Year 11 Chemistry Module 2 ⏱ ~40 min Lesson 17 of 20

Back Calculations
& Unknown Concentrations

In quantitative analysis chemists often know the product and need to work backwards — using a precipitate mass or titration result to find the concentration of an unknown solution. This appears in nearly every HSC Chemistry exam.

🔍
📐

Back Calculation Pathway

Titration: n(standard) = c × V → ÷ ratio → n(unknown) → c = n ÷ V(flask)

Gravimetric: m(precipitate) → n = m ÷ MM → ÷ ratio → n(unknown) → c = n ÷ V

Average titre: discard rough; average concordant results (within 0.10 mL)
⚠️ Volume in the flask (aliquot) is used to find c(unknown). Volume from the burette (titre) is used to find n(standard). Swapping these is the most common error in this lesson.
📖 Know

Key Facts

  • Back-calc starts with known product/standard, finds unknown
  • Concordant = titres within 0.10 mL of each other
  • Rough titre is always discarded
  • Primary standard: known mass → standardise unknown solution
💡 Understand

Concepts

  • Why excess reagent guarantees complete reaction in gravimetric analysis
  • The flask-vs-burette volume distinction
  • How non-1:1 ratios affect back calculations
✅ Can Do

Skills

  • Find c(unknown) from titration data — 1:1 and non-1:1 ratios
  • Find c(unknown) from precipitate mass in gravimetric data
  • Calculate average concordant titre correctly

📚 Core Content

🔍

The Back Calculation Method

A back calculation works in the opposite direction to a forward stoichiometry problem. Instead of "given reactant, find product", you work "given product, find unknown reactant".

Known
V & c
standard
→
Calculate
n(standard)
= c × V
→
Apply ratio
n(unknown)
= n ÷ ratio
→
Find
c(unknown)
= n ÷ V(flask)

Concordant Titres

Multiple titrations are performed to ensure reliability. The first trial (rough) is always discarded. Remaining results within 0.10 mL of each other are concordant — average those only.

Example: Rough = 21.8 mL, T1 = 22.4 mL, T2 = 22.5 mL, T3 = 22.3 mL
T1 and T3 differ by 0.1 mL ✓; T2 and T3 differ by 0.2 mL — marginal. Use T1, T2, T3 if all within 0.2 mL and question doesn't specify stricter criteria.
Average = (22.4 + 22.5 + 22.3) ÷ 3 = 22.4 mL

Primary Standard Back Calculation

A known mass of a pure primary standard is dissolved to a known volume, giving a precisely known concentration. This standard is then titrated against the unknown solution.

Steps: m(primary standard) ÷ MM → n(total) ÷ V(total) → c(standard) → n(aliquot) = c × V(aliquot) → apply ratio → n(unknown) → c = n ÷ V(flask)

🧮 Worked Examples

Worked Example 1 — Titration back-calc, 1:1 ratio

Standard → [Unknown]
25.0 mL of NaOH (unknown concentration) is titrated against 0.100 mol/L HCl. Concordant titres: 22.3, 22.4, 22.3 mL. Find [NaOH]. HCl + NaOH → NaCl + H₂O.
  1. 1
    Average concordant titre
    V(HCl) = (22.3+22.4+22.3)÷3 = 22.33 mL = 0.02233 L
  2. 2
    n(HCl) — the known burette solution
    n(HCl) = 0.100 × 0.02233 = 2.233×10⁻³ mol
  3. 3
    Ratio 1:1 → n(NaOH) = 2.233×10⁻³ mol
  4. 4
    c(NaOH) using flask volume = 25.0 mL
    c(NaOH) = 2.233×10⁻³ ÷ 0.0250 = 0.0893 mol/L
✓ Answerc(NaOH) = 0.0893 mol/L

Worked Example 2 — H₂SO₄ vs NaOH, non-1:1 ratio

Non-1:1 Ratio
25.0 mL of H₂SO₄ (unknown) is titrated with 0.200 mol/L NaOH. Average titre = 18.50 mL. Find [H₂SO₄]. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
  1. 1
    n(NaOH)
    n(NaOH) = 0.200 × 0.01850 = 3.700×10⁻³ mol
  2. 2
    Ratio NaOH:H₂SO₄ = 2:1
    n(H₂SO₄) = 3.700×10⁻³ ÷ 2 = 1.850×10⁻³ mol
  3. 3
    c(H₂SO₄) in 25.0 mL flask
    c = 1.850×10⁻³ ÷ 0.0250 = 0.0740 mol/L
✓ Answerc(H₂SO₄) = 0.0740 mol/L

Worked Example 3 — Primary standard (Na₂CO₃ vs HCl)

Primary Standard
0.530 g of Na₂CO₃ is dissolved to 100.0 mL. 25.0 mL aliquots are titrated against HCl. Average titre = 24.5 mL. Find [HCl]. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. (Na = 22.990, C = 12.011, O = 15.999)
  1. 1
    n(Na₂CO₃) in full solution
    MM = 105.99; n = 0.530 ÷ 105.99 = 5.001×10⁻³ mol
  2. 2
    n in 25.0 mL aliquot
    c = 5.001×10⁻³ ÷ 0.100 = 0.05001 mol/L
    n(aliquot) = 0.05001 × 0.0250 = 1.250×10⁻³ mol
  3. 3
    Ratio 1:2 → n(HCl)
    n(HCl) = 1.250×10⁻³ × 2 = 2.500×10⁻³ mol
  4. 4
    c(HCl) from titre
    c(HCl) = 2.500×10⁻³ ÷ 0.0245 = 0.102 mol/L
✓ Answerc(HCl) = 0.102 mol/L

Worked Example 4 — Gravimetric back calculation

Precipitate → [unknown]
25.0 mL of BaCl₂ solution reacts with excess Na₂SO₄. 0.699 g of BaSO₄ precipitate forms. Find [BaCl₂]. BaCl₂ + Na₂SO₄ → BaSO₄↓ + 2NaCl. (Ba = 137.33, S = 32.06, O = 15.999)
  1. 1
    n(BaSO₄)
    MM(BaSO₄) = 233.39; n = 0.699 ÷ 233.39 = 2.995×10⁻³ mol
  2. 2
    Ratio 1:1 → n(BaCl₂) = 2.995×10⁻³ mol
  3. 3
    c(BaCl₂)
    c = 2.995×10⁻³ ÷ 0.0250 = 0.120 mol/L
✓ Answerc(BaCl₂) = 0.120 mol/L
⚠️

Common Mistakes

Using titre volume instead of flask volume for c(unknown)
The unknown is in the flask (25.0 mL aliquot). Its concentration = n ÷ V(flask). The titre volume from the burette belongs to the standard solution. Swapping these gives an answer off by the ratio of the two volumes.
✓ Fix: Label clearly — Flask = unknown. Burette = standard. c(unknown) = n ÷ V(flask). Never use titre volume for c(unknown).
Forgetting the non-1:1 ratio
H₂SO₄ + 2NaOH is the classic trap. If you skip the 2:1 ratio and assume 1:1, your answer for [H₂SO₄] is exactly double the correct value. This is reliably tested every few years in the HSC.
✓ Fix: Write the balanced equation and circle the ratio before calculating. n(H₂SO₄) = n(NaOH) ÷ 2, not n(NaOH) × 1.
Including the rough titre in the average
The rough titre is always larger than the true endpoint because students deliberately overshoot on the first run to find the approximate endpoint range. It must always be discarded. Including it biases the average upward, giving a systematic error in c(unknown).
✓ Fix: Identify and discard the rough titre first. Average only concordant results (within 0.10 mL of each other).

📓 Copy Into Your Books

▼

🔍 Back Calc Steps

  • 1. n(standard) = c × V(titre)
  • 2. Apply mole ratio
  • 3. c(unknown) = n ÷ V(flask)
  • Always use V in litres

🧪 Concordant Titres

  • Discard rough (first) titre always
  • Concordant = within 0.10 mL
  • Average concordant results only
  • Use average in calculation

⚗️ Primary Standard Path

  • m ÷ MM → n(total) ÷ V = c
  • n(aliquot) = c × V(aliquot)
  • Apply ratio → n(HCl)
  • c(HCl) = n ÷ V(titre)

⚠️ Volume Rules

  • Flask volume → c(unknown)
  • Titre volume → n(standard)
  • Total volume → c(product in mixture)
  • NEVER mix these up

📝 How are you completing this lesson?

🧪 Activities

🔍 Activity 1 — Back Calculation Drill

Titration and Gravimetric Calculations

Identify flask vs burette for each problem before calculating.

1 25.0 mL NaOH titrated with 0.150 mol/L HCl. Average titre = 20.0 mL. Find [NaOH].

n(HCl) = 0.150 × 0.0200 = 3.00×10⁻³; ratio 1:1; n(NaOH) = 3.00×10⁻³c(NaOH) = 3.00×10⁻³ ÷ 0.0250 = 0.120 mol/L

2 20.0 mL H₂SO₄ titrated with 0.100 mol/L NaOH. Average titre = 16.0 mL. Find [H₂SO₄]. (H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O)

n(NaOH) = 0.100 × 0.0160 = 1.60×10⁻³; ratio 2:1n(H₂SO₄) = 1.60×10⁻³ ÷ 2 = 8.00×10⁻⁴c(H₂SO₄) = 8.00×10⁻⁴ ÷ 0.0200 = 0.0400 mol/L

3 Titres: Rough = 21.0, T1 = 22.5, T2 = 22.6, T3 = 22.4 mL. (a) Which are concordant? (b) Calculate average titre.

(a) T1, T2, T3 are concordant (Rough discarded). T1–T3: 22.4–22.6, range = 0.2 mL. T1 and T3 are within 0.1 mL; T2 is 0.1–0.2 from T3/T1.
(b) Average = (22.5 + 22.6 + 22.4) ÷ 3 = 22.5 mL

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✏️ Complete in workbook

❓ Multiple Choice

🎯

Test Your Knowledge

1. In a titration to find [NaOH], which volume is used to calculate c(NaOH)?

A
The average titre from the burette
B
The volume of NaOH in the conical flask (aliquot)
C
The total volume of both solutions combined
D
The volume of standard solution prepared

2. 25.0 mL NaOH titrated with 0.200 mol/L HCl requires 15.0 mL. What is [NaOH]?

A
0.200 mol/L
B
0.300 mol/L
C
0.120 mol/L
D
0.0600 mol/L

3. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. 25.0 mL H₂SO₄ titrated with 0.100 mol/L NaOH (titre = 20.0 mL). What is [H₂SO₄]?

A
0.0400 mol/L
B
0.0800 mol/L
C
0.100 mol/L
D
0.200 mol/L

4. Titres: Rough = 22.1, T1 = 23.4, T2 = 23.5, T3 = 23.3 mL. Correct average titre is:

A
23.175 mL (all four)
B
22.1 mL (rough only)
C
23.45 mL (T1 and T2 only)
D
23.4 mL (average T1, T2, T3)

5. 0.212 g of Na₂CO₃ (MM = 105.99) is dissolved to 250 mL. What is c(Na₂CO₃)?

A
0.0200 mol/L
B
0.00800 mol/L
C
0.0800 mol/L
D
0.00200 mol/L

✍️ Short Answer

📝

Extended Questions

6. 0.424 g of Na₂CO₃ (MM = 105.99) is dissolved to 100.0 mL. 25.0 mL aliquots are titrated against HCl. Titres: Rough = 19.8, T1 = 21.3, T2 = 21.4, T3 = 21.2 mL. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. (a) Calculate the average concordant titre. (b) Find [HCl]. 5 MARKS

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7. 40.0 mL of AgNO₃ solution reacts with excess NaCl. 1.148 g of AgCl precipitate forms. (a) Find [AgNO₃]. (b) Explain why NaCl must be in excess. AgNO₃ + NaCl → AgCl↓ + NaNO₃. (Ag = 107.87, Cl = 35.453) 4 MARKS

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✅ Comprehensive Answers

▼

❓ Multiple Choice

1. B — c(unknown) = n ÷ V(flask). The flask volume is the aliquot of the unknown solution.

2. C — n(HCl) = 0.200 × 0.0150 = 0.00300; ratio 1:1; c(NaOH) = 0.00300 ÷ 0.0250 = 0.120 mol/L.

3. A — n(NaOH) = 0.100 × 0.020 = 0.00200; ratio 2:1; n(H₂SO₄) = 0.00100; c = 0.00100 ÷ 0.025 = 0.0400 mol/L.

4. D — Discard rough (22.1). Average T1, T2, T3 = (23.4+23.5+23.3)÷3 = 23.4 mL.

5. B — n = 0.212 ÷ 105.99 = 2.000×10⁻³; c = 2.000×10⁻³ ÷ 0.250 = 0.00800 mol/L.

📝 Short Answer Model Answers

Q6 (5 marks):

(a) Discard rough (19.8). T1=21.3, T2=21.4, T3=21.2 — all within 0.2 mL, concordant. Average = (21.3+21.4+21.2)÷3 = 21.3 mL

(b) n(Na₂CO₃) = 0.424 ÷ 105.99 = 4.001×10⁻³; c = 4.001×10⁻³ ÷ 0.100 = 0.04001 mol/L n(aliquot) = 0.04001 × 0.0250 = 1.000×10⁻³; ratio 1:2; n(HCl) = 2.000×10⁻³ c(HCl) = 2.000×10⁻³ ÷ 0.02130 = 0.0939 mol/L

Q7 (4 marks):

(a) MM(AgCl) = 143.32; n = 1.148 ÷ 143.32 = 8.010×10⁻³; ratio 1:1; n(AgNO₃) = 8.010×10⁻³ c(AgNO₃) = 8.010×10⁻³ ÷ 0.0400 = 0.200 mol/L

(b) If NaCl is not in excess, some Ag⁺ ions would remain in solution unreacted. The precipitate mass would be less than the amount corresponding to all the AgNO₃ present. The calculated n(AgNO₃) would be too small, giving an underestimate of c(AgNO₃). Excess NaCl ensures all Ag⁺ is precipitated, so the mass of AgCl accurately reflects the full quantity of AgNO₃.

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