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In this worksheet, students will learn how to solve titration problems where there isn't a 1:1 ratio of acid neutralising alkali, or where the concentrations are given in terms of mass, not moles. Key stage:  KS 4

GCSE Boards:   OCR 21st Century, Eduqas

Curriculum topic:   Chemical Analysis, Chemistry of Acids

Curriculum subtopic:   How are the Amounts of Chemicals in Solution Measured?, Chemistry of Acids

Difficulty level:   ### QUESTION 1 of 10

Titration is a really important method in chemistry. It lets us control the amounts of reactants in a reaction carefully, and we can use the results to work out the concentration of an unknown solution. There are some tricky aspects of titration, which examiners may ask you about to get marks corresponding to the very top grades. (If you're not sure about the basics of titration, try the level 1 or 2 activities on titration first, even if you don't normally do them. There's too much to fit in one activity for this topic.)

Recap

Once you have done a titration experiment, you know that a definite volume of acid neutralises a definite volume of alkali. If you know the concentration of one of these solutions, you can work out how many moles there are of that chemical. That tells you how many moles there are of the other chemical, so you can work out its concentration as well.

First difficulty: what if 1 mole of acid doesn't neutralise 1 mole of alkali?

If you neutralise HCl with NaOH, 1 mole of alkali neutralises 1 mole of acid: HCl + NaOH → NaCl + H2O.

The equation for neutralising sulfuric acid with NaOH is H2SO4 + 2 NaOH → Na2SO4 + 2 H2O. That means you need two moles of sodium hydroxide to neutralise one mole of acid.

Example: 25.0 cm3 of NaOH (unknown concentration) is neutralised by 10.5 cm3 of H2SO4 (0.1 mols per dm3 concentration). What is the concentration of the sodium hydroxide?

Amount of acid = (10.5 / 1000) x 0.1 = 0.001 05 mol.

From the equation, there is twice as much alkali as acid, so amount of alkali = 0.001 05 x 2 = 0.002 10 mol.

The concentration of NaOH = 0.002 10 ÷ (25.0 / 1000) = 0.0084 mol per dm3 (also written mol dm-3).

Second difficulty: what if the concentration is in grams per dm3, not moles per dm3?

If the concentration is in grams per dm3, we need to convert grams to moles before doing the titration calculations. We do that using relative formula mass.

Example: 25.0 cm3 of NaOH (unknown concentration) is neutralised by 20.0 cm3 of HCl (7.3 g per dm3 concentration). What is the concentration of the sodium hydroxide in g dm-3?

Work out the number of moles of HCl:

Relative formula mass of HCl = 1 + 35.5 = 36.5. So 1 mol = 36.5 g.

7.3  ÷  36.5 = 0.2, so 7.3 g dm-3 = 0.2 mol dm-3

Total number of moles of HCl = (20.0 / 1000) x 0.2 = 0.004 mol.

Work out the number of moles of NaOH:

Neutralising 1 mol of NaOH needs 1 mol of HCl, so moles of NaOH = 0.004 mol.

Convert the moles of NaOH into grams:

Relative formula mass of NaOH = 23 + 16 + 1 = 40. So 1 mol = 40 g.

So the mass of NaOH = 0.004 x 40 = 0.16 g.

Work out the concentration:

Concentration = mass ÷ volume = 0.16 ÷ (25 / 1000) = 6.4 g dm-3

None of the steps in the calculation are too hard; the difficult thing is doing all of them in the right order to get to the answer. It's a lot easier to do this if you add words to the number. Writing down what each calculation is trying to do will make it much easier to focus on doing the right things in the right order. it will also help someone reading your work (like an examiner) understand what you have done.

If we neutralise sulfuric acid with potassium hydroxide, the reaction equation is this:

H2SO4 (aq) + 2 KOH (aq) → K2SO4 (aq) + 2 H2O (l)

In a titration experiment, 28.3 cm3 of sulfuric acid (concentration 0.2 mol dm-3) neutralises 25.0 cm3 of potassium hydroxide.

How many moles of sulfuric acid were used in this experiment, and how many moles of potassium hydroxide?

If we neutralise sulfuric acid with potassium hydroxide, the reaction equation is this:

H2SO4 (aq) + 2 KOH (aq) → K2SO4 (aq) + 2 H2O (l)

In a titration experiment, 28.3 cm3 of sulfuric acid (concentration 0.2 mol dm-3) neutralises 25.0 cm3 of potassium hydroxide.

In the last question, you worked out that there were 0.01132 moles of potassium hydroxide.

What was the concentration of potassium hydroxide? Give your answer to 3 significant figures.

Sodium oxide (Na2O) is soluble in water, making an alkali.

If we neutralise hydrochloric acid (HCl)  with sodium oxide, what is the balanced equation? Pick one half from each list.

If we neutralise hydrochloric acid (HCl)  with sodium oxide (Na2O), the reaction equation is this:

2 HCl (aq) + Na2O (aq) → 2 NaCl (aq) + 2 H2O (l)

In a titration experiment, 17.4 cm3 of hydrochloric acid (concentration 0.15 mol dm-3) neutralises 25.0 cm3 of sodium oxide.

How many moles of hydrochloric acid were used in this experiment, and how many moles of sodium oxide?

If we neutralise hydrochloric acid (HCl)  with sodium oxide (Na2O), the reaction equation is this:

2 HCl (aq) + Na2O (aq) → 2 NaCl (aq) + 2 H2O (l)

In a titration experiment, 17.4 cm3 of hydrochloric acid (concentration 0.15 mol dm-3) neutralises 25.0 cm3 of sodium oxide.

In the last question, you worked out that there was 0.001305 mol of sodium oxide, so what is the concentration of the alkali?

If we dissolve 5.85 g of NaCl in 1 dm3 water, what is its concentration in mol dm-3?

The equation for neutralising hydrochloric acid with sodium hydroxide is

HCl (aq) + NaOH (aq) → NaCl (aq) + H<sub>2</sub>O (l).

We have hydrochloric acid made by dissolving 7.3 g in 500 cm3 water.

We have 25 cm3 sodium hydroxide, with unknown concentration.

What is the concentration of the hydrochloric acid, in mol dm-3?

The equation for neutralising hydrochloric acid with sodium hydroxide is

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l).

We have hydrochloric acid made by dissolving 7.3 g in 500 cm3 water. In the last question, you worked out that this concentration was 0.4 mol dm-3.

We have 25 cm3 sodium hydroxide, with unknown concentration.

In a titration experiment, the volume of hydrochloric acid needed to neutralise the sodium hydroxide was 26.9 cm3.

How many moles of sodium hydroxide were in the conical flask?

The equation for neutralising hydrochloric acid with sodium hydroxide is

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l).

We have hydrochloric acid made by dissolving 7.3 g in 500 cm3 water. Earlier on, you worked out that this concentration was 0.4 mol dm-3.

We have 25 cm3 sodium hydroxide, with unknown concentration.

In a titration experiment, the volume of hydrochloric acid needed to neutralise the sodium hydroxide was 26.9 cm3.

In the last question, you worked out that there was 0.01076 mol of sodium hydroxide.

What is the concentration of sodium hydroxide in mol dm-3?

The equation for neutralising hydrochloric acid with sodium hydroxide is

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l).

We have hydrochloric acid made by dissolving 7.3 g in 500 cm3 water. Earlier on, you worked out that this concentration was 0.4 mol dm-3.

We have 25 cm3 sodium hydroxide, with unknown concentration.

In a titration experiment, the volume of hydrochloric acid needed to neutralise the sodium hydroxide was 26.9 cm3.

Earlier on, you worked out that there was 0.01076 mol of sodium hydroxide.

What is the concentration of sodium hydroxide in mol dm-3?

In the last question, you worked out that the concentration was 0.4304 mol dm-3.

What is the concentration of sodium hydroxide in g dm-3? Give your answer to three significant figures.

• Question 1

If we neutralise sulfuric acid with potassium hydroxide, the reaction equation is this:

H2SO4 (aq) + 2 KOH (aq) → K2SO4 (aq) + 2 H2O (l)

In a titration experiment, 28.3 cm3 of sulfuric acid (concentration 0.2 mol dm-3) neutralises 25.0 cm3 of potassium hydroxide.

How many moles of sulfuric acid were used in this experiment, and how many moles of potassium hydroxide?

EDDIE SAYS
number of moles of sulfuric acid = 0.2 x (28.3 / 1000) = 0.00566 1 mol of sulfuric acid neutralises 2 mol potassium hydroxide, so number of moles of potassium hydroxide = 2 x 0.00566 = 0.01132.
• Question 2

If we neutralise sulfuric acid with potassium hydroxide, the reaction equation is this:

H2SO4 (aq) + 2 KOH (aq) → K2SO4 (aq) + 2 H2O (l)

In a titration experiment, 28.3 cm3 of sulfuric acid (concentration 0.2 mol dm-3) neutralises 25.0 cm3 of potassium hydroxide.

In the last question, you worked out that there were 0.01132 moles of potassium hydroxide.

What was the concentration of potassium hydroxide? Give your answer to 3 significant figures.

EDDIE SAYS
number of moles of potassium hydroxide = 0.01132. concentration = 0.01132 ÷ (25.0 / 1000) = 0.4528 mol dm-3. To 3 significant figures, this is 0.453 mol dm-3.
• Question 3

Sodium oxide (Na2O) is soluble in water, making an alkali.

If we neutralise hydrochloric acid (HCl)  with sodium oxide, what is the balanced equation? Pick one half from each list.

EDDIE SAYS
Start by balancing the sodiums; to use both Na on the reactant side, we need to make 2 NaCl on the product side. Now balance the chlorines; to have enough chlorine to make 2 NaCl, we need 2 HCl on the reactant side. Now the oxygen and hydrogen balance as well, so everything balances.
• Question 4

If we neutralise hydrochloric acid (HCl)  with sodium oxide (Na2O), the reaction equation is this:

2 HCl (aq) + Na2O (aq) → 2 NaCl (aq) + 2 H2O (l)

In a titration experiment, 17.4 cm3 of hydrochloric acid (concentration 0.15 mol dm-3) neutralises 25.0 cm3 of sodium oxide.

How many moles of hydrochloric acid were used in this experiment, and how many moles of sodium oxide?

EDDIE SAYS
number of moles of hydrochloric acid = 0.15 x (17.4 / 1000) = 0.00261 2 mol of hydrochloric acid neutralises 1 mol sodium oxide, so number of moles of sodium oxide = 0.5 x 0.00261 = 0.001305.
• Question 5

If we neutralise hydrochloric acid (HCl)  with sodium oxide (Na2O), the reaction equation is this:

2 HCl (aq) + Na2O (aq) → 2 NaCl (aq) + 2 H2O (l)

In a titration experiment, 17.4 cm3 of hydrochloric acid (concentration 0.15 mol dm-3) neutralises 25.0 cm3 of sodium oxide.

In the last question, you worked out that there was 0.001305 mol of sodium oxide, so what is the concentration of the alkali?

EDDIE SAYS
0.001305 mol in 25 cm3 gives concentration = 0.001305 ÷ (25.0 / 1000) = 0.0522
• Question 6

If we dissolve 5.85 g of NaCl in 1 dm3 water, what is its concentration in mol dm-3?

EDDIE SAYS
The relative formula mass for NaCl = 23 + 35.5 = 58.5 So 1 mol of NaCl has mass = 58.5 g. 5.85 / 58.5 = 0.1, so 5.85 g is 0.1 mol. If this is dissolved in 1 dm3 water, the concentration is 0.1 mol dm-3.
• Question 7

The equation for neutralising hydrochloric acid with sodium hydroxide is

HCl (aq) + NaOH (aq) → NaCl (aq) + H<sub>2</sub>O (l).

We have hydrochloric acid made by dissolving 7.3 g in 500 cm3 water.

We have 25 cm3 sodium hydroxide, with unknown concentration.

What is the concentration of the hydrochloric acid, in mol dm-3?

EDDIE SAYS
The molar mass of HCl = 1 + 35.5 = 36.5. 7.3 ÷ 36.5 = 0.2, so 7.3 g is 0.2 mol. Concentration = amount ÷ liquid volume = 0.2 ÷ 0.5 = 0.4 mol dm-3
• Question 8

The equation for neutralising hydrochloric acid with sodium hydroxide is

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l).

We have hydrochloric acid made by dissolving 7.3 g in 500 cm3 water. In the last question, you worked out that this concentration was 0.4 mol dm-3.

We have 25 cm3 sodium hydroxide, with unknown concentration.

In a titration experiment, the volume of hydrochloric acid needed to neutralise the sodium hydroxide was 26.9 cm3.

How many moles of sodium hydroxide were in the conical flask?

EDDIE SAYS
Amount of HCl = (26.9 /1000) x 0.4 = 0.01076 mol. In this reaction, 1 mole of HCl neutralises 1 mole of NaOH.
• Question 9

The equation for neutralising hydrochloric acid with sodium hydroxide is

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l).

We have hydrochloric acid made by dissolving 7.3 g in 500 cm3 water. Earlier on, you worked out that this concentration was 0.4 mol dm-3.

We have 25 cm3 sodium hydroxide, with unknown concentration.

In a titration experiment, the volume of hydrochloric acid needed to neutralise the sodium hydroxide was 26.9 cm3.

In the last question, you worked out that there was 0.01076 mol of sodium hydroxide.

What is the concentration of sodium hydroxide in mol dm-3?

EDDIE SAYS
Concentration = 0.01076 ÷ (25.0 / 1000) Concentration = 0.4304 mol dm-3
• Question 10

The equation for neutralising hydrochloric acid with sodium hydroxide is

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l).

We have hydrochloric acid made by dissolving 7.3 g in 500 cm3 water. Earlier on, you worked out that this concentration was 0.4 mol dm-3.

We have 25 cm3 sodium hydroxide, with unknown concentration.

In a titration experiment, the volume of hydrochloric acid needed to neutralise the sodium hydroxide was 26.9 cm3.

Earlier on, you worked out that there was 0.01076 mol of sodium hydroxide.

What is the concentration of sodium hydroxide in mol dm-3?

In the last question, you worked out that the concentration was 0.4304 mol dm-3.

What is the concentration of sodium hydroxide in g dm-3? Give your answer to three significant figures.

EDDIE SAYS
The molar mass of NaOH = 23 + 16 + 1 = 40, so the mass of 1 mole of NaOH = 40 g. The mass of 0.4304 moles of NaOH = 0.4304 x 40 = 17.216 g To 3 significant figures, this is 17.2 g.
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