Titration calculations — finding concentration — Chemistry

Before this

This leads to

A volume that is really a count

The last lecture ended with a number in your notebook: a mean titre of 25.87 cm³. It came from careful work. You rinsed, read the meniscus at eye level, and repeated until your results agreed. But 25.87 cm³ is a volume, and the question was a concentration. The last lecture promised to turn one into the other, using $n = c V$ and the mole ratio. Here is how.

Suppose the burette held hydrochloric acid of concentration 0.100 mol/dm³. The flask held 25.0 cm³ of sodium hydroxide solution of unknown concentration. At the end point, 25.87 cm³ of acid had reacted exactly with all the alkali.

Think about what that means. The acid we added contains a number of moles we can work out, because we know both its concentration and its volume. The equation tells us how many moles of alkali those moles of acid react with. And the alkali was all in 25.0 cm³. So the titre is not just a volume. It is a count of particles, written in cm³. The calculation only unpacks it.

The chain: five links

Every titration calculation follows the same chain. Each link uses one idea you already have.

The chain of a titration calculation. Start with the solution whose concentration and volume you both know. Its moles, through the mole ratio, give the moles of the other solution. Divide by the volume of that solution to get its concentration.

Write the balanced equation, with state symbols. It gives the mole ratio.
Find the moles of the solution you know completely, the one whose concentration and volume are both known: $n = c \times V$ , with $V$ in dm³.
Use the mole ratio to find the moles of the other solution.
Find its concentration: $c = n \div V$ , with $V$ the volume of that solution, in dm³.
State the answer with its unit, mol/dm³, to a sensible number of significant figures.

Notice the wording of link 2. It does not say "the solution in the burette". Usually the standard solution is in the burette, but not always. Sometimes a question puts the known solution in the flask and the unknown in the burette. The chain does not care where each solution sits. It only asks which one you know completely.

Now the hook, link by link.

Link 1: $HCl(aq) + NaOH(aq) \to NaCl(aq) + H_{2} O(l)$ . The ratio is 1 : 1.

Link 2: the acid is known completely. Its volume is $25.87 \div 1000 = 0.02587$ dm³. So $n (HCl) = 0.100 \times 0.02587 = 2.587 \times 1 0^{- 3}$ mol.

Link 3: $n (NaOH) = n (HCl) \times \frac{1}{1} = 2.587 \times 1 0^{- 3}$ mol.

Link 4: the alkali was in $25.0 \div 1000 = 0.0250$ dm³. So $c (NaOH) = 2.587 \times 1 0^{- 3} \div 0.0250 = 0.10348$ mol/dm³.

Link 5: $c (NaOH) \approx 0.103$ mol/dm³, to 3 significant figures (3 s.f.).

Why divide by 1000, and why only once

A concentration in mol/dm³ means "moles in each cubic decimetre". A cubic decimetre is 1000 cm³. Your burette and pipette measure in cm³. So before you multiply, you must say how many cubic decimetres you have: 25.87 cm³ is 0.02587 dm³. This is the single most common lost mark in titration questions, so write the conversion as its own line, every time.

Divide only once. Students who have just been warned about the 1000 sometimes divide twice, or divide the concentration as well. A quick check catches both errors. A typical titration uses about $2 \times 1 0^{- 3}$ to $5 \times 1 0^{- 3}$ mol. If your answer to link 2 is 2.587 mol, you forgot to divide. If it is $2.587 \times 1 0^{- 6}$ mol, you divided twice.

Stop and think

Near the end of the titration, a student washes the inside of the flask with about 20 cm³ of distilled water. In link 4, should she divide by 25.0 cm³ or by the larger volume now in the flask?

Show the answer

By 25.0 cm³. The question asks for the concentration of the sodium hydroxide solution she pipetted, and the moles in the flask all came from that 25.0 cm³ portion. Adding water changed the volume in the flask but not the number of moles, so it did not change the titre either. That is why the last lecture said washing with water has no effect. The volume in link 4 is always the volume of the measured portion of the solution you are finding, not the volume of whatever ended up in the flask.

When the ratio is not 1 : 1

Many acids give more than one hydrogen ion. Sulfuric acid, H₂SO₄, gives two. So it needs two hydroxide ions, and two NaOH, for every H₂SO₄. Before you see the calculation, test your instinct.

Predict — the same concentration, a different acid

25.0 cm³ of 0.100 mol/dm³ hydrochloric acid needs exactly 25.00 cm³ of 0.100 mol/dm³ sodium hydroxide. Now the flask holds 25.0 cm³ of 0.100 mol/dm³ sulfuric acid instead. How much of the same sodium hydroxide will it need?

One unit of each acid, and the sodium hydroxide it needs.

Each H₂SO₄ gives two hydrogen ions, so it needs two hydroxide ions. The same number of moles of sulfuric acid needs twice the moles of NaOH, so twice the volume: 50.00 cm³.

The point to take away: equal moles of acid do not need equal moles of base. The equation decides. This is why link 3 is a line of its own, written the same way every time:

n(wanted) = n(known) \times (coefficient of wanted \div coefficient of known)

Written like this, the ratio cannot be turned upside down by accident. For sulfuric acid found from sodium hydroxide, it is $n (H_{2} SO_{4}) = n (NaOH) \times \frac{1}{2}$ . For sodium hydroxide found from sulfuric acid, it is $\times \frac{2}{1}$ . That is it. That is the whole difficulty of the calculation, and it lives in one line.

Worked examples

Worked Example 1 — The known solution in the flask

25.0 cm³ of sodium hydroxide, 0.100 mol/dm³, is pipetted into a flask. Hydrochloric acid of unknown concentration is run in from the burette. The mean titre is 20.00 cm³. Find the concentration of the acid.

1. $NaOH(aq) + HCl(aq) \to NaCl(aq) + H_{2} O(l)$ . Ratio 1 : 1.

2. Here the solution known completely is in the flask. $V = 25.0 \div 1000 = 0.0250$ dm³, so $n (NaOH) = 0.100 \times 0.0250 = 2.50 \times 1 0^{- 3}$ mol.

3. $n (HCl) = 2.50 \times 1 0^{- 3} \times \frac{1}{1} = 2.50 \times 1 0^{- 3}$ mol.

4. The acid's volume is the titre: $20.00 \div 1000 = 0.02000$ dm³. $c (HCl) = 2.50 \times 1 0^{- 3} \div 0.02000 = 0.125$ mol/dm³.

Check: it took less acid than alkali to react, so the acid must be more concentrated than 0.100 mol/dm³. 0.125 is. ✓

Worked Example 2 — A 1 : 2 ratio

25.0 cm³ of sodium hydroxide, 0.200 mol/dm³, is titrated with sulfuric acid. The mean titre is 12.50 cm³. Find the concentration of the sulfuric acid.

1. $H_{2} SO_{4} (aq) + 2 NaOH(aq) \to Na_{2} SO_{4} (aq) + 2 H_{2} O(l)$ .

2. $n (NaOH) = 0.200 \times 0.0250 = 5.00 \times 1 0^{- 3}$ mol.

3. $n (H_{2} SO_{4}) = 5.00 \times 1 0^{- 3} \times \frac{1}{2} = 2.50 \times 1 0^{- 3}$ mol.

4. $c (H_{2} SO_{4}) = 2.50 \times 1 0^{- 3} \div 0.01250 = 0.200$ mol/dm³.

Check: the acid and the alkali have the same concentration, yet only half the volume of acid was needed. That fits: each H₂SO₄ deals with two NaOH. ✓

Worked Example 3 — From raw results to an answer

20.0 cm³ of potassium hydroxide solution is titrated with nitric acid, 0.150 mol/dm³. The titres are: rough 18.65, then 18.30, 18.25 and 18.35 cm³. Find the concentration of the potassium hydroxide.

The mean titre first. The rough titre is left out. The three accurate titres range from 18.25 to 18.35, a spread of 0.10 cm³, so all three are concordant. Mean $= (18.30 + 18.25 + 18.35) \div 3 = 18.30$ cm³.

1. $KOH(aq) + HNO_{3} (aq) \to KNO_{3} (aq) + H_{2} O(l)$ . Ratio 1 : 1.

2. $n (HNO_{3}) = 0.150 \times 0.01830 = 2.745 \times 1 0^{- 3}$ mol.

3. $n (KOH) = 2.745 \times 1 0^{- 3}$ mol.

4. $c (KOH) = 2.745 \times 1 0^{- 3} \div 0.0200 = 0.13725$ mol/dm³.

5. The data with the fewest significant figures, 0.150 and 20.0, have 3. So the answer is $c (KOH) \approx 0.137$ mol/dm³ (3 s.f.).

Check: less acid (18.30 cm³) than alkali (20.0 cm³) was needed in a 1 : 1 reaction, so the alkali is weaker than the acid: 0.137 is less than 0.150. ✓ Keep all the digits until the last line. Rounding the moles early to $2.7 \times 1 0^{- 3}$ would give 0.135, which is wrong in the third figure.

Here is the same kind of calculation with one wrong line. Find it before you read on.

Spot the error — a sulfuric acid titration

25.0 cm³ of sulfuric acid is titrated with 0.150 mol/dm³ sodium hydroxide. The mean titre is 21.40 cm³. A student finds the concentration of the acid. Tap the line where the working first goes wrong.

A student's working for the concentration of sulfuric acid (25.0 cm³, titrated with 0.150 mol/dm³ NaOH, mean titre 21.40 cm³), with the wrong line marked.

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l) (correct)
n(NaOH) = 0.150 × 21.40 ÷ 1000 = 3.21 × 10⁻³ mol (correct)
The error: n(H₂SO₄) = 3.21 × 10⁻³ × 2 = 6.42 × 10⁻³ mol. The ratio is upside down: one H₂SO₄ reacts with two NaOH, so n(H₂SO₄) = 3.21 × 10⁻³ × ½ = 1.605 × 10⁻³ mol.
c(H₂SO₄) = 6.42 × 10⁻³ ÷ 0.0250 ≈ 0.257 mol/dm³ (follows from the wrong line)

The correct answer is 1.605 × 10⁻³ ÷ 0.0250 ≈ 0.0642 mol/dm³, four times smaller.

Notice how far the mistake travelled. Multiplying instead of dividing by 2 made the answer four times too big. A check would have caught it. Less alkali (21.40 cm³) than acid (25.0 cm³) was needed, yet each H₂SO₄ uses two NaOH. So the acid must be much weaker than 0.150 mol/dm³, not stronger.

Worked Example 4 — Planning: what titre should I expect?

A student plans to titrate 25.0 cm³ of 0.0400 mol/dm³ sulfuric acid with 0.100 mol/dm³ sodium hydroxide. Is a 50 cm³ burette large enough?

Run the chain from the acid. $n (H_{2} SO_{4}) = 0.0400 \times 0.0250 = 1.00 \times 1 0^{- 3}$ mol. $n (NaOH) = 1.00 \times 1 0^{- 3} \times \frac{2}{1} = 2.00 \times 1 0^{- 3}$ mol. Volume of NaOH $= 2.00 \times 1 0^{- 3} \div 0.100 = 0.0200$ dm³ $= 20.00$ cm³.

Yes. The expected titre of about 20 cm³ fits easily in one filling. Chemists plan like this before they start: a titre between about 15 and 40 cm³ keeps the reading error small compared with the titre, while still needing only one filling.

Check: $0.100 \times 0.0200 = 2.00 \times 1 0^{- 3}$ mol NaOH, and half of that is the $1.00 \times 1 0^{- 3}$ mol of acid. ✓

Where the "known" concentration comes from

Every titration rests on one solution whose concentration is known accurately. How is that solution known? Someone weighed a solid, dissolved it, and made it up to an exact volume in a volumetric flask. That only works if the solid is a primary standard: a substance pure enough, and stable enough, that its weighed mass is all the substance you think it is.

A good primary standard is very pure. It does not absorb water or carbon dioxide from the air, so its mass does not change while you weigh it. It has a fairly large molar mass, so a small weighing error is a small fraction of the whole. Anhydrous sodium carbonate, Na₂CO₃, is a common primary standard for acids.

Sodium hydroxide is not a primary standard. Solid sodium hydroxide absorbs water from the air, and it reacts with carbon dioxide. A sample weighed as 4.00 g contains less than 4.00 g of NaOH. So a sodium hydroxide solution is first standardised: titrated against a primary standard to find its true concentration. Then it can be used as a standard solution itself. Try it yourself, question 3, does exactly this.

Three classic mistakes

Trap 1 — Volumes left in cm³

$n = c V$ needs $V$ in dm³. Using 25.87 instead of 0.02587 makes the moles 1000 times too big. Write the ÷ 1000 as its own line, and check that the moles come out in the thousandths of a mole.

Trap 2 — The ratio turned upside down

For a 1 : 2 reaction, multiplying when you should divide makes the answer four times wrong once it reaches the concentration. Write n(wanted) = n(known) × (coefficient of wanted ÷ coefficient of known), and read the coefficients from the equation.

Trap 3 — The wrong volume in the last step

In link 4, divide by the volume of the solution you are finding: the pipetted portion or the titre, whichever belongs to it. Not the other solution's volume, and not the total volume in the flask.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

25.0 cm³ of potassium hydroxide solution is neutralised by a mean titre of 22.50 cm³ of 0.120 mol/dm³ hydrochloric acid. Find the concentration of the potassium hydroxide.

Hint 1
Which solution is known completely? Start the chain there.

Hint 2
$n (HCl) = 0.120 \times 0.02250$ . The ratio KOH : HCl is 1 : 1.

Full answer
$KOH(aq) + HCl(aq) \to KCl(aq) + H_{2} O(l)$ . $n (HCl) = 0.120 \times 0.02250 = 2.70 \times 1 0^{- 3}$ mol $= n (KOH)$ . $c (KOH) = 2.70 \times 1 0^{- 3} \div 0.0250 = 0.108$ mol/dm³. Check: less acid than alkali was used, so the alkali is weaker than 0.120: 0.108 ✓.
You will titrate 20.0 cm³ of 0.0750 mol/dm³ sulfuric acid with 0.120 mol/dm³ sodium hydroxide. What titre do you expect?

Hint 1
Here the acid is known completely. Find its moles first.

Hint 2
$n (NaOH) = n (H_{2} SO_{4}) \times \frac{2}{1}$ . Then $V = n \div c$ .

Full answer
$n (H_{2} SO_{4}) = 0.0750 \times 0.0200 = 1.50 \times 1 0^{- 3}$ mol. $n (NaOH) = 1.50 \times 1 0^{- 3} \times \frac{2}{1} = 3.00 \times 1 0^{- 3}$ mol. $V = 3.00 \times 1 0^{- 3} \div 0.120 = 0.0250$ dm³ $= 25.00$ cm³. Check: the acid supplies $2 \times 1.50 \times 1 0^{- 3}$ mol of hydrogen ions, and 25.00 cm³ of NaOH supplies $0.120 \times 0.0250 = 3.00 \times 1 0^{- 3}$ mol of hydroxide ions ✓.
Standardising an acid. 1.325 g of anhydrous sodium carbonate ( $M = 106$ g/mol) is dissolved and made up to 250 cm³ in a volumetric flask. 25.0 cm³ of this solution needs a mean titre of 24.50 cm³ of hydrochloric acid. Find the concentration of the acid. The equation is $Na_{2} CO_{3} (aq) + 2 HCl(aq) \to 2 NaCl(aq) + H_{2} O(l) + CO_{2} (g)$ .

Hint 1
First find the concentration of the sodium carbonate solution: moles from mass, then $c = n \div V$ with 0.250 dm³.

Hint 2
The 25.0 cm³ portion holds one tenth of the moles. Then $n (HCl) = n (Na_{2} CO_{3}) \times \frac{2}{1}$ .

Full answer
$n (Na_{2} CO_{3}) = 1.325 \div 106 = 0.0125$ mol in 0.250 dm³, so $c = 0.0500$ mol/dm³. In 25.0 cm³: $0.0500 \times 0.0250 = 1.25 \times 1 0^{- 3}$ mol. $n (HCl) = 2.50 \times 1 0^{- 3}$ mol. $c (HCl) = 2.50 \times 1 0^{- 3} \div 0.02450 \approx 0.102$ mol/dm³ (3 s.f.). Check: 24.50 cm³ is close to 25.0 cm³, and the acid's concentration (0.102) is close to twice the carbonate's (0.0500), as the 1 : 2 ratio predicts ✓.
Vinegar contains ethanoic acid, CH₃COOH, which reacts 1 : 1 with sodium hydroxide. A student dilutes 25.0 cm³ of vinegar to exactly 250 cm³ with distilled water. 25.0 cm³ of the diluted solution needs a mean titre of 20.80 cm³ of 0.100 mol/dm³ NaOH. Find the concentration of ethanoic acid in the original vinegar, in mol/dm³ and in g/dm³. (Invented results; $M (CH_{3} COOH) = 60$ g/mol.)

Hint 1
Run the chain to find the concentration of the diluted solution first.

Hint 2
The vinegar was diluted from 25.0 to 250 cm³, ten times. So the original is ten times more concentrated. For g/dm³, multiply mol/dm³ by the molar mass.

Full answer
$n (NaOH) = 0.100 \times 0.02080 = 2.08 \times 1 0^{- 3}$ mol $= n (CH_{3} COOH)$ in 25.0 cm³ of diluted solution. Diluted: $2.08 \times 1 0^{- 3} \div 0.0250 = 0.0832$ mol/dm³. Original: $10 \times 0.0832 = 0.832$ mol/dm³. In g/dm³: $0.832 \times 60 \approx 49.9$ g/dm³, about 5 g of ethanoic acid in every 100 cm³ of vinegar. Check: the student diluted first because undiluted vinegar would need about 208 cm³ of the NaOH, far more than one burette ✓. The indicator must be phenolphthalein, because ethanoic acid is weak.

The bigger idea

Look back at the chain. Nothing in it is new. Moles from concentration and volume came from one lecture. The mole ratio came from another. The titration only supplies the measurements. This is how chemistry works at every level: a few simple ideas, joined in the right order, answer a question that looks hard.

That same chain runs in laboratories every day. Food and medicine makers check the strength of what they produce, and water companies check what they supply. Some of those titrations use other reactions instead of acids and bases, such as reactions with iodine, or with metal ions in hard water. Many are done by automatic machines. The calculation is still n = cV, the mole ratio, and c = n ÷ V.

Sometimes the chain cannot start directly. A solid that does not dissolve, such as the calcium carbonate in an eggshell or in an antacid tablet, cannot be titrated in a flask. The next lecture shows the trick chemists use instead: add a known excess of acid, then titrate what is left over.

Key concepts in this lesson

A titre is a count of particles written as a volume, and the calculation unpacks it.
Start with the solution whose concentration and volume are both known, and find its moles with n = cV, with V in dm³.
The mole ratio from the balanced equation gives the moles of the other solution, so equal moles of acid do not always need equal moles of base.
The concentration of the other solution is its moles divided by its own measured volume in dm³, not the total volume in the flask.
Answers are given in mol/dm³ to the significant figures of the least precise data, usually 3, with no rounding until the last line.
A standard solution is made from a primary standard, or standardised against one, because a substance such as sodium hydroxide cannot be weighed accurately.