A solid that will not sit still for a titration
The last lecture ended with a problem. A solid that does not dissolve, such as the calcium carbonate in an eggshell or in a piece of limestone, cannot be titrated in a flask. It promised a trick that chemists use instead: add a known excess of acid, then titrate what is left over. This lecture is that trick.
First, see why the direct method fails. Put some powdered limestone in a conical flask with an indicator, and add acid from a burette. The acid attacks the surface of each grain. The grains react slowly, the mixture fizzes, and some grains hide under others. The indicator changes colour when the liquid has a little extra acid in it, even though solid is still sitting in the flask. Wait a minute, and more solid reacts, and the colour changes back. There is no sharp end point. The burette reading means nothing.
So the question changes. Instead of "how much acid does the solid need?", we ask "how much acid is left after the solid has had all it wants?"
The idea: measure the change, not the price
You already use this idea when you pay in a shop. Suppose you do not know the price of your shopping. You hand over a $50 note, and you get $12 back in change. Then the shopping cost $50 − $12 = $38. You never measured the price directly. You knew what you gave, and you counted what came back.
A back titration works the same way.
- What you give: a known amount of acid, more than the sample can use. This is the known excess. Known means you measured its volume and concentration exactly. Excess means some is certain to be left over.
- The change: the acid left over after the reaction is complete. This is a solution, so you can titrate it, with a standard alkali, by the method of the last two lectures.
- The price: the acid the sample used, found by subtraction.
moles of reagent that reacted with the sample = moles added − moles left over
The moles left over come from the titration. Then the mole ratio of the first reaction turns the moles that reacted into moles of the sample.
Notice that a back titration involves two reactions. The first is between the sample and the excess reagent. The second is the titration of the leftover reagent. Each reaction has its own balanced equation and its own mole ratio. Most mistakes in this topic come from using the ratio of one reaction in the step that belongs to the other.
When a direct titration is not possible
A direct titration needs three things. The substance must be in solution, its reaction with the titrant must be fast, and there must be a sharp end point. A back titration helps when one of these fails.
- The substance does not dissolve. Calcium carbonate in limestone or eggshell, and insoluble bases such as magnesium hydroxide in some antacid tablets. A known excess of acid dissolves all of it.
- The reaction is slow. Aspirin is an example, as you will see in Worked Example 2. Part of its reaction with alkali only finishes after heating, and you cannot heat and titrate at the same time.
- The substance escapes as a gas. Ammonia is volatile. To measure the nitrogen in a fertiliser, the ammonia is driven out by heating and bubbled into a known excess of acid, which traps all of it.
Practise spotting the situations below.
Put each analysis in the right box. For each one, ask: is the substance in solution, does it react quickly, and does it stay in the flask?
Eight analyses, sorted by the method that suits them.
| Analysis | Method | Reason |
|---|---|---|
| Concentration of a hydrochloric acid solution | Direct | A solution, fast reaction, sharp end point. |
| Ethanoic acid in vinegar | Direct | A solution that reacts quickly with alkali. |
| Concentration of a sodium carbonate solution | Direct | Already dissolved; reacts quickly with acid. |
| Calcium carbonate in an eggshell | Back | An insoluble solid. |
| Magnesium hydroxide in an antacid tablet | Back | An insoluble base. |
| Magnesium oxide powder | Back | A base that does not dissolve in water. |
| Nitrogen in a fertiliser, released as ammonia | Back | A gas, trapped in excess acid. |
| Aspirin, including the slow reaction of its ester group | Back | The reaction needs heating to finish. |
Choosing the excess
"Add an excess" sounds easy, but the size of the excess matters. If you add too little acid, some solid is left unreacted, and the whole method fails. If you add only a little more than the sample needs, the leftover is tiny. The titre will then be very small, and the usual reading uncertainty of a burette becomes a large fraction of it. If you add far too much, the titre may be larger than one burette full. Worse, the answer becomes a small difference between two large numbers, and small errors in each grow large in the difference.
So a chemist plans. They estimate the largest amount of sample that could be present, add enough reagent for that, and add a sensible extra on top. Try it below with the limestone from Worked Example 1.
A 2.50 g sample of limestone will react with 1.00 mol/dm³ hydrochloric acid. The leftover acid will be titrated with 0.500 mol/dm³ sodium hydroxide from a 50 cm³ burette. Choose how much acid to add. Then change the true purity of the sample: a plan must still work if the sample turns out to be pure.
The expected titre of 0.500 mol/dm³ NaOH for a 2.50 g limestone sample, for different volumes of 1.00 mol/dm³ HCl added. An 80% pure sample uses 0.0400 mol HCl; a pure sample uses 0.0500 mol.
| HCl added | Titre if 80% pure | Titre if pure | Verdict |
|---|---|---|---|
| 40 cm³ | 0 cm³ | not enough acid | Fails: no excess. |
| 45 cm³ | 10.0 cm³ | not enough acid | Fails if the sample is pure. |
| 60 cm³ | 40.0 cm³ | 20.0 cm³ | A good plan. |
| 70 cm³ | 60.0 cm³ | 40.0 cm³ | Titre may need more than one burette. |
What should you have seen? With an 80% pure sample, 50 cm³ of acid seems fine: the titre is 20.0 cm³. But if the sample were pure, the same 50 cm³ would be used up exactly, with nothing left to titrate. A safe plan, such as 60 cm³, works whatever the purity turns out to be. This is why the excess must be chosen for the worst case, before you know the answer.
Worked examples
A 2.50 g sample of impure limestone is added to 50.0 cm³ of 1.00 mol/dm³ hydrochloric acid. When the fizzing stops, the remaining acid is titrated with 0.500 mol/dm³ sodium hydroxide. The titre is 20.0 cm³. Assume the impurities do not react with acid. Find the percentage of calcium carbonate in the limestone.
The two reactions:
First reaction: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
Second reaction (the titration): HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Moles of HCl added (the solution you know completely):
dm³, so mol.
Moles of NaOH in the titre: dm³, so mol.
Ratio line for the second reaction: n(HCl left over) = n(NaOH) × 1/1 = 0.0100 mol.
Subtraction: n(HCl reacted) = 0.0500 − 0.0100 = 0.0400 mol.
Ratio line for the first reaction: n(CaCO₃) = n(HCl reacted) × 1/2 = 0.0200 mol.
Mass and purity: (CaCO₃) = 40 + 12 + 3 × 16 = 100 g/mol. Mass = 0.0200 × 100 = 2.00 g. Purity = 2.00 ÷ 2.50 × 100% = 80.0%.
Check: the answer is below 100%, and the subtraction gave a positive number, as it must. The 2.00 g of carbonate needs 0.0400 mol of acid, and 0.0400 + 0.0100 = 0.0500 mol, the amount added. ✓
In Worked Example 1, the student forgot that the solution was still full of dissolved carbon dioxide, which is slightly acidic. Some of the sodium hydroxide reacted with it, so the titre came out a little too large. Does this make the calculated purity too high or too low?
Show the answer
Too low. A larger titre means more acid appears to be left over. Then the subtraction gives less acid used by the sample, so less calcium carbonate is calculated. In a back titration, an error in the titre moves the answer in the opposite direction. This is why some methods boil the solution briefly before titrating, to drive out the carbon dioxide.
Aspirin, C₉H₈O₄, is an acid, and it also contains an ester group. When it is heated with excess sodium hydroxide, both parts react. So each mole of aspirin uses 2 mol of NaOH:
C₉H₈O₄(s) + 2NaOH(aq) → CH₃COONa(aq) + HOC₆H₄COONa(aq) + H₂O(l)
The ester reaction is slow at room temperature, so the mixture must be heated. That is why this is a back titration. An aspirin tablet labelled 300 mg is heated gently with 50.0 cm³ of 0.100 mol/dm³ NaOH. After cooling, the leftover NaOH needs 16.85 cm³ of 0.100 mol/dm³ HCl. Find the mass of aspirin in the tablet. Take (aspirin) = 180 g/mol.
Moles of NaOH added: mol.
Moles of HCl in the titre: mol. HCl and NaOH react 1 : 1, so n(NaOH left over) = mol.
Subtraction: n(NaOH reacted) = mol.
Ratio line for the first reaction: n(aspirin) = mol.
Mass: g, which is about 298 mg, or about 99.5% of the label.
Check: this ratio of 2 is not a guess. In one official method, from the United States Pharmacopeia, aspirin is boiled gently with a measured excess of sodium hydroxide for 10 minutes, and the excess is titrated with acid. The method states that each cm³ of 0.5 mol/dm³ NaOH is equivalent to 45.04 mg of aspirin. Each cm³ holds 0.5 × 10⁻³ mol of NaOH, and 45.04 mg of aspirin is 0.25 × 10⁻³ mol (with = 180.16). That is exactly 2 mol of NaOH for each mole of aspirin. ✓
A 0.500 g sample of ammonium sulfate fertiliser is heated with excess sodium hydroxide. All the ammonia gas that is released is bubbled into 50.0 cm³ of 0.200 mol/dm³ HCl, where it reacts: NH₃(g) + HCl(aq) → NH₄Cl(aq). The leftover HCl needs 24.80 cm³ of 0.100 mol/dm³ NaOH. Find the percentage by mass of nitrogen in the fertiliser.
Moles of HCl added: mol.
Moles of HCl left over: n(NaOH) = mol, and the ratio is 1 : 1.
Subtraction: n(HCl reacted) = mol, so n(NH₃) = mol (ratio 1 : 1).
Nitrogen: each NH₃ has one N atom, so the mass of N = g. Percentage = 0.105 ÷ 0.500 × 100% ≈ 21.1%.
Check: pure ammonium sulfate, (NH₄)₂SO₄, has and contains 2 × 14 = 28 of nitrogen, which is 28 ÷ 132 × 100% ≈ 21.2%. The fertiliser is almost pure. ✓
Three classic mistakes
The moles from the titration are the moles left over, not the moles that reacted with the sample. Always write the subtraction line: moles reacted = moles added − moles left over. If the result is negative, either the subtraction is the wrong way round or the reagent was not in excess.
A back titration has two equations. In Worked Example 1, HCl reacts with NaOH 1 : 1, but with CaCO₃ 2 : 1. Label each ratio line with the reaction it comes from, and use it only in that step.
Often the leftover solution is made up to 250.0 cm³, and only 25.0 cm³ portions are titrated. The titration then measures only one tenth of the leftover reagent. Multiply by 250.0 ÷ 25.0 = 10 before you subtract.
Work each one on paper first. Open a hint only when you are stuck. Use H = 1, O = 16, N = 14, Mg = 24, Ca = 40, C = 12.
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An antacid tablet of mass 1.00 g contains magnesium hydroxide. It is added to 50.0 cm³ of 0.500 mol/dm³ HCl: Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l). The leftover acid needs 27.50 cm³ of 0.200 mol/dm³ NaOH. Find the percentage by mass of magnesium hydroxide in the tablet.
Hint 1
Find the moles of HCl added and the moles left over. Subtract.
Hint 2
HCl reacted = 0.0250 − 0.00550 = 0.0195 mol. The ratio Mg(OH)₂ : HCl is 1 : 2. (Mg(OH)₂) = 58 g/mol.
Full answer
HCl added = 0.500 × 0.0500 = 0.0250 mol. NaOH = 0.200 × 0.02750 = 5.50 × 10⁻³ mol = HCl left over. HCl reacted = 0.0195 mol. n(Mg(OH)₂) = 0.0195 × ½ = 9.75 × 10⁻³ mol. Mass = 9.75 × 10⁻³ × 58 ≈ 0.566 g, so about 56.6% of the tablet. Check: 0.566 g needs 2 × 9.75 × 10⁻³ = 0.0195 mol acid, and 0.0195 + 0.0055 = 0.0250 mol, as added. ✓
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A 1.25 g sample of limestone is dissolved in 100.0 cm³ of 0.500 mol/dm³ HCl (an excess). The solution is made up to 250.0 cm³ with water. A 25.0 cm³ portion needs a mean titre of 26.00 cm³ of 0.100 mol/dm³ NaOH. Find the percentage purity of the limestone.
Hint 1
The titration measures the leftover acid in one tenth of the solution. Scale up before subtracting.
Hint 2
Left over in the portion: 2.60 × 10⁻³ mol. In the whole flask: 0.0260 mol.
Full answer
HCl added = 0.500 × 0.1000 = 0.0500 mol. NaOH in the titre = 0.100 × 0.02600 = 2.60 × 10⁻³ mol, which is the HCl in 25.0 cm³. In 250.0 cm³: × 10 = 0.0260 mol left over. HCl reacted = 0.0500 − 0.0260 = 0.0240 mol. n(CaCO₃) = 0.0240 × ½ = 0.0120 mol. Mass = 0.0120 × 100 = 1.20 g. Purity = 1.20 ÷ 1.25 × 100% = 96.0%. Check: without the × 10, you would get 0.0500 − 0.0026 = 0.0474 mol reacted and a purity of 190%, which is impossible, so the scale-up is needed. ✓
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Ammonia from a sample is absorbed in 25.0 cm³ of 0.100 mol/dm³ sulfuric acid: 2NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(aq). The leftover acid needs 18.00 cm³ of 0.100 mol/dm³ NaOH: H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l). Find the mass of ammonia absorbed.
Hint 1
Both reactions have a ratio of 2 here, but in different directions. Write each ratio line carefully.
Hint 2
NaOH = 1.80 × 10⁻³ mol, so H₂SO₄ left over = 0.90 × 10⁻³ mol.
Full answer
H₂SO₄ added = 0.100 × 0.0250 = 2.50 × 10⁻³ mol. NaOH = 0.100 × 0.01800 = 1.80 × 10⁻³ mol, so H₂SO₄ left over = 1.80 × 10⁻³ × ½ = 0.90 × 10⁻³ mol. H₂SO₄ reacted = 2.50 × 10⁻³ − 0.90 × 10⁻³ = 1.60 × 10⁻³ mol. n(NH₃) = 1.60 × 10⁻³ × 2 = 3.20 × 10⁻³ mol. Mass = 3.20 × 10⁻³ × 17 ≈ 0.0544 g, or 54.4 mg. Check: the NH₃ (3.20 mmol) must be twice the acid it used (1.60 mmol). ✓ A common error is to use the ratio 2 in the wrong direction at one step.
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The official method for aspirin uses about 1.5 g of aspirin and 50.0 cm³ of 0.5 mol/dm³ NaOH, boiled gently, with the excess titrated by 0.25 mol/dm³ sulfuric acid. Predict the titre for 1.50 g of pure aspirin ( = 180 g/mol). Then explain why the method uses this much NaOH, and why its writers chose a sample of about 1.5 g rather than 0.15 g.
Hint 1
Aspirin uses 2 mol NaOH per mol. Sulfuric acid uses 2 mol NaOH per mol too.
Hint 2
n(aspirin) ≈ 8.33 × 10⁻³ mol, so it uses ≈ 0.0167 mol NaOH. NaOH added = 0.0250 mol.
Full answer
n(aspirin) = 1.50 ÷ 180 ≈ 8.33 × 10⁻³ mol. NaOH used = 2 × 8.33 × 10⁻³ ≈ 0.0167 mol. NaOH added = 0.5 × 0.0500 = 0.0250 mol, so NaOH left over ≈ 0.0250 − 0.0167 = 8.33 × 10⁻³ mol. H₂SO₄ needed = ½ × 8.33 × 10⁻³ ≈ 4.17 × 10⁻³ mol, so the titre = 4.17 × 10⁻³ ÷ 0.25 ≈ 0.0167 dm³, about 16.7 cm³. Check: 16.7 cm³ fits easily in one burette. ✓ The NaOH is a clear excess: even pure aspirin uses only two thirds of it, so a sample slightly above 1.5 g still leaves plenty. A sample of 0.15 g would use only a tenth as much NaOH. Then the answer would be a small difference between two large numbers (0.0250 and about 0.0233 mol). Small reading errors in the titre would cause a large percentage error in the aspirin.
A back titration is an example of a very general strategy: when you cannot measure something directly, measure what is left after it has acted. The cashier's change, the fuel left in a tank after a journey and the acid left after a limestone sample has dissolved all tell you about something you never measured directly.
The strategy has a cost, and it is worth remembering. Every back titration ends in a subtraction, so its errors behave differently from a direct titration's. An error in the titre moves the answer in the opposite direction, and a small difference between two large numbers magnifies the uncertainty. Good analysts plan the excess so that the difference stays large. The same thinking about errors returns in the next unit on energetics, where many enthalpy changes are found indirectly, from the difference between measured quantities.
- A back titration adds a known excess of reagent to the sample, then titrates the reagent that is left over.
- It is used when the substance is an insoluble solid, reacts too slowly, or escapes as a gas.
- The moles that reacted with the sample are the moles added minus the moles left over.
- A back titration involves two reactions, and each ratio line must use the equation for its own reaction.
- The excess must be large enough for the purest possible sample, but small enough to give a titre that fits in one burette.
- If only a portion of the leftover solution is titrated, the result must be scaled up before subtracting.