Chemistry • Year 12 • Module 6 • Lesson 18

Back Titration & Conductometric Titration

Synthesise, evaluate, and justify: two extended Master questions that demand evidence-based reasoning and source critique at Band 5–6 standard.

Master · Band 5–6 · Extended Response

1. Data + scenario evaluation — wastewater acid neutralisation (NSW EPA)

A NSW EPA environmental audit team analyses acidic wastewater from a mining leachate pond (pH ≈2.0) at a copper mine in the Cobar region. The wastewater must be neutralised with CaCO3 (ground limestone) before discharge. The EPA requires <0.5 mmol/L residual acidity (expressed as equivalent H+ concentration) in the treated water. The audit team uses back titration to verify residual acidity after limestone treatment. 8 marks

Experimental data — three treated wastewater samples

Sample Volume treated (mL) c(NaOH) added (mol/L) V(NaOH) added (mL) c(HCl) back-titrated (mol/L) V(HCl) back-titre (mL)
1100.00.10010.000.05008.40
2100.00.10010.000.05006.20
3100.00.10010.000.05009.80

Note: In this back titration, excess NaOH is added to the treated wastewater to neutralise all residual acid. The excess NaOH is then back-titrated with standard HCl. n(residual acid) = n(NaOH added) − n(NaOH excess). n(NaOH excess) = n(HCl back-titre) since HCl + NaOH is 1:1.

In your response you must:

  1. Calculate the residual acid concentration (mmol/L) for each sample, showing full working for Sample 1.
  2. Determine whether any sample fails the EPA limit of <0.5 mmol/L residual acidity.
  3. Explain why back titration (rather than direct titration) is more appropriate for this analysis, giving at least two specific reasons related to the nature of the wastewater.
  4. Evaluate one limitation of using back titration for EPA compliance monitoring in a field context.
Hint: For Sample 1, n(NaOH added) = 0.100 × 0.01000 mol; n(NaOH excess) = n(HCl back-titre) = 0.0500 × V(HCl). c(residual acid) = n(residual acid) ÷ V(sample). Remember to convert to mmol/L.

2. Source critique — evaluate a student’s laboratory report claim

Read the excerpt below from a Year 12 chemistry student’s laboratory report on back titration of limestone samples. Identify the scientific error(s), explain the correct chemistry, and describe how the error would be detected experimentally. 7 marks

Student report excerpt (contains errors):

“After adding excess 0.500 mol/L HCl to the crushed limestone sample and allowing the tablet to react, I immediately began back-titrating the solution with 0.250 mol/L NaOH. I used the volume of NaOH added divided by the volume of HCl originally added to determine the fraction of acid that had reacted. I found that 38.0 mL of NaOH was required to neutralise the excess, confirming that the back titration was complete when the indicator turned pink. Since the two solutions have a 2:1 concentration ratio, the moles of CaCO3 equalled the moles of NaOH used in the back-titration.”

In your response you must:

  1. Identify all scientific errors in the excerpt (there are at least three distinct errors).
  2. For each error, state the correct chemistry or procedure.
  3. Explain how one of these errors would produce an incorrect numerical result, and state whether it causes an over- or underestimate of % CaCO3.
  4. Describe one experimental observation that would alert the student that the procedure was flawed.
Hint: Look for errors in (a) procedure timing, (b) the calculation method (volumes vs moles), and (c) the mole-ratio logic. The 2:1 concentration ratio between HCl and NaOH does NOT mean 2:1 in moles unless the volumes are equal.
Answers — Do not peek before attempting

Q1 — Wastewater back titration (M1, 8 marks)

Requirement 1 — Calculations (3 marks):

Sample 1 (full working): n(NaOH added) = 0.100 × 0.01000 = 1.000 × 10−3 mol. n(NaOH excess) = n(HCl back-titre) = 0.0500 × 0.008400 = 4.200 × 10−4 mol. n(residual acid) = 1.000 × 10−3 − 4.200 × 10−4 = 5.800 × 10−4 mol. c(residual acid) = 5.800 × 10−4 ÷ 0.1000 L = 5.80 × 10−3 mol/L = 5.80 mmol/L. [1]

Sample 2: n(excess) = 0.0500 × 0.00620 = 3.100 × 10−4. n(residual) = 6.900 × 10−4. c = 6.90 mmol/L. [0.5]

Sample 3: n(excess) = 0.0500 × 0.00980 = 4.900 × 10−4. n(residual) = 5.100 × 10−4. c = 5.10 mmol/L. [0.5]

Requirement 2 — EPA compliance (1 mark): All three samples exceed the <0.5 mmol/L EPA limit (5.80, 6.90, 5.10 mmol/L respectively). The limestone treatment has not reduced residual acidity to compliant levels; the discharge cannot proceed. [1]

Requirement 3 — Why back titration (2 marks): (i) Mining leachate typically contains coloured metal ions (Fe3+, Cu2+) that would obscure an indicator colour change, making direct titration endpoints unreliable. [1] (ii) The wastewater may contain dissolved CO2, solid particulates, or species that react slowly; adding excess NaOH ensures complete neutralisation of all acidic species before the back-titration begins, giving a more accurate residual-acid value than attempting to find an endpoint in an inhomogeneous system. [1]

Requirement 4 — Limitation (1 mark): Back titration in a field context requires two standard solutions (NaOH and HCl) of accurately known concentration, and the standard solutions must be protected from CO2 absorption and temperature changes during transport. Any degradation of the NaOH standard (from CO2 absorption forming Na2CO3) introduces systematic error into the back-titration result. [1]

Q2 — Source critique (M2, 7 marks)

Requirement 1+2 — Errors and corrections (3 marks):

Error 1 — Immediate back-titration without boiling: "I immediately began back-titrating." The CO2 produced by CaCO3 + 2HCl must be expelled by gentle boiling before the back-titration starts. If CO2 remains, it dissolves to form H2CO3, which reacts with NaOH, inflating the apparent volume of NaOH needed and overestimating n(HCl)excess. Correct procedure: boil gently for 2–3 minutes, cool, then back-titrate. [1]

Error 2 — Volume fraction instead of mole calculation: "the volume of NaOH added divided by the volume of HCl originally added" is dimensionally wrong and chemically meaningless as a route to n(HCl)reacted. The correct method uses moles: n(HCl)reacted = n(HCl)total − n(HCl)excess, where n(HCl)excess = c(NaOH) × V(NaOH) for a 1:1 HCl:NaOH mole ratio. [1]

Error 3 — Moles of CaCO3 equalled moles of NaOH: This ignores the balanced equation and mole ratio. The back-titration NaOH reacts with excess HCl, not with CaCO3. n(CaCO3) must be found from n(HCl)reacted ÷ 2, not from n(NaOH back-titre). Using n(CaCO3) = n(NaOH) directly gives a value twice or otherwise wrong depending on concentrations. [1]

Requirement 3 — Numerical impact of Error 1 (2 marks): Not boiling off CO2 causes H2CO3 + 2NaOH → Na2CO3 + 2H2O, consuming extra NaOH. This makes V(NaOH back-titre) appear larger, inflating n(HCl)excess and therefore reducing n(HCl)reacted. Since n(CaCO3) = n(HCl)reacted÷2, it is underestimated. The calculated % CaCO3 is therefore lower than the true value — an underestimate. [2]

Requirement 4 — Experimental alert (1 mark): If CO2 has not been expelled, the student would observe persistent effervescence (bubbling) even after the indicator has changed colour, indicating the reaction is not yet complete and the endpoint is unreliable. Alternatively, a consistently low % CaCO3 result compared to the certified reference value would signal systematic error. [1]

HSCScience • Chemistry • Year 12 • Module 6 • Lesson 18 • Worksheet 3 (Master) • Band 5–6