Chemistry • Year 11 • Module 2 • Lesson 17
Back Calculations & Unknown Concentrations
Apply the back-calculation method to real titration data, data tables, and cause-and-effect reasoning about experimental errors — including the classic flask-vs-burette mix-up.
1. Interpret experimental data — comparing student titration results
Three students each performed a titration of NaOH (unknown) against 0.100 mol/L HCl (standard). Each placed 25.00 mL of NaOH in the conical flask. Their titre results are shown below. 9 marks
| Student | Rough titre (mL) | Titre 1 (mL) | Titre 2 (mL) | Titre 3 (mL) | Concordant titres | Average titre (mL) | c(NaOH) (mol/L) |
|---|---|---|---|---|---|---|---|
| A | 18.2 | 20.1 | 20.2 | 20.1 | |||
| B | 19.5 | 20.4 | 20.8 | 20.5 | |||
| C | 20.3 | 20.4 | 20.3 | 20.5 |
1.1 Complete the “Concordant titres”, “Average titre”, and “c(NaOH)” columns for each student. Show your working for Student A below. Balanced equation: HCl + NaOH → NaCl + H2O. 6 marks (1 per completed row of last 3 columns)
1.2 Student B’s concordant titres span 0.4 mL. Explain why strict concordance (within 0.10 mL) matters for the reliability of c(NaOH), and identify what systematic error could cause a wide spread of titre results. 2 marks
1.3 Student C’s rough titre (20.3 mL) is close to the other titres. A classmate suggests including it in the average. Evaluate this suggestion. 1 mark
2. Interpret graph — acid-base titration at ANSTO’s Lucas Heights facility
A quality-control chemist at ANSTO (Lucas Heights, NSW) standardises a batch of NaOH(aq) using 0.0500 mol/L H2SO4 as the standard. She records the pH of the NaOH solution in the flask as H2SO4 is added from the burette. The graph below shows her data. 8 marks
Figure 2. pH of NaOH(aq) in conical flask vs volume of 0.0500 mol/L H2SO4(aq) added from burette. Flask volume = 20.00 mL NaOH. Balanced equation: H2SO4 + 2NaOH → Na2SO4 + 2H2O. Illustrative data; adapted from typical HSC lab practice.
2.1 Using the graph, identify the equivalence point volume and explain what is occurring chemically at this point. 2 marks
2.2 Use the equivalence point volume from the graph to calculate c(NaOH). Show full working, including the mole ratio. 3 marks
2.3 A student reads the equivalence point as 26.0 mL instead of 25.0 mL. Calculate the percentage error this introduces in the final c(NaOH) and classify whether the error makes the result too high or too low. 3 marks
3. Cause-and-effect chain — tracing a volume mix-up error
A student performs a back-calculation titration but uses the titre volume (24.0 mL) in the formula c(unknown) = n ÷ V instead of the flask volume (25.0 mL). Trace the consequences by filling in the empty “effect” boxes. 5 marks
| Cause | Effect (fill in) |
|---|---|
| Student writes V = 24.0 mL = 0.0240 L instead of 0.0250 L in c = n ÷ V. | |
| The denominator V is smaller than the correct value. | |
| The calculated c(unknown) is larger than the true value. | |
| The student reports c(unknown) = 0.1042 mol/L instead of the true 0.1000 mol/L. |
Overall outcome — How does this systematic error affect the accuracy of the reported concentration, and how could a student detect it without redoing the experiment? 1 mark
4. Compare titration vs gravimetric back-calculation
Complete the two-column table below. For each feature, describe how it applies to each type of back-calculation. 8 marks (1 per cell)
| Feature | Titration back-calculation | Gravimetric back-calculation |
|---|---|---|
| What quantity is measured experimentally? | ||
| What formula converts that measurement to moles? | ||
| How is n(unknown) found from the measurement? | ||
| What precaution ensures all the unknown reacts? |
Q1.1 — Concordant titres and c(NaOH) table
Student A: Concordant: T1, T2, T3 (range 20.1–20.2 = 0.1 mL — borderline; accept T1 and T3 as strictly within 0.10 mL, with average 20.10 mL, or all three with average 20.13 mL).
Average titre = (20.1 + 20.2 + 20.1) ÷ 3 = 20.13 mL = 0.02013 L.
n(HCl) = 0.100 × 0.02013 = 2.013 × 10−3 mol. Ratio 1:1. n(NaOH) = 2.013 × 10−3 mol.
c(NaOH) = 2.013 × 10−3 ÷ 0.02500 = 0.0805 mol/L.
Student B: T1 (20.4) and T3 (20.5) are within 0.10 mL — these are concordant. T2 (20.8) is outside. Average of T1 and T3 = (20.4 + 20.5) ÷ 2 = 20.45 mL.
n(HCl) = 0.100 × 0.02045 = 2.045 × 10−3. c(NaOH) = 2.045 × 10−3 ÷ 0.02500 = 0.0818 mol/L.
Student C: T1 (20.4), T2 (20.3), T3 (20.5) — all within 0.20 mL; T2 and T1 within 0.10 mL; T1 and T3 within 0.10 mL. Accept all three as concordant (standard HSC criterion). Average = (20.4 + 20.3 + 20.5) ÷ 3 = 20.40 mL.
n(HCl) = 0.100 × 0.02040 = 2.040 × 10−3. c(NaOH) = 2.040 × 10−3 ÷ 0.02500 = 0.0816 mol/L.
Q1.2 — Why concordance matters (2 marks)
Concordant titres within 0.10 mL ensure the average titre is reproducible and not biased by outliers caused by endpoint-detection errors or inconsistent technique [1]. A wide spread (as in Student B’s 0.4 mL range) suggests a systematic error such as inconsistent indicator use, reading the burette at different eye levels, or variable addition speed near the endpoint, all of which reduce the reliability of the computed c(NaOH) [1].
Q1.3 — Including Student C’s rough titre (1 mark)
The suggestion is incorrect. The rough titre must always be discarded regardless of its numerical value, because its purpose was to locate the approximate endpoint — the student may have added acid rapidly, overshoot slightly, and stopped at a slightly different colour. Even if the value is close, its methodological origin means it cannot be assumed to have been obtained with the same care as subsequent titres [1].
Q2.1 — Equivalence point from graph (2 marks)
Equivalence point volume ≈ 25.0 mL (the inflection point of the steep pH drop, where pH = 7) [1]. At this point, the moles of H2SO4 added exactly equals the amount required to neutralise all the NaOH in the flask; no acid or base is in excess and the solution is neutral [1].
Q2.2 — Calculate c(NaOH) (3 marks)
n(H2SO4) = 0.0500 × 0.02500 = 1.250 × 10−3 mol [1].
Ratio H2SO4:NaOH = 1:2, so n(NaOH) = 1.250 × 10−3 × 2 = 2.500 × 10−3 mol [1].
c(NaOH) = 2.500 × 10−3 ÷ 0.02000 = 0.125 mol/L [1].
Q2.3 — Error analysis (3 marks)
If V = 26.0 mL is used: n(H2SO4) = 0.0500 × 0.02600 = 1.300 × 10−3 mol; n(NaOH) = 2.600 × 10−3 mol; c(NaOH) = 2.600 × 10−3 ÷ 0.02000 = 0.130 mol/L [1].
% error = |0.130 − 0.125| ÷ 0.125 × 100 = 4.0% [1].
The result is too high, because a larger titre volume gives a larger n(H2SO4) and hence a larger n(NaOH), making the calculated c(NaOH) an overestimate [1].
Q3 — Cause-and-effect chain
Row 1: The denominator in c = n ÷ V is 0.0240 L instead of 0.0250 L — 4.2% smaller than the correct value.
Row 2: Since c = n ÷ V, a smaller V gives a larger calculated c. The result is a factor of 0.0250/0.0240 ≈ 1.042 too high.
Row 3: The student concludes the solution is more concentrated than it actually is, which could lead to incorrect dosing if this concentration is used in a subsequent calculation or preparation.
Row 4: The reported 0.1042 mol/L is higher than the true 0.1000 mol/L. Using V = 0.0240 L (the titre volume) in c = n ÷ V instead of the flask volume (0.0250 L) gives c = 2.500×10−3 ÷ 0.0240 = 0.1042 mol/L, which is 4.2% above the true value. The smaller denominator inflates the calculated concentration.
Overall: The systematic error causes a 4.2% overestimate of the concentration (too high), because using the smaller titre volume (0.0240 L) instead of the correct flask volume (0.0250 L) in c = n ÷ V inflates the result by a factor of 0.0250/0.0240 ≈ 1.042. A student could detect it by checking whether the volume used for c = n ÷ V matches the flask (pipette) volume recorded in the method.
Q4 — Compare and contrast table
What is measured: Titration: volume of standard solution (titre, mL). Gravimetric: mass of precipitate (g).
Formula to find moles: Titration: n = c × V (moles of standard). Gravimetric: n = m ÷ MM (moles of precipitate).
How n(unknown) is found: Titration: apply mole ratio from balanced equation to n(standard). Gravimetric: apply mole ratio from balanced equation to n(precipitate).
Precaution to ensure complete reaction: Titration: titrate slowly near endpoint; multiple concordant titres ensure the equivalence point is accurately located. Gravimetric: the precipitating reagent is added in excess to ensure all of the unknown ion is completely precipitated.