Chemistry • Year 11 • Module 2 • Lesson 17
Back Calculations & Unknown Concentrations
Cement the core vocabulary, the back-calculation pathway steps, and the concordant titre rule before moving to harder problems.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: back calculation, concordant titres, average titre, rough titre, primary standard, excess reagent method, flask volume, titre volume, mole ratio, significant figures. 10 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | A multi-step calculation that works backwards from experimental data (e.g. a precipitate mass or a titre volume) to find an unknown quantity such as concentration. | |
| 1.2 | Titre results that agree within 0.10 mL of each other; only these are averaged to find the final titre used in calculations. | |
| 1.3 | The mean of the concordant titre results; this value is used in the back-calculation formula n = c × V. | |
| 1.4 | The first trial titration, always discarded because students deliberately overshoot the endpoint to locate the approximate endpoint range. | |
| 1.5 | A substance of precisely known purity and composition used to prepare a solution of accurately known concentration, which then standardises the unknown solution. | |
| 1.6 | A technique where a known excess of one reagent is added; the unreacted excess is then titrated with a second standard solution to find the original unknown amount by subtraction. | |
| 1.7 | The fixed volume of the unknown solution placed in the conical flask (typically from a pipette); this is the volume used in c(unknown) = n ÷ V. | |
| 1.8 | The volume of standard solution dispensed from the burette to reach the endpoint; used to calculate moles of the standard via n = c × V. | |
| 1.9 | The ratio of coefficients between two substances in a balanced equation; used to convert moles of standard solution to moles of unknown solution. | |
| 1.10 | The number of meaningful digits in a measurement; intermediate values should not be rounded, only the final answer. |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
2.1 In a back-calculation titration, the volume of the unknown solution from the flask is used to calculate n(unknown) via n = c × V. T / F
2.2 The rough titre is always discarded because it is likely to be less accurate than subsequent titres. T / F
2.3 If the balanced equation is HCl + NaOH → NaCl + H2O, then n(NaOH) = n(HCl) regardless of volumes used. T / F
2.4 For the reaction H2SO4 + 2NaOH → Na2SO4 + 2H2O, n(H2SO4) = n(NaOH) ÷ 2. T / F
2.5 Concordant titres are results that are within 0.10 mL of each other; including a result outside this range would introduce a systematic error. T / F
2.6 In a gravimetric back-calculation, the unknown concentration is found by dividing the moles of precipitate by the volume of the solution that formed the precipitate. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)
Word bank:
burette · concordant · discard · flask · mole ratio · precipitate · standard · titre
In a back-calculation titration, the solution of known concentration (the ___________ solution) is placed in the ___________, and the unknown solution is placed in the conical ___________. The volume of standard solution used to reach the endpoint is called the ___________. The first titration is called the rough trial and must always be ___________. Only results that are within 0.10 mL of each other are called ___________ results and may be averaged. After finding moles of the standard solution, the ___________ is applied to find moles of the unknown. In gravimetric back-calculations, the mass of the ___________ replaces the titre volume as the starting point.
4. Function recall
Answer each question in 1–2 sentences using precise terms from the lesson. 8 marks (2 each)
4.1 Why must volumes be converted to litres before substituting into n = c × V in a back calculation?
4.2 What is the function of a primary standard in a back-calculation experiment, and why must it have high purity?
4.3 In a gravimetric back-calculation, why is the reagent that causes precipitation added in excess?
4.4 State the back-calculation rule for which volume is used to find c(unknown), and explain why using the other volume gives an incorrect answer.
5. Build a concept map
Draw labelled arrows between the six terms below to show how they connect in a back-calculation. Each arrow must carry a linking phrase (e.g. “is used to find”, “requires”, “is divided by”). Aim for at least 6 labelled arrows. 6 marks (1 per valid labelled arrow)
Supplied terms: average concordant titre · n(standard) · mole ratio · n(unknown) · flask volume · c(unknown).
Q1 — Term–definition match
1.1 back calculation • 1.2 concordant titres • 1.3 average titre • 1.4 rough titre • 1.5 primary standard • 1.6 excess reagent method • 1.7 flask volume • 1.8 titre volume • 1.9 mole ratio • 1.10 significant figures.
Q2 — True / false with correction
2.1 False. The flask volume is used to find c(unknown) via c = n ÷ V(flask). The formula n = c × V is used to find n(standard) by substituting the titre volume of the standard solution from the burette.
2.2 True. The rough titre is discarded; however, the reason is more precisely that students deliberately overshoot on the first run to locate the approximate endpoint, making the result systematically larger than the true titre. Award the mark for “less accurate” as well.
2.3 True. In a 1:1 reaction, n(NaOH) = n(HCl) at the equivalence point, regardless of the volumes used to reach that point.
2.4 True. The balanced equation shows a 2:1 ratio NaOH:H2SO4, so n(H2SO4) = n(NaOH) ÷ 2. This is correct.
2.5 True. Including a result outside the 0.10 mL range would increase the spread of the average, biasing the calculated concentration away from the true value.
2.6 True. In a gravimetric back-calculation, moles of precipitate is found from its mass and molar mass (n = m ÷ MM), then via the mole ratio n(unknown) is found, and finally c(unknown) = n(unknown) ÷ V(solution volume).
Q3 — Cloze paragraph
In order: standard / burette / flask / titre / discard / concordant / mole ratio / precipitate.
Q4.1 — Units in n = c × V
Concentration c is expressed in mol/L, so V must be in litres for the units to cancel correctly (mol = mol/L × L). If V is left in mL, the calculated n will be 1000 times too large, and every subsequent step in the back calculation will be wrong.
Q4.2 — Function of a primary standard
A primary standard provides a solution of precisely known concentration, which serves as the reference in the titration. High purity is essential because any impurity reduces the effective amount of active substance; if the mass weighed out includes impurities, the calculated moles and hence the computed c(unknown) will be incorrect.
Q4.3 — Why excess precipitating reagent is used
The precipitating reagent must be in excess to ensure all of the unknown ion is completely precipitated. If it is not in excess, some unknown ion remains in solution unreacted, the precipitate mass is less than it should be, and the calculated c(unknown) will be an underestimate.
Q4.4 — Volume rule for c(unknown)
The flask volume (the fixed pipette volume of the unknown solution, e.g. 25.0 mL) is used: c(unknown) = n(unknown) ÷ V(flask). Using the titre volume instead would give the concentration of the unknown if it had been in the burette — this is not what was measured — and would produce an answer off by the ratio V(flask)/V(titre).
Q5 — Sample concept map
Correct maps should include arrows such as:
- average concordant titre — is multiplied by c(standard) to find → n(standard)
- n(standard) — is multiplied/divided by → mole ratio
- mole ratio — applied to n(standard) gives → n(unknown)
- n(unknown) — divided by → flask volume
- flask volume — divides n(unknown) to give → c(unknown)
- c(unknown) — is the answer found from → n(unknown)
Award 1 mark per valid labelled arrow (minimum 6, maximum 6 marked).