The shape of the lesson
By the end of the lesson, learners can
- Two units of concentration: Core: know that concentration is measured in g/dm 3 and in mol/dm 3.
- Inside a titration: Run a titration in the simulator: fill the burette with acid, pipette 25.0 cm 3 of alkali into the flask, add indicator, and open the tap until the colour just changes.
- When the ratio is not 1: 1: Sulfuric acid has two hydrogen ions per formula, so it reacts with sodium hydroxide in a 1: 2 ratio.
- cm3 ÷ 1000 = dm3 before any calculation.
- moles = concentration × volume in dm3.
- Ratio from the balanced equation; H2SO4 reacts 1 : 2 with NaOH.
- g/dm3 = mol/dm3 × Mr.
One picture every idea returns to
Concentration can be given in g/dm3 or in mol/dm3. Moles of solute = concentration in mol/dm3 × volume in dm3, and to change cm3 into dm3 you divide by 1000. In a titration you find the moles of the solution you know completely, use the equation's ratio, and then divide by the volume of the other solution in dm3.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, How much is dissolved.. Concentration tells you how much solute is in each cubic decimetre of solution. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Two units of concentration | Core: know that concentration is measured in g/dm 3 and in mol/dm 3. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Inside a titration | Run a titration in the simulator: fill the burette with acid, pipette 25.0 cm 3 of alkali into the flask, add indicator, and open the tap until the colour just changes. Project tab 6 of the lab (Stoichiometry lab, titration) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: When the ratio is not 1: 1 | Sulfuric acid has two hydrogen ions per formula, so it reacts with sodium hydroxide in a 1: 2 ratio. | Numbered Heads Together |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 16 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in When the ratio is not 1: 1, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Burettes and pipettes measure in cm 3, but concentrations are per dm 3. | Forgetting to divide by 1000 makes 25 cm 3 of 0.2 mol/dm 3 look like 5 mol instead of 0.005 mol. |
| Do not skip the ratio. | Treating the sulfuric acid titration as 1: 1 gives 0.0025 ÷ 0.0125 = 0.20 mol/dm 3, double the right answer. Write the balanced equation first and read the ratio from it. |
Concentration and titrations: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
State two units in which the concentration of a solution can be given, and say what each one measures.
Extended. Calculate the mass of sodium hydroxide needed to make 250 cm3 of a 0.2 mol/dm3 solution. (Ar: H 1, O 16, Na 23)
Extended. 25.0 cm3 of potassium hydroxide solution is neutralised by 22.5 cm3 of 0.100 mol/dm3 hydrochloric acid. KOH + HCl → KCl + H2O. Calculate the concentration of the potassium hydroxide in mol/dm3 and in g/dm3. (Ar: H 1, O 16, K 39)
Extended. 20.0 cm3 of 0.050 mol/dm3 sulfuric acid exactly neutralises 25.0 cm3 of sodium hydroxide solution. H2SO4 + 2NaOH → Na2SO4 + 2H2O. Calculate the concentration of the sodium hydroxide.
A student records these titres: rough 23.40 cm3, then 22.60 cm3, 22.95 cm3 and 22.65 cm3. Choose the titres to use and calculate the mean titre.
Extended. 0.40 g of sodium hydroxide is dissolved in water. Calculate the volume of 0.50 mol/dm3 hydrochloric acid, in cm3, needed to neutralise it. (Ar: H 1, O 16, Na 23)
Answer key and marking notes
- g/dm3: the mass of solute in grams dissolved in each 1 dm3 of solution.
- mol/dm3: the number of moles of solute dissolved in each 1 dm3 of solution.
- Moles = 0.250 dm3 × 0.2 mol/dm3 = 0.05 mol.
- Mr(NaOH) = 40, so mass = 0.05 × 40 = 2.0 g.
- Moles of HCl = 0.0225 × 0.100 = 0.00225 mol.
- Ratio 1 : 1, so KOH = 0.00225 mol.
- Concentration = 0.00225 ÷ 0.0250 = 0.090 mol/dm3.
- Mr(KOH) = 56, so 0.090 × 56 = 5.04 g/dm3.
- Moles of H2SO4 = 0.0200 × 0.050 = 0.0010 mol.
- Ratio 1 : 2, so NaOH = 2 × 0.0010 = 0.0020 mol.
- Concentration = 0.0020 ÷ 0.0250 = 0.080 mol/dm3.
- Use the concordant titres, those within 0.10 cm3 of each other: 22.60 and 22.65 cm3. Ignore the rough titre and 22.95 cm3.
- Mean = (22.60 + 22.65) ÷ 2 = 22.625, which is 22.63 cm3 to 2 decimal places.
- Moles of NaOH = 0.40 ÷ 40 = 0.010 mol.
- Ratio 1 : 1, so 0.010 mol of HCl is needed.
- Volume = 0.010 ÷ 0.50 = 0.020 dm3 = 20 cm3.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
A bottle is labelled "sodium chloride solution, 20 g/dm3". What does this mean?
- AEach 1 dm3 of the solution contains 20 g of dissolved sodium chloride
- B20 g of sodium chloride was dissolved in 1 dm3 of water
- CEach 1 dm3 of the solution has a mass of 20 g
- DEach 1 dm3 of the solution contains 20 mol of sodium chloride
In a titration, which piece of apparatus is used to measure exactly 25.0 cm3 of alkali into the conical flask?
- Aa volumetric pipette
- Ba burette
- Ca measuring cylinder
- Da beaker
Solution P contains 0.5 mol of solute in 1 dm3. Solution Q contains 0.5 mol of the same solute in 2 dm3. Which statement is correct?
- AP is twice as concentrated as Q
- BP and Q have the same concentration because they contain the same number of moles
- CQ is more concentrated because it has a larger volume
- DQ is twice as concentrated as P
A solution of hydrochloric acid contains 7.3 g/dm3 of HCl. What is its concentration in mol/dm3? (Ar: H 1, Cl 35.5)
- A0.20 mol/dm3
- B266 mol/dm3
- C5.0 mol/dm3
- D0.020 mol/dm3
How many moles of solute are in 150 cm3 of a 0.40 mol/dm3 solution?
- A0.060 mol
- B60 mol
- C2.67 mol
- D0.00267 mol
25.0 cm3 of 0.050 mol/dm3 sodium carbonate solution is neutralised by 20.0 cm3 of hydrochloric acid. Na2CO3 + 2HCl → 2NaCl + H2O + CO2. What is the concentration of the hydrochloric acid?
- A0.125 mol/dm3
- B0.0625 mol/dm3
- C0.0313 mol/dm3
- D0.080 mol/dm3
What volume of 0.20 mol/dm3 sulfuric acid exactly neutralises 30.0 cm3 of 0.10 mol/dm3 potassium hydroxide? H2SO4 + 2KOH → K2SO4 + 2H2O
- A7.5 cm3
- B15.0 cm3
- C30.0 cm3
- D0.0075 cm3
50.0 cm3 of 2.0 mol/dm3 hydrochloric acid is put in a volumetric flask and made up to 250 cm3 with distilled water. What is the concentration of the diluted acid?
- A0.40 mol/dm3
- B0.50 mol/dm3
- C10 mol/dm3
- D2.0 mol/dm3
Quiz key
g/dm3 is the mass of solute in each 1 dm3 of solution. Dissolving 20 g in 1 dm3 of water does not give exactly 1 dm3 of solution, so the volume must be that of the solution, not the water. The unit is grams of solute, not the mass of the solution, and mol/dm3 would be needed for moles.
A volumetric pipette delivers one fixed volume, such as 25.0 cm3, very accurately. The burette holds the other solution and measures the variable volume added (the titre). A measuring cylinder and a beaker are far less accurate.
P has 0.5 mol in each dm3; Q spreads the same 0.5 mol through 2 dm3, so it has 0.25 mol in each dm3. P is twice as concentrated. The same number of moles in a larger volume means a lower concentration, not the same or a higher one.
Mr(HCl) = 36.5. mol/dm3 = g/dm3 ÷ Mr = 7.3 ÷ 36.5 = 0.20 mol/dm3. 266 multiplies by Mr instead of dividing; 5.0 divides the Mr by the concentration; 0.020 slips a factor of 10.
150 cm3 = 150 ÷ 1000 = 0.150 dm3. moles = 0.40 × 0.150 = 0.060 mol. 60 mol forgets to divide by 1000; 2.67 divides the concentration by the volume in dm3; 0.00267 divides it by the volume in cm3.
Na2CO3 = 0.0250 × 0.050 = 0.00125 mol. Each mol needs 2 mol of HCl, so HCl = 0.0025 mol. Concentration = 0.0025 ÷ 0.0200 = 0.125 mol/dm3. 0.0625 treats the ratio as 1 : 1; 0.0313 halves instead of doubling; 0.080 comes from scaling the concentration by the volume ratio the wrong way (0.050 × 20.0 ÷ 25.0, then doubled).
KOH = 0.0300 × 0.10 = 0.0030 mol. Ratio H2SO4 : KOH = 1 : 2, so H2SO4 = 0.0015 mol. Volume = 0.0015 ÷ 0.20 = 0.0075 dm3 = 7.5 cm3. 15.0 cm3 ignores the ratio; 30.0 cm3 uses it the wrong way round; 0.0075 is the volume in dm3, not cm3.
Moles of HCl = 0.0500 × 2.0 = 0.10 mol, and this does not change. Final volume 250 cm3 = 0.250 dm3, so concentration = 0.10 ÷ 0.250 = 0.40 mol/dm3. 0.50 divides by the 200 cm3 of water added instead of the total volume; 10 multiplies by the dilution factor instead of dividing; 2.0 forgets that the solution has been diluted.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.