Chemistry • Year 12 • Module 6 • Lesson 18
Back Titration & Conductometric Titration
Lock in the key vocabulary, the four-step calculation logic, and the conductometric curve shapes before attempting higher-band questions.
1. Term–definition match
Match each term (left column) to its correct definition (right column) by writing the letter in the Answer column. Each letter is used once. 9 marks
| # | Term | Definition | Answer |
|---|---|---|---|
| 1 | Back titration | A. The minimum on a strong acid/strong base conductometric titration curve; where all H+ has been neutralised. | |
| 2 | Excess reagent | B. A titration method where electrical conductance of the solution is recorded as titrant is added. | |
| 3 | n(analyte) = n(total) − n(excess) | C. A technique in which a known excess of one reagent is added to the sample, and the unreacted excess is then titrated. | |
| 4 | Conductometric titration | D. The step in a back titration calculation that finds moles of reagent consumed by the analyte. | |
| 5 | Equivalence point (conductometric) | E. The standard solution added in deliberate surplus to ensure the analyte fully reacts before the back-titration begins. | |
| 6 | Molar conductivity (H+) | F. Why conductance falls before the EP: each Na+ (κ≈50) replaces an H+ (κ≈350). | |
| 7 | Sevenfold conductance drop | G. κ≈350 S·cm2·mol−1; highest of any common ion because proton hopping (Grotthuss mechanism) allows faster migration than classical ionic drift. | |
| 8 | CaCO3 + 2HCl → CaCl2 + H2O + CO2 | H. The balanced equation that gives the 1:2 mole ratio for limestone, antacid and eggshell back titrations. | |
| 9 | V-shape conductance curve | I. The graph produced during a strong acid + strong base conductometric titration: conductance decreases to a minimum then increases. |
2. True or false — with correction
Circle T or F. If the statement is false, write the corrected version on the line below. 10 marks (1 mark T/F + 1 mark correction where false)
2.1 Back titration is required when the analyte is insoluble, volatile, or reacts too slowly for a direct titration. T / F
2.2 In a back titration of CaCO3, the mole ratio of CaCO3 to HCl is 1:1. T / F
2.3 The correct subtraction in a back titration is: n(reacted) = n(total added) − n(excess remaining). T / F
2.4 In a conductometric titration of HCl with NaOH, conductance increases continuously because NaOH adds Na+ and OH− ions. T / F
2.5 The equivalence point in a strong acid + strong base conductometric titration is identified as the minimum conductance on the curve. T / F
3. Fill-in-the-blank paragraph
Use the word bank below to complete the passage. Each word is used once. 10 marks
Back titration is used when a direct titration is impossible because the analyte is or reacts too to produce a clean endpoint. In a back titration, a precisely measured of standard acid is added to the sample. After complete reaction, the remaining acid is titrated with standard base. The moles of acid that reacted with the analyte are found by subtracting: n(excess) is from n(total). The moles of analyte are then found using the from the balanced equation.
In a conductometric titration, the electrical of the solution is measured as titrant is added. For a strong acid + strong base system, conductance falls before the point because highly conductive ions (κ≈350) are replaced by less conductive Na+ ions (κ≈50). The conductance reaches a at the equivalence point, then rises as excess OH− is added, producing a characteristic graph.
4. Function recall — explain each in 1–2 sentences
Answer each using precise lesson vocabulary. 8 marks (2 each)
4.1 Why must the initial acid be in excess in a back titration?
4.2 Why is a conductometric titration preferred when analysing a coloured or turbid solution?
4.3 What does the minimum conductance on a V-shaped curve indicate?
4.4 Why is it necessary to gently boil a CaCO3 back titration mixture before titrating the excess acid?
Q1 — Term–definition match
1→C • 2→E • 3→D • 4→B • 5→A • 6→G • 7→F • 8→H • 9→I
Q2 — True / false with correction
2.1 True.
2.2 False. Correction: the mole ratio is 1:2 (one mole CaCO3 reacts with two moles HCl). Therefore n(CaCO3) = n(HCl)reacted ÷ 2.
2.3 True.
2.4 False. Correction: conductance decreases before the equivalence point because each H+ ion (κ≈350) is replaced by Na+ (κ≈50) — a sevenfold reduction in conductance per ion replaced. Only after the EP, when excess OH− is added, does conductance rise.
2.5 True.
Q3 — Cloze paragraph
In order: insoluble • slow • excess • subtracted • mole ratio • conductance • equivalence • H+ • minimum • V-shape
Q4.1 — Why excess acid
The acid must be in excess to guarantee that 100% of the analyte (e.g. insoluble CaCO3) dissolves and fully reacts. If the acid were not in excess, unreacted analyte would remain and there would be no leftover acid to back-titrate, making the calculation impossible.
Q4.2 — Conductometric for coloured/turbid solutions
Indicators rely on visible colour changes at the endpoint. In coloured or turbid solutions the indicator colour change is obscured. A conductometric titration measures electrical conductance, which is independent of solution colour or turbidity, so the equivalence point can still be determined precisely.
Q4.3 — Minimum conductance
The minimum conductance on the V-shaped curve is the equivalence point: all highly conductive H+ ions have been neutralised and converted to water, leaving only Na+ and Cl− (both low molar conductivity). Adding any more NaOH after this point introduces OH− (κ≈198) and conductance rises.
Q4.4 — Why boil the CaCO3 mixture
Boiling drives off dissolved CO2 gas as vapour. If CO2 remained in solution it would form H2CO3, which would react with NaOH during the back-titration (H2CO3 + 2NaOH → Na2CO3 + 2H2O), consuming extra NaOH. This would overestimate n(excess HCl) and underestimate n(CaCO3).