The shape of the lesson
By the end of the lesson, learners can
- Concentration and dilution: Every solution calculation starts with n = c × V.
- The titration: Run the titration yourself: add acid from the burette, slow to single drops as the indicator starts to change, and record the titre.
- Back titration: When a solid reacts too slowly or is insoluble, add a known excess of acid, let it react completely, then titrate the acid that is left over.
- cm3 → dm3: divide by 1000.
- Known solution first: moles, then the equation's ratio.
- Scale up if the titrated sample was a fraction of a larger volume.
- Mean of concordant titres only, within 0.10 cm3.
One picture every idea returns to
Concentration c = n ÷ V, in mol dm−3 with V in dm3. So n = c × V, remembering that V in cm3 must be divided by 1000. Concentration in g dm−3 = c × M. In a titration: find the moles of the solution of known concentration, use the mole ratio from the equation, then divide by the volume of the other solution. Use the mean of concordant titres, those within 0.10 cm3 of each other.
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, Moles in a measured volume.. Concentration tells you how many moles are 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: Concentration and dilution | Every solution calculation starts with n = c × V. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The titration | Run the titration yourself: add acid from the burette, slow to single drops as the indicator starts to change, and record the titre. 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: Back titration | When a solid reacts too slowly or is insoluble, add a known excess of acid, let it react completely, then titrate the acid that is left over. | 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, 18 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 Back titration, 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 |
|---|---|
| Two slips cost the most marks: using a volume in cm 3 as if it were in dm 3 (answers 1000 times too big), and ignoring the mole ratio when the acid is diprotic, as with H 2 SO 4. | Also leave out the rough titre and any titre not within 0.10 cm 3 of the others when you take the mean. |
Solutions and titrations: worksheet
6 questions, 18 marks. Written in the style of the papers; no past paper question is reproduced.
5.85 g of sodium chloride is dissolved in water and made up to 500 cm3. Calculate the concentration in mol dm−3. (Mr: NaCl 58.5)
25.0 cm3 of 0.0500 mol dm−3 sodium carbonate solution is titrated with hydrochloric acid; the mean titre is 22.70 cm3. Na2CO3 + 2HCl → 2NaCl + H2O + CO2. Calculate the concentration of the acid.
1.43 g of hydrated sodium carbonate, Na2CO3·xH2O, is dissolved and made up to 250 cm3. 25.0 cm3 portions need 20.00 cm3 of 0.0500 mol dm−3 HCl. Na2CO3 + 2HCl → 2NaCl + H2O + CO2. Find x. (Mr: Na2CO3 106.0, H2O 18.0)
1.50 g of impure calcium carbonate is added to 40.0 cm3 of 1.00 mol dm−3 hydrochloric acid (an excess). The acid remaining needs 15.0 cm3 of 1.00 mol dm−3 sodium hydroxide. Calculate the percentage by mass of calcium carbonate in the sample. (Mr: CaCO3 100.1)
What volume of 0.100 mol dm−3 sodium hydroxide exactly neutralises 20.0 cm3 of 0.150 mol dm−3 sulfuric acid? H2SO4 + 2NaOH → Na2SO4 + 2H2O.
Explain why, in a titration, (a) the burette is rinsed with the acid before it is filled, (b) the acid is added dropwise near the end point, and (c) the titration is repeated until concordant titres are obtained.
Answer key and marking notes
- n = 5.85 ÷ 58.5 = 0.100 mol.
- c = 0.100 ÷ 0.500 = 0.200 mol dm−3.
- n(Na2CO3) = 0.0500 × 25.0/1000 = 1.25 × 10−3 mol.
- n(HCl) = 2 × 1.25 × 10−3 = 2.50 × 10−3 mol.
- c(HCl) = 2.50 × 10−3 ÷ 0.02270 = 0.110 mol dm−3.
- n(HCl) = 0.0500 × 0.02000 = 1.00 × 10−3 mol, so n(Na2CO3) in 25.0 cm3 = 5.00 × 10−4 mol.
- In 250 cm3: 10 × 5.00 × 10−4 = 5.00 × 10−3 mol.
- M = 1.43 ÷ 5.00 × 10−3 = 286 g mol−1.
- 18.0x = 286 − 106.0 = 180, so x = 10.
- HCl added = 1.00 × 0.0400 = 0.0400 mol; HCl left = n(NaOH) = 1.00 × 0.0150 = 0.0150 mol.
- HCl that reacted with CaCO3 = 0.0400 − 0.0150 = 0.0250 mol.
- CaCO3 + 2HCl, so n(CaCO3) = 0.0125 mol = 0.0125 × 100.1 = 1.25 g.
- Percentage = 1.25 ÷ 1.50 × 100 = 83.4%.
- n(H2SO4) = 0.150 × 0.0200 = 3.00 × 10−3 mol, so n(NaOH) = 6.00 × 10−3 mol.
- V = 6.00 × 10−3 ÷ 0.100 = 0.0600 dm3 = 60.0 cm3.
- (a) Any water left in the burette would dilute the acid and increase the titre.
- (b) Adding single drops means the end point is not overshot, so the titre is accurate.
- (c) Concordant results show the titre is reliable; their mean reduces the effect of random errors.
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.
25.0 cm3 of sodium hydroxide solution is exactly neutralised by 18.75 cm3 of 0.100 mol dm−3 hydrochloric acid. NaOH + HCl → NaCl + H2O. What is the concentration of the sodium hydroxide?
- A0.0750 mol dm−3
- B0.133 mol dm−3
- C0.0375 mol dm−3
- D75.0 mol dm−3
What volume of 2.00 mol dm−3 sulfuric acid must be diluted with water to make 500 cm3 of 0.100 mol dm−3 sulfuric acid?
- A25.0 cm3
- B50.0 cm3
- C0.0250 cm3
- D475 cm3
3.16 g of potassium manganate(VII), KMnO4 (Mr 158.0), is dissolved in water and made up to 500 cm3. What is the concentration of the solution?
- A0.0400 mol dm−3
- B0.0200 mol dm−3
- C0.0100 mol dm−3
- D4.00 × 10−5 mol dm−3
25.0 cm3 portions of 0.0500 mol dm−3 sodium carbonate are titrated with hydrochloric acid. Na2CO3 + 2HCl → 2NaCl + H2O + CO2. The burette readings are shown. What is the concentration of the hydrochloric acid?
- A0.111 mol dm−3
- B0.0553 mol dm−3
- C0.109 mol dm−3
- D0.0452 mol dm−3
In an acid-alkali titration, which piece of apparatus is used to transfer exactly 25.0 cm3 of the alkali into the conical flask?
- Aa volumetric pipette with a filler
- Ba 50 cm3 measuring cylinder
- Ca 250 cm3 volumetric flask
- Da beaker marked at 25 cm3
What volume of 0.200 mol dm−3 hydrochloric acid reacts exactly with 0.530 g of anhydrous sodium carbonate, Na2CO3 (Mr 106.0)? Na2CO3 + 2HCl → 2NaCl + H2O + CO2
- A50.0 cm3
- B25.0 cm3
- C12.5 cm3
- D0.0500 cm3
0.500 g of impure magnesium oxide is added to 50.0 cm3 of 0.500 mol dm−3 hydrochloric acid (an excess). MgO + 2HCl → MgCl2 + H2O. The acid left over needs 21.60 cm3 of 0.250 mol dm−3 sodium hydroxide for neutralisation. The impurities do not react. What is the percentage by mass of MgO in the sample? (Ar: O 16.0, Mg 24.3)
- A79.0%
- B158%
- C21.8%
- D101%
3.15 g of hydrated ethanedioic acid, H2C2O4·xH2O, is dissolved and made up to 250 cm3. A 25.0 cm3 portion needs 25.00 cm3 of 0.200 mol dm−3 sodium hydroxide. H2C2O4 + 2NaOH → Na2C2O4 + 2H2O. What is the value of x? (Ar: H 1.0, C 12.0, O 16.0)
- A2
- B7
- C65
- D30
Quiz key
n(HCl) = 0.100 × 18.75/1000 = 1.875 × 10−3 mol = n(NaOH) (1 : 1). c = 1.875 × 10−3 ÷ 0.0250 = 0.0750 mol dm−3. It must be less than 0.100 because a smaller volume of acid neutralised a larger volume of alkali. 0.133 mol dm−3 puts the volumes the wrong way round. 0.0375 mol dm−3 halves the answer as if the acid were diprotic. 75.0 mol dm−3 converts one volume to dm3 but not the other (18.75 × 0.100 ÷ 0.0250).
Moles needed = 0.100 × 0.500 = 0.0500 mol. Volume of stock = 0.0500 ÷ 2.00 = 0.0250 dm3 = 25.0 cm3. (Or c1V1 = c2V2: 2.00 × V = 0.100 × 500.) 50.0 cm3 doubles the answer because sulfuric acid is diprotic, but dilution does not involve a reaction ratio. 0.0250 cm3 is the volume in dm3 with the wrong unit. 475 cm3 is the volume of water added, not the volume of acid.
n = 3.16 ÷ 158.0 = 0.0200 mol; c = 0.0200 ÷ 0.500 = 0.0400 mol dm−3. 0.0200 mol dm−3 is the amount in moles, forgetting that the volume is only half a dm3. 0.0100 mol dm−3 multiplies by 0.500 instead of dividing. 4.00 × 10−5 mol dm−3 divides by 500 without converting cm3 to dm3.
Titres 1 and 2 (22.65 and 22.55 cm3) are concordant; the rough titre and titre 3 are not. Mean = 22.60 cm3. n(Na2CO3) = 0.0500 × 0.0250 = 1.25 × 10−3 mol, so n(HCl) = 2.50 × 10−3 mol and c = 2.50 × 10−3 ÷ 0.02260 = 0.111 mol dm−3. 0.0553 mol dm−3 ignores the 1 : 2 ratio. 0.109 mol dm−3 averages all four titres, including the rough one. 0.0452 mol dm−3 ignores the ratio and puts the volumes the wrong way round (0.0500 × 22.60 ÷ 25.0).
A volumetric (bulb) pipette is calibrated to deliver one fixed volume, here 25.0 cm3, with high accuracy. A measuring cylinder is far less precise. A volumetric flask is used to make up a standard solution to a fixed total volume, not to deliver a portion. Beaker markings are only rough guides.
n(Na2CO3) = 0.530 ÷ 106.0 = 5.00 × 10−3 mol; n(HCl) = 1.00 × 10−2 mol; V = 0.0100 ÷ 0.200 = 0.0500 dm3 = 50.0 cm3. 25.0 cm3 uses a 1 : 1 ratio. 12.5 cm3 applies the ratio the wrong way (halving instead of doubling). 0.0500 cm3 is the value in dm3 with the wrong unit.
Acid added = 0.500 × 0.0500 = 0.0250 mol. Acid left = n(NaOH) = 0.250 × 0.02160 = 0.00540 mol. Acid used by MgO = 0.0250 − 0.00540 = 0.0196 mol, so n(MgO) = 0.00980 mol, mass = 0.00980 × 40.3 = 0.395 g, and 0.395 ÷ 0.500 × 100 = 79.0%. 158% forgets the 1 : 2 ratio. 21.8% works from the excess acid as if it had reacted with the oxide. 101% uses all the acid added and ignores the back titration; a result above 100% signals the error.
n(NaOH) = 0.200 × 0.02500 = 5.00 × 10−3 mol, so n(acid) in 25.0 cm3 = 2.50 × 10−3 mol and in 250 cm3 = 0.0250 mol. M = 3.15 ÷ 0.0250 = 126 g mol−1. Anhydrous H2C2O4 = 90.0, so water = 36.0 and x = 36.0 ÷ 18.0 = 2. 7 divides the whole 126 by 18.0 without first subtracting the anhydrous acid. 65 forgets to scale from 25.0 cm3 to 250 cm3 (M = 1260). 30 forgets both the 1 : 2 ratio and the scaling (M = 630).
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.