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]