.

Before and after

Needs first: Concentration in mol/dm³; Neutralisation and naming salts

Leads to: the next skill in this area.

Materials: Teacher demo or video, Indicator color chart, Mini whiteboards, A diagram or short video clip of the idea, Printed student sheet (below).

Warm-up · 4 min

  • Quick retrieval from earlier learning (Concentration in mol/dm³; Neutralisation and naming salts). Students answer on mini whiteboards.
  • Show answers together and note who needs support in the main task.
  1. 0.025 mol of hydrochloric acid is in 250 cm³ of solution. What is the concentration, in mol/dm³? → 0.1 mol/dm³Prompt: concentration = moles ÷ volume in dm³
  2. Name the salt: calcium carbonate + sulfuric acid → ___ + water + carbon dioxide (calcium chloride / calcium nitrate / calcium oxide / calcium sulfate) → calcium sulfatePrompt: The first part of the salt's name comes from the metal.
  3. How many moles of sodium hydroxide are in 25 cm³ of a 1 mol/dm³ solution? → 0.025 molPrompt: moles = concentration × volume in dm³

Hook · 3 min

  • Drop an antacid tablet in water (or show a clip) and ask whether a new substance is being made.
  • Share the objective: Titration calculations. Read the success criteria together.

I do: model it · 7 min

  • Work through the example out loud, saying why you do each step, not just what you do.
  • Example: 20 cm³ of 0.5 mol/dm³ sodium hydroxide is neutralised by 12.5 cm³ of hydrochloric acid. HCl + NaOH → NaCl + Hâ‚‚O. What is the concentration of the acid, in mol/dm³? Give 3 decimal places.
  • Step 1: n(NaOH) = 0.5 × 0.02 = 0.01
  • Step 2: n(HCl) = 0.01
  • Step 3: 0.01 ÷ 0.0125 = 0.8
  • Step 4: So the answer is: 0.8 mol/dm³
  • Answer: 0.8 mol/dm³
  • Pause after each step and ask a student to say what comes next.

We do: guided practice · 6 min

  • Do these together. Students try each step on whiteboards before you reveal it.
  • Use the hints as prompts rather than giving the answer.
  1. 20 cm³ of 0.5 mol/dm³ sodium hydroxide is neutralised by 12.5 cm³ of hydrochloric acid. HCl + NaOH → NaCl + H₂O. What is the concentration of the acid, in mol/dm³? Give 3 decimal places. → 0.8 mol/dm³Prompt: Find moles of sodium hydroxide: concentration × volume in dm³.
  2. Titres (cm³): rough 22.95, run 1 22.2, run 2 22.25. Ignoring the rough titre, calculate the mean titre from the concordant results (within 0.10 cm³), in cm³. Give 2 decimal places. → 22.23 cm³Prompt: The rough titre is only a guide.

You do: independent practice · 7 min

  • Students work on the student sheet. Circulate and check the first two answers of every student early.
  • Students who finish: write a question of their own for a partner, with the answer.

The 6 questions are on the student sheet below, with answers in the key.

Exit ticket · 3 min

  • Two questions on their own. Sort answers into got it / nearly / not yet to plan the next lesson.
  1. 25 cm³ of 0.2 mol/dm³ sodium hydroxide is neutralised by 22.5 cm³ of sulfuric acid. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. What is the concentration of the acid, in mol/dm³? Give 3 decimal places. → 0.111 mol/dm³Prompt: Find moles of sodium hydroxide: concentration × volume in dm³.
  2. Titres (cm³): rough 23.01, run 1 22.26, run 2 22.31. Ignoring the rough titre, calculate the mean titre from the concordant results (within 0.10 cm³), in cm³. Give 2 decimal places. → 22.29 cm³Prompt: The rough titre is only a guide.

Watch out for

  • Everyday meanings of words getting in the way of the science meaning: define key words explicitly.
  • Remembering facts without the reason: ask "why?" and "what is the evidence?"

Support and stretch

Support

  • Review first: Concentration in mol/dm³.
  • Use concrete materials: teacher demo or video.
  • Give the first step of each question, or a partly worked example to finish.
  • Pair with a confident partner for the guided practice.

Stretch

  • Explain a common mistake in writing and how to fix it.
  • Write a word problem that uses today's skill, then solve a partner's.

Homework and more practice

Free online practice with hints and printable worksheets for this skill: eurekami.com/science/grade-12.

Free lesson plan from eurekami.com/lesson-plan-builder

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Titration calculations

Name

Date

Score/ 6

  1. Question 1

    Titres (cm³): rough 23.99, run 1 23.39, run 2 23.44. Ignoring the rough titre, calculate the mean titre from the concordant results (within 0.10 cm³), in cm³. Give 2 decimal places.

    Answercm³

  2. Question 2

    Why is acid added dropwise near the end point?

    • A. to make the indicator brighter
    • B. to keep the burette full
    • C. to avoid adding too much past the end point
    • D. because drops react faster

  3. Question 3

    20 cm³ of 0.5 mol/dm³ sodium hydroxide is neutralised by 15 cm³ of hydrochloric acid. HCl + NaOH → NaCl + H₂O. What is the concentration of the acid, in mol/dm³? Give 3 decimal places.

    Answermol/dm³

  4. Question 4

    Titres (cm³): rough 22.76, run 1 21.71, run 2 21.61. Ignoring the rough titre, calculate the mean titre from the concordant results (within 0.10 cm³), in cm³. Give 2 decimal places.

    Answercm³

  5. Question 5

    Titres (cm³): rough 24.93, run 1 24.38, run 2 24.43. Ignoring the rough titre, calculate the mean titre from the concordant results (within 0.10 cm³), in cm³. Give 2 decimal places.

    Answercm³

  6. Question 6

    25 cm³ of 0.2 mol/dm³ sodium hydroxide is neutralised by 15 cm³ of hydrochloric acid. HCl + NaOH → NaCl + H₂O. What is the concentration of the acid, in mol/dm³? Give 3 decimal places.

    Answermol/dm³

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Answer key: Titration calculations

  1. 1. 23.42 cm³

  2. 2. C. to avoid adding too much past the end point

  3. 3. 0.667 mol/dm³

  4. 4. 21.66 cm³

  5. 5. 24.41 cm³

  6. 6. 0.333 mol/dm³