Back Calculations & Unknown Concentrations
In quantitative analysis chemists often know the product and need to work backwards — using a precipitate mass or titration result to find the concentration of an unknown solution. This appears in nearly every HSC Chemistry exam.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
In a titration, the standard solution is in the burette and the unknown solution is in the conical flask. If you know the concentration of the standard and the volume it takes to reach the endpoint, how would you work backwards to find the concentration of the unknown — and which volume goes into the formula c = n ÷ V for each substance?
Titration: n(standard) = c × V → ÷ ratio → n(unknown) → c = n ÷ V(flask)
Gravimetric: m(precipitate) → n = m ÷ MM → ÷ ratio → n(unknown) → c = n ÷ V
Average titre: discard rough; average concordant results (within 0.10 mL)
⚠️ Volume in the flask (aliquot) is used to find c(unknown). Volume from the burette (titre) is used to find n(standard). Swapping these is the most common error in this lesson.
Key facts
- Back-calc starts with known product/standard, finds unknown
- Concordant = titres within 0.10 mL of each other
- Rough titre is always discarded
- Primary standard: known mass → standardise unknown solution
Concepts
- Why excess reagent guarantees complete reaction in gravimetric analysis
- The flask-vs-burette volume distinction
- How non-1:1 ratios affect back calculations
Skills
- Find c(unknown) from titration data — 1:1 and non-1:1 ratios
- Find c(unknown) from precipitate mass in gravimetric data
- Calculate average concordant titre correctly
A back calculation works in the opposite direction to a forward stoichiometry problem. Instead of "given reactant, find product", you work "given product, find unknown reactant".
Concordant titres
Multiple titrations are performed to ensure reliability. The first trial (rough) is always discarded. Remaining results within 0.10 mL of each other are concordant — average those only.
Primary standard back calculation
A known mass of a pure primary standard is dissolved to a known volume, giving a precisely known concentration. This standard is then titrated against the unknown solution.
Back-calculation uses a known starting material (primary standard or weighed product) to work back to an unknown concentration. Pathway: m ÷ MM → n(standard) → × mole ratio → n(unknown) → c = n ÷ V(flask). Flask volume gives c(unknown); titre volume gives n(standard) — never swap. Discard the rough titre; average only concordant titres (within 0.10 mL).
Pause — copy the highlighted pathway into your book before moving on.
Did you get this? True or false: the flask (aliquot) volume is used to calculate c(unknown), while the titre (burette) volume is used to find n(standard).
Quick check: Why is the rough titre always discarded from the average?
Step through concordant titres, mole ratio, and unknown concentration.
We just saw the titration back-calculation — starting from a known standard to find an unknown concentration. That raises a question: what if there is no solution to titrate against — only a weighed solid? This card answers it → gravimetric back-calculation uses precipitate mass as the starting point.
In a gravimetric back calculation, you start from the mass of a precipitate. Add a large excess of precipitating agent so that all of the target ion is captured in the solid. Then weigh the dry precipitate, convert mass → moles, apply the mole ratio, and divide by the original flask volume to get the unknown concentration.
The pathway: m(precipitate) → n = m ÷ MM → ÷ ratio → n(unknown) → c = n ÷ V(flask).
Gravimetric back-calculation: m(precipitate) ÷ MM → n(precipitate) → × mole ratio → n(unknown) → c = n ÷ V(flask). Precipitating agent must be in excess to capture all target ions. Dry the precipitate to constant mass before weighing. Use the volume of the original unknown solution, not the total mixed volume.
Pause — copy the highlighted pathway into your book before moving on.
Fill the blanks: drag each token into the matching blank.
Gravimetric back calc starts from the ___ of dry precipitate. The precipitating agent must be in ___ so all target ion reacts. Convert m → n, apply the ___, then divide by the ___ volume to get c(unknown).
Worked examples · reveal as you go
25.0 mL of NaOH (unknown concentration) is titrated against 0.100 mol/L HCl. Concordant titres: 22.3, 22.4, 22.3 mL. Find [NaOH]. HCl + NaOH → NaCl + H₂O.
25.0 mL of H₂SO₄ (unknown) is titrated with 0.200 mol/L NaOH. Average titre = 18.50 mL. Find [H₂SO₄]. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
0.530 g of Na₂CO₃ is dissolved to 100.0 mL. 25.0 mL aliquots are titrated against HCl. Average titre = 24.5 mL. Find [HCl]. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. (Na = 22.990, C = 12.011, O = 15.999)
25.0 mL of BaCl₂ solution reacts with excess Na₂SO₄. 0.699 g of BaSO₄ precipitate forms. Find [BaCl₂]. BaCl₂ + Na₂SO₄ → BaSO₄↓ + 2NaCl. (Ba = 137.33, S = 32.06, O = 15.999)
Two truths, one lie — about back-calculating an unknown concentration from a titration. Pick the lie.
Spot the slip-up — a student finds c(NaOH) from a titration with 0.1000 mol L⁻¹ HCl. The 25.00 mL NaOH aliquot needed an average titre of 22.50 mL of the acid. One line has an error — click it.
- Mole ratio: HCl : NaOH = 1 : 1 (from balanced equation)
- n(HCl) = 0.1000 × 0.02250 = 2.250 × 10⁻³ mol
- n(NaOH) = 2.250 × 10⁻³ mol (1:1 ratio)
- c(NaOH) = 2.250 × 10⁻³ ÷ 0.02250 = 0.1000 mol L⁻¹
- ∴ c(NaOH) ≈ 0.1000 mol L⁻¹
Common errors · the 3 traps that cost marks
Using titre volume instead of flask volume for c(unknown)
The unknown is in the flask (25.0 mL aliquot). Its concentration = n ÷ V(flask). The titre volume from the burette belongs to the standard solution. Swapping these gives an answer off by the ratio of the two volumes.
Fix: Label clearly — Flask = unknown. Burette = standard. c(unknown) = n ÷ V(flask). Never use titre volume for c(unknown).
Forgetting the non-1:1 ratio
H₂SO₄ + 2NaOH is the classic trap. If you skip the 2:1 ratio and assume 1:1, your answer for [H₂SO₄] is exactly double the correct value. This is reliably tested every few years in the HSC.
Fix: Write the balanced equation and circle the ratio before calculating. n(H₂SO₄) = n(NaOH) ÷ 2, not n(NaOH) × 1.
Including the rough titre in the average
The rough titre is always larger than the true endpoint because students deliberately overshoot on the first run to find the approximate endpoint range. It must always be discarded. Including it biases the average upward, giving a systematic error in c(unknown).
Fix: Identify and discard the rough titre first. Average only concordant results (within 0.10 mL of each other).
Quick-fire practice · 5 reps +2 XP per reveal
25.0 mL NaOH titrated with 0.150 mol/L HCl. Average titre = 20.0 mL. Find [NaOH].
20.0 mL H₂SO₄ titrated with 0.100 mol/L NaOH. Average titre = 16.0 mL. Find [H₂SO₄]. (H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O)
Titres: Rough = 21.0, T1 = 22.5, T2 = 22.6, T3 = 22.4 mL. (a) Which are concordant? (b) Calculate average titre.
30.0 mL of AgNO₃ reacts with excess NaCl. 0.860 g of AgCl precipitate forms. Find [AgNO₃]. AgNO₃ + NaCl → AgCl↓ + NaNO₃. (MM(AgCl) = 143.32)
0.265 g of Na₂CO₃ (MM = 105.99) is dissolved to 250 mL. 25.0 mL aliquots are titrated against HCl. Average titre = 22.1 mL. Find [HCl]. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂.
At the start of this lesson, you thought about how to work backwards from a titration to find the concentration of an unknown solution, and which volume to use for each substance.
The key is: n(standard) = c(standard) × V(titre, from burette). Apply the mole ratio. Then c(unknown) = n(unknown) ÷ V(aliquot, from flask). The burette titre is used for the standard solution; the flask (aliquot) volume is used for the unknown. Swapping these is the most common error. In gravimetric back-calculations, start from the precipitate mass instead of a titre.
Pick your answer, then rate your confidence — that tells the system what to drill next.
Q1. 0.424 g of Na₂CO₃ (MM = 105.99) is dissolved to 100.0 mL. 25.0 mL aliquots are titrated against HCl. Titres: Rough = 19.8, T1 = 21.3, T2 = 21.4, T3 = 21.2 mL. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. (a) Calculate the average concordant titre. (b) Find [HCl].
Q2. 40.0 mL of AgNO₃ solution reacts with excess NaCl. 1.148 g of AgCl precipitate forms. (a) Find [AgNO₃]. (b) Explain why NaCl must be in excess. AgNO₃ + NaCl → AgCl↓ + NaNO₃. (Ag = 107.87, Cl = 35.453)
Q3. A student standardises an NaOH solution using oxalic acid dihydrate (H₂C₂O₄·2H₂O, MM = 126.07) as a primary standard. They dissolve 0.756 g in 250.0 mL water. 25.0 mL aliquots are titrated against NaOH. The equation is: H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O. Titres: Rough = 24.5, T1 = 25.8, T2 = 25.7, T3 = 25.9 mL. (a) Identify the concordant titres and calculate the average. (b) Find c(NaOH). (c) State one reason why oxalic acid dihydrate is suitable as a primary standard.
Q4. A student performs a back-calculation titration but accidentally reads the flask volume (25.0 mL) as 0.250 mL when inputting into their calculation. They use this value as V(unknown) in c = n ÷ V. (a) Describe the effect this error has on their calculated concentration — will it be too high or too low, and by what factor? (b) Explain how a chemist could detect this error without redoing the experiment.
📖 Comprehensive answers (click to reveal)
Multiple choice — drill bank
1. B — c(unknown) = n ÷ V(flask). The flask volume is the aliquot of the unknown solution.
2. C — n(HCl) = 0.200 × 0.0150 = 0.00300; ratio 1:1; c(NaOH) = 0.00300 ÷ 0.0250 = 0.120 mol/L.
3. A — n(NaOH) = 0.100 × 0.020 = 0.00200; ratio 2:1; n(H₂SO₄) = 0.00100; c = 0.00100 ÷ 0.025 = 0.0400 mol/L.
4. D — Discard rough (22.1). Average T1, T2, T3 = (23.4+23.5+23.3)÷3 = 23.4 mL.
5. B — n = 0.212 ÷ 105.99 = 2.000×10⁻³; c = 2.000×10⁻³ ÷ 0.250 = 0.00800 mol/L.
6. C — The rough titre (18.4) must be discarded. T1 = 19.8, T2 = 19.6, T3 = 20.1. Range = 20.1 − 19.6 = 0.5 mL, which exceeds 0.1 mL — not strictly concordant. In practice, students should ideally repeat to get truly concordant results. If forced to average T1–T3: (19.8+19.6+20.1)÷3 = 19.83 mL. The student's inclusion of the rough titre in their average is an error regardless.
7. A — In a back-calculation titration, the unknown solution (acetic acid) goes in the flask (aliquot) and the standard solution (standardised NaOH) goes in the burette. The titre of NaOH gives n(NaOH), then the mole ratio gives n(acetic acid), and dividing by the flask volume gives c(acetic acid). Option B has the burette and flask swapped.
Short answer model answers
Q1 (5 marks):
(a) Discard rough (19.8). T1=21.3, T2=21.4, T3=21.2 — all within 0.2 mL, concordant. Average = (21.3+21.4+21.2)÷3 = 21.3 mL.
(b) n(Na₂CO₃) = 0.424 ÷ 105.99 = 4.001×10⁻³; c = 4.001×10⁻³ ÷ 0.100 = 0.04001 mol/L. n(aliquot) = 0.04001 × 0.0250 = 1.000×10⁻³; ratio 1:2; n(HCl) = 2.000×10⁻³. c(HCl) = 2.000×10⁻³ ÷ 0.02130 = 0.0939 mol/L.
Q2 (4 marks):
(a) MM(AgCl) = 143.32; n = 1.148 ÷ 143.32 = 8.010×10⁻³; ratio 1:1; n(AgNO₃) = 8.010×10⁻³. c(AgNO₃) = 8.010×10⁻³ ÷ 0.0400 = 0.200 mol/L.
(b) If NaCl is not in excess, some Ag⁺ ions would remain in solution unreacted. The precipitate mass would be less than the amount corresponding to all the AgNO₃ present. The calculated n(AgNO₃) would be too small, giving an underestimate of c(AgNO₃). Excess NaCl ensures all Ag⁺ is precipitated, so the mass of AgCl accurately reflects the full quantity of AgNO₃.
Q3 (6 marks):
(a) Discard rough (24.5). T1 = 25.8, T2 = 25.7, T3 = 25.9. Range = 25.9 − 25.7 = 0.2 mL. These are within 0.2 mL — concordant. Average = (25.8+25.7+25.9) ÷ 3 = 25.8 mL.
(b) n(H₂C₂O₄) = 0.756 ÷ 126.07 = 5.997×10⁻³ mol; c = 5.997×10⁻³ ÷ 0.250 = 0.02399 mol/L. n(aliquot) = 0.02399 × 0.0250 = 5.997×10⁻⁴; ratio 1:2; n(NaOH) = 1.199×10⁻³. c(NaOH) = 1.199×10⁻³ ÷ 0.02580 = 0.04647 mol/L ≈ 0.0465 mol/L.
(c) Oxalic acid dihydrate is suitable as a primary standard because it has a high molar mass (126.07 g/mol), which reduces the relative error in weighing; it is stable, non-hygroscopic, available in high purity, and does not absorb CO₂ or water vapour from the air appreciably during weighing.
Q4 (4 marks):
(a) If 0.250 mL is used instead of 25.0 mL, the denominator is 100 times too small (0.000250 L instead of 0.025 L). Since c = n ÷ V, a smaller V gives a larger c. The calculated concentration will be 100× too high (a factor of 100 error).
(b) A chemist could detect this error by: (i) checking the units — 0.250 mL is implausibly small for a pipette measurement; a standard pipette delivers 25.0 mL not 0.250 mL; (ii) comparing the calculated concentration with expected values for a typical solution (e.g., a concentration of 40 mol/L for NaOH is physically impossible, as NaOH is only soluble to about 25 mol/L); (iii) checking the recorded experimental data against the flask volume stated in the method.
Five timed questions on back calculations and unknown concentrations. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
⚔ Enter the arenaClimb platforms, hit checkpoints, and answer back-calc questions. Quick recall, lighter than the boss.
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