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.
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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?
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📚 Core Content
Wrong: A catalyst increases the yield of products in an equilibrium reaction.
Right: A catalyst speeds up both forward and reverse reactions equally; it does not change equilibrium yield.
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".
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.
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.
🧮 Worked Examples
🧪 Activities
1 25.0 mL NaOH titrated with 0.150 mol/L HCl. Average titre = 20.0 mL. Find [NaOH].
2 20.0 mL H₂SO₄ titrated with 0.100 mol/L NaOH. Average titre = 16.0 mL. Find [H₂SO₄]. (H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O)
3 Titres: Rough = 21.0, T1 = 22.5, T2 = 22.6, T3 = 22.4 mL. (a) Which are concordant? (b) Calculate average titre.
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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.
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✍️ Short Answer
6. 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]. 5 MARKS
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7. 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) 4 MARKS
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8. 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. 6 MARKS
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9. 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. 4 MARKS
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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.
Q6 (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/LQ7 (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.
Q8 (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 (using standard HSC criterion of ≤0.2 mL). 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.
Q9 (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.
Put your knowledge of back-titration calculations and determining unknown concentrations to the test. Answer correctly to deal damage — get it wrong and the boss hits back. Pool: lessons 1–17.
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