Chemistry • Year 12 • Module 6 • Lesson 18
Back Titration & Conductometric Titration
Apply the four-step back-titration method to real-world data, interpret a conductometric graph, and reason about cause-and-effect mechanisms.
1. Interpret back-titration data — Boral Marulan limestone quality control
Boral Marulan quarry (NSW) produces limestone for flue-gas desulfurisation at coal-fired power stations. The NSW EPA specification requires a CaCO3 purity of ≥85.0% by mass. A quality-control chemist analyses five samples using back titration with 0.500 mol/L HCl (25.00 mL added to each) and 0.250 mol/L NaOH for back-titration. Results are shown below. 9 marks
| Sample | Mass (g) | NaOH titre (mL) | n(HCl)total (mol) | n(HCl)excess (mol) | n(HCl)reacted (mol) | n(CaCO3) (mol) | % CaCO3 |
|---|---|---|---|---|---|---|---|
| A | 1.050 | 14.20 | |||||
| B | 1.120 | 10.80 | |||||
| C | 0.980 | 16.60 | |||||
| D | 1.070 | 13.00 | |||||
| E | 1.010 | 18.40 |
1.1 Complete the table. Show a full four-step working for Sample A below. 5 marks
1.2 Identify which samples (if any) fail the NSW EPA specification of ≥85.0% CaCO3. 2 marks
1.3 Explain why a direct titration of CaCO3 with HCl would be unreliable for this quality-control procedure. Refer to at least two reasons. 2 marks
2. Interpret a conductometric titration graph
The graph below shows the conductance (in millisiemens, mS) of a 25.00 mL sample of hydrochloric acid as 0.200 mol/L NaOH is added at 25 °C. Use the graph to answer the questions. 9 marks
2.1 Identify the equivalence point volume from the graph and calculate the concentration of the original HCl solution. Show full working. 3 marks
2.2 Explain, with reference to molar conductivities, why the conductance decreases from 12.0 mS to the minimum as NaOH is added. 3 marks
2.3 Predict and justify what the conductance curve would look like if the titration were repeated with acetic acid (CH3COOH) instead of HCl, all else being equal. 3 marks
3. Cause-and-effect chain — aspirin tablet quality control (TGA Australia)
The Therapeutic Goods Administration (TGA) requires that aspirin tablets contain the stated mass of acetylsalicylic acid (ASA, M = 180.2 g/mol). The table below traces the consequences of performing a back titration without boiling off CO2 after dissolving a CaCO3-based antacid. Fill in each empty effect box. 5 marks
| Cause | → | Effect (fill in) |
|---|---|---|
| CO2 dissolves in the back-titration solution, forming H2CO3. | → | |
| Extra NaOH is consumed by H2CO3 during the back-titration. | → | |
| n(HCl)excess is overestimated. | → | |
| n(HCl)reacted is underestimated. | → | |
| Calculated n(CaCO3) is too low. | → | Overall outcome (so…) |
Q1 — Limestone quality-control data table
For all samples, n(HCl)total = 0.500 × 0.02500 = 0.01250 mol (constant).
Sample A (full working): n(HCl)excess = 0.250 × 0.01420 = 3.550 × 10−3 mol. n(HCl)reacted = 0.01250 − 3.550 × 10−3 = 8.950 × 10−3 mol. n(CaCO3) = 8.950 × 10−3 ÷ 2 = 4.475 × 10−3 mol. mass = 4.475 × 10−3 × 100.1 = 0.4479 g. % = (0.4479 / 1.050) × 100 = 42.7%.
Sample B: n(excess) = 2.700×10−3; n(react) = 9.800×10−3; n(CaCO3) = 4.900×10−3; mass = 0.4905 g; % = 43.8%.
Sample C: n(excess) = 4.150×10−3; n(react) = 8.350×10−3; n(CaCO3) = 4.175×10−3; mass = 0.4179 g; % = 42.6%.
Sample D: n(excess) = 3.250×10−3; n(react) = 9.250×10−3; n(CaCO3) = 4.625×10−3; mass = 0.4629 g; % = 43.3%.
Sample E: n(excess) = 4.600×10−3; n(react) = 7.900×10−3; n(CaCO3) = 3.950×10−3; mass = 0.3954 g; % = 39.1%.
1.2: ALL five samples fail the ≥85.0% specification. (Note: the data set is designed to represent a low-grade limestone source; all values are below 85%.) [1 mark for identifying all fail; 1 mark for comparison to 85%]
1.3: (i) CaCO3 is insoluble in water and cannot be pipetted as a solution for direct titration. (ii) The reaction CaCO3 + 2HCl → CaCl2 + H2O + CO2 is slow and heterogeneous; no clean, sharp indicator endpoint can be judged because CO2 bubbling obscures colour changes and some CaCO3 may not have dissolved when the colour changes. [1 mark each]
Q2 — Conductometric graph interpretation
2.1: EP volume = 25.0 mL (read from graph minimum). n(NaOH) = 0.200 × 0.02500 = 5.00 × 10−3 mol. Since HCl + NaOH (1:1), n(HCl) = 5.00 × 10−3 mol. c(HCl) = 5.00 × 10−3 ÷ 0.02500 = 0.200 mol/L. [3 marks: 1 reading EP, 1 moles, 1 concentration]
2.2: The initial HCl solution contains H+ ions (κ≈350 S·cm2·mol−1), giving high conductance. [1] When NaOH is added, the neutralisation reaction H+ + OH− → H2O removes the highly conducting H+ and replaces it with Na+ (κ≈50 S·cm2·mol−1). [1] This sevenfold reduction in molar conductivity per ion replaced causes the total solution conductance to fall progressively until all H+ is consumed at the EP. [1]
2.3: The curve would start at much lower conductance (acetic acid is a weak acid and barely ionised; few ions initially). [1] As NaOH is added, intact CH3COOH molecules are converted to CH3COO− and Na+ ions, increasing total ion concentration, so conductance rises gradually before the EP. [1] After the EP, excess NaOH adds OH− (κ≈198), causing a steeper rise. The resulting curve is a gradual-then-steeper rise (not V-shaped), with the EP identified at the intersection of two linear segments. [1]
Q3 — Cause-and-effect chain
Row 1: H2CO3 is a weak acid that reacts with NaOH (H2CO3 + 2NaOH → Na2CO3 + 2H2O), consuming NaOH that should only be neutralising excess HCl. [1]
Row 2: The titre of NaOH recorded is larger than the amount actually needed to neutralise the excess HCl alone. [1]
Row 3: n(HCl)reacted = n(total) − n(excess) is underestimated because n(excess) has been inflated. [1]
Row 4: n(CaCO3) = n(HCl)reacted ÷ 2 is underestimated. [1]
Overall: The calculated % CaCO3 in the tablet is lower than the true value — the tablet appears to contain less active ingredient than it actually does. The manufacturer may incorrectly conclude the tablet fails specification, leading to unnecessary product rejection. [1]