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The titration calculation, step by step

Covers AS91910

The route from titre to answer

Every titration calculation follows the same five steps, whatever the chemistry. Learn the route and you never have to think about where to start.

1. Moles of titrant — from the burette: $n = c \times V$ .
2. Moles of the substance — from the mole ratio in the balanced equation.
3. Concentration in the flask — $c = n / V$ , using the aliquot volume.
4. Scale back — apply the dilution factor to get the original product.
5. Convert if needed — into g L⁻¹, % by mass, or mg per tablet.

The relationships

$n = c \times V \qquad c = \frac{n}{V} \qquad n = \frac{m}{M} \qquad m = n \times M$

$n$ — amount of substance, in mol
$c$ — concentration, in mol L⁻¹
$V$ — volume, in L (always convert from mL by dividing by 1000)
$m$ — mass, in g
$M$ — molar mass, in g mol⁻¹
The single most common arithmetic error in this standard is forgetting to convert mL to L. Every volume, every time.

The mole ratio step

Read the ratio straight from the balanced equation — the coefficients are the ratio.
For $2\text{HCl} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$ , the ratio HCl : Na₂CO₃ is 2 : 1.
- So $n(\text{HCl}) = 2 \times n(\text{Na}_2\text{CO}_3)$ , and $n(\text{Na}_2\text{CO}_3) = \tfrac{1}{2} n(\text{HCl})$ .
Decide which way to multiply by asking which substance you have more of. If the ratio is 2 : 1 and you are going from the "1" to the "2", the number gets bigger.
Getting this step upside down gives an answer wrong by exactly a factor of 2 or 5, which is why it is worth pausing on.

Applying the dilution factor

The dilution factor is $\dfrac{\text{final volume}}{\text{volume of product taken}}$ .
- 25.0 mL made up to 250.0 mL → factor 10.
- 10.0 mL made up to 500.0 mL → factor 50.
The concentration you calculate from the titre is that of the diluted solution. Multiply by the dilution factor to get the original.
Multiply, never divide. The original product is more concentrated than the diluted one, so the number must get bigger.

Converting to useful units

Consumer product labels rarely use mol L⁻¹, so the last step usually converts:

Wanted	How
g L⁻¹	$c(\text{mol L}^{-1}) \times M$
% by mass (m/v)	$\dfrac{\text{g L}^{-1}}{10}$ , for dilute aqueous solutions where 1 L ≈ 1000 g
mass per tablet	$n$ in the whole flask $\times M$ , divided by the number of tablets
mg per 100 mL	$\text{g L}^{-1} \times 100$

Worked ExampleA complete titration calculation with a non-1:1 ratio

A student determines the concentration of ethanoic acid in vinegar. 25.0 mL of vinegar was diluted to 250.0 mL in a volumetric flask. 20.0 mL aliquots of the diluted vinegar were titrated with 0.1009 mol L⁻¹ sodium hydroxide, giving a mean titre of 16.35 mL. Calculate the concentration of ethanoic acid in the original vinegar, in mol L⁻¹ and in g L⁻¹. (M(CH₃COOH) = 60.05 g mol⁻¹)

Step 1 — Moles of titrant used

Convert the titre to litres first: $16.35 \text{ mL} = 0.01635 \text{ L}$ .

$n(\text{NaOH}) = c \times V = 0.1009 \times 0.01635 = 1.64971... \times 10^{-3} \text{ mol}$

Step 2 — Moles of ethanoic acid, via the mole ratio

The equation is:

$\text{CH}_3\text{COOH}(aq) + \text{NaOH}(aq) \rightarrow \text{CH}_3\text{COONa}(aq) + \text{H}_2\text{O}(l)$

The ratio is 1 : 1, so:

$n(\text{CH}_3\text{COOH}) = 1.64971... \times 10^{-3} \text{ mol}$

Step 3 — Concentration in the diluted vinegar

The aliquot was 20.0 mL = 0.0200 L:

$c = \frac{n}{V} = \frac{1.64971... \times 10^{-3}}{0.0200} = 0.0824857... \text{ mol L}^{-1}$

Step 4 — Scale back to the original vinegar

The dilution factor is:

$\frac{250.0}{25.0} = 10$

$c(\text{vinegar}) = 0.0824857... \times 10 = 0.824857... \text{ mol L}^{-1}$

$c(\text{vinegar}) = 0.825 \text{ mol L}^{-1}$

Step 5 — Convert to g L⁻¹

$\text{mass concentration} = c \times M = 0.824857... \times 60.05 = 49.532... \text{ g L}^{-1}$

$= 49.5 \text{ g L}^{-1}$

Step 6 — Sanity check

Vinegar is sold as about 4–5% ethanoic acid, which for a dilute aqueous solution is roughly 40–50 g L⁻¹. The answer sits at the top of that range, which is consistent with a stronger vinegar.

Answer: 0.825 mol L⁻¹, or 49.5 g L⁻¹ — about 4.95% ethanoic acid by mass.

Note that this example uses a 1 : 1 ratio for clarity. Your own investigation must include at least one calculation with a non-1:1 ratio — see the back titration worked example on the next page.

Worked ExampleA calculation with a 2 : 1 mole ratio

A drain-cleaning product contains sulfuric acid. 10.0 mL of the product was diluted to 500.0 mL. 25.0 mL aliquots of the diluted solution required a mean titre of 21.40 mL of 0.1052 mol L⁻¹ sodium hydroxide. Calculate the concentration of sulfuric acid in the original product.

Step 1 — Moles of titrant

$n(\text{NaOH}) = 0.1052 \times 0.02140 = 2.25128 \times 10^{-3} \text{ mol}$

Step 2 — Apply the mole ratio

Sulfuric acid is diprotic, so one mole of it neutralises two moles of hydroxide:

$\text{H}_2\text{SO}_4(aq) + 2\text{NaOH}(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(l)$

The ratio H₂SO₄ : NaOH is 1 : 2, so there is half as much acid as base:

$n(\text{H}_2\text{SO}_4) = \frac{2.25128 \times 10^{-3}}{2} = 1.12564 \times 10^{-3} \text{ mol}$

Check the direction: we are going from the "2" to the "1", so the number should get smaller — and it did.

Step 3 — Concentration in the diluted solution

$c = \frac{1.12564 \times 10^{-3}}{0.0250} = 0.0450256 \text{ mol L}^{-1}$

Step 4 — Scale back

The dilution factor is $500.0 / 10.0 = 50$ :

$c(\text{product}) = 0.0450256 \times 50 = 2.25128 \text{ mol L}^{-1}$

$c(\text{product}) = 2.25 \text{ mol L}^{-1}$

Step 5 — Significant figures

The titre (21.40 mL, 4 sf), the titrant concentration (0.1052, 4 sf) and the aliquot (25.0 mL, 3 sf) are the measured quantities. The least precise is 3 significant figures, so the answer is given to 3 sf.

Answer: the drain cleaner contains sulfuric acid at 2.25 mol L⁻¹.

The titration calculation, step by step

Covers AS91910

The route from titre to answer

Every titration calculation follows the same five steps, whatever the chemistry. Learn the route and you never have to think about where to start.

1. Moles of titrant — from the burette: $n = c \times V$ .
2. Moles of the substance — from the mole ratio in the balanced equation.
3. Concentration in the flask — $c = n / V$ , using the aliquot volume.
4. Scale back — apply the dilution factor to get the original product.
5. Convert if needed — into g L⁻¹, % by mass, or mg per tablet.

The relationships

$n = c \times V \qquad c = \frac{n}{V} \qquad n = \frac{m}{M} \qquad m = n \times M$

$n$ — amount of substance, in mol
$c$ — concentration, in mol L⁻¹
$V$ — volume, in L (always convert from mL by dividing by 1000)
$m$ — mass, in g
$M$ — molar mass, in g mol⁻¹
The single most common arithmetic error in this standard is forgetting to convert mL to L. Every volume, every time.

The mole ratio step

Read the ratio straight from the balanced equation — the coefficients are the ratio.
For $2\text{HCl} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$ , the ratio HCl : Na₂CO₃ is 2 : 1.
- So $n(\text{HCl}) = 2 \times n(\text{Na}_2\text{CO}_3)$ , and $n(\text{Na}_2\text{CO}_3) = \tfrac{1}{2} n(\text{HCl})$ .
Decide which way to multiply by asking which substance you have more of. If the ratio is 2 : 1 and you are going from the "1" to the "2", the number gets bigger.
Getting this step upside down gives an answer wrong by exactly a factor of 2 or 5, which is why it is worth pausing on.

Applying the dilution factor

The dilution factor is $\dfrac{\text{final volume}}{\text{volume of product taken}}$ .
- 25.0 mL made up to 250.0 mL → factor 10.
- 10.0 mL made up to 500.0 mL → factor 50.
The concentration you calculate from the titre is that of the diluted solution. Multiply by the dilution factor to get the original.
Multiply, never divide. The original product is more concentrated than the diluted one, so the number must get bigger.

Converting to useful units

Consumer product labels rarely use mol L⁻¹, so the last step usually converts:

Wanted	How
g L⁻¹	$c(\text{mol L}^{-1}) \times M$
% by mass (m/v)	$\dfrac{\text{g L}^{-1}}{10}$ , for dilute aqueous solutions where 1 L ≈ 1000 g
mass per tablet	$n$ in the whole flask $\times M$ , divided by the number of tablets
mg per 100 mL	$\text{g L}^{-1} \times 100$

Worked ExampleA complete titration calculation with a non-1:1 ratio

Step 1 — Moles of titrant used

Convert the titre to litres first: $16.35 \text{ mL} = 0.01635 \text{ L}$ .

$n(\text{NaOH}) = c \times V = 0.1009 \times 0.01635 = 1.64971... \times 10^{-3} \text{ mol}$

Step 2 — Moles of ethanoic acid, via the mole ratio

The equation is:

$\text{CH}_3\text{COOH}(aq) + \text{NaOH}(aq) \rightarrow \text{CH}_3\text{COONa}(aq) + \text{H}_2\text{O}(l)$

The ratio is 1 : 1, so:

$n(\text{CH}_3\text{COOH}) = 1.64971... \times 10^{-3} \text{ mol}$

Step 3 — Concentration in the diluted vinegar

The aliquot was 20.0 mL = 0.0200 L:

$c = \frac{n}{V} = \frac{1.64971... \times 10^{-3}}{0.0200} = 0.0824857... \text{ mol L}^{-1}$

Step 4 — Scale back to the original vinegar

The dilution factor is:

$\frac{250.0}{25.0} = 10$

$c(\text{vinegar}) = 0.0824857... \times 10 = 0.824857... \text{ mol L}^{-1}$

$c(\text{vinegar}) = 0.825 \text{ mol L}^{-1}$

Step 5 — Convert to g L⁻¹

$\text{mass concentration} = c \times M = 0.824857... \times 60.05 = 49.532... \text{ g L}^{-1}$

$= 49.5 \text{ g L}^{-1}$

Step 6 — Sanity check

Answer: 0.825 mol L⁻¹, or 49.5 g L⁻¹ — about 4.95% ethanoic acid by mass.

Worked ExampleA calculation with a 2 : 1 mole ratio

Step 1 — Moles of titrant

$n(\text{NaOH}) = 0.1052 \times 0.02140 = 2.25128 \times 10^{-3} \text{ mol}$

Step 2 — Apply the mole ratio

Sulfuric acid is diprotic, so one mole of it neutralises two moles of hydroxide:

$\text{H}_2\text{SO}_4(aq) + 2\text{NaOH}(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(l)$

The ratio H₂SO₄ : NaOH is 1 : 2, so there is half as much acid as base:

$n(\text{H}_2\text{SO}_4) = \frac{2.25128 \times 10^{-3}}{2} = 1.12564 \times 10^{-3} \text{ mol}$

Check the direction: we are going from the "2" to the "1", so the number should get smaller — and it did.

Step 3 — Concentration in the diluted solution

$c = \frac{1.12564 \times 10^{-3}}{0.0250} = 0.0450256 \text{ mol L}^{-1}$

Step 4 — Scale back

The dilution factor is $500.0 / 10.0 = 50$ :

$c(\text{product}) = 0.0450256 \times 50 = 2.25128 \text{ mol L}^{-1}$

$c(\text{product}) = 2.25 \text{ mol L}^{-1}$

Step 5 — Significant figures

Answer: the drain cleaner contains sulfuric acid at 2.25 mol L⁻¹.

The titration calculation, step by step

The route from titre to answer

The relationships

The mole ratio step

Applying the dilution factor

Converting to useful units

Step 1 — Moles of titrant used

Step 2 — Moles of ethanoic acid, via the mole ratio

Step 3 — Concentration in the diluted vinegar

Step 4 — Scale back to the original vinegar

Step 5 — Convert to g L−1

Step 6 — Sanity check

Step 1 — Moles of titrant

Step 2 — Apply the mole ratio

Step 3 — Concentration in the diluted solution

Step 4 — Scale back

Step 5 — Significant figures

The titration calculation, step by step

The route from titre to answer

The relationships

The mole ratio step

Applying the dilution factor

Converting to useful units

Step 1 — Moles of titrant used

Step 2 — Moles of ethanoic acid, via the mole ratio

Step 3 — Concentration in the diluted vinegar

Step 4 — Scale back to the original vinegar

Step 5 — Convert to g L−1

Step 6 — Sanity check

Step 1 — Moles of titrant

Step 2 — Apply the mole ratio

Step 3 — Concentration in the diluted solution

Step 4 — Scale back

Step 5 — Significant figures

Step 5 — Convert to g L⁻¹

Step 5 — Convert to g L⁻¹