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CHEMISTRY

Titration calculationsHIGHER This whole page covers Higher material

Calculating a concentrationIn a titration, you will know the concentration and volume of one of the reagents. You will only know the volume of the other reagent – its concentration will have to be calculated using the mean volume from the titration. The worked example below shows you the steps needed to do this.

25.00 cm3 of sodium hydroxide solution was pipetted into a conical fl ask. It was titrated against 0.10 mol dm–3 hydrochloric acid. The mean volume of acid needed was 24.00 cm3.

(a) Write the balanced equation for the reaction.

HCl + NaOH → NaCl + H2O

(b) Calculate the concentration of sodium hydroxide used in the titration above.

Volume of HCl = 24.00 cm3 ÷ 1000 = 0.024 dm3

Number of moles of HCl = concentration × volume = 0.10 mol dm–3 × 0.024 dm3 = 0.0024 molFrom the equation, 1 mol of HCl reacts with 1 mol of NaOH, so there will be 0.0024 mol of NaOHVolume of NaOH = 25.00 cm3 ÷ 1000 = 0.025 dm3

Concentration of NaOH = number of moles ÷ volume = 0.0024 mol ÷ 0.025 dm3 = 0.096 mol dm–3

A titration calculation like this one is usually worth three marks in the examination. In a recent examination, fewer than one in ten of students gained full marks and over half only gained one mark. Remember that your working out often earns marks, even if the fi nal answer is wrong.

Students have struggled with this topic in recent exams - be prepared!

Divide by 1000 to convert

from cm3 to dm3.

Notice that you do not need to

use any relative atomic masses

or relative formula masses in

these calculations.

1. 25.00 cm3 of potassium hydroxide, KOH, solution was pipetted into a conical fl ask. It was titrated against 0.20 mol dm–3 nitric acid, HNO3.

(a) A group of students carried out the experiment three times and the volume of acid used was 25.5cm3, 29.0 cm3 and 27.5 cm3. Calculate the mean volume of the acid used. (2 marks)

(b) With another group of students the mean volume of the acid used was 28.00 cm3. Calculate the concentration of potassium hydroxide solution. (3 marks)

target

A*-C

This whole page covers Higher material

A titration should be repeated to

identify any anomalous values. Calculate

the volume of solution added from

the burette by subtracting the start

reading from the end reading in each

run. Any anomalous volumes should

be left out when calculating the mean

volume added.

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CHEMISTRY

More calculationsfrom equations

HIGHER This whole page covers Higher material

Calculating a volumeYou can use mole calculations to predict the volume of one reagent that will neutralise a measured volume of another reagent. To do this, you need to know the concentrations of both reactants.

(a) Write the balanced equation for the reaction between sulfuric acid and sodium hydroxide.

H2SO4 + 2NaOH → Na2SO4 + 2H2O

(b) What volume of 1.0 mol dm–3 sulfuric acid will be needed to neutralise 25.00 cm3 of 0.8 mol dm–3 sodium hydroxide solution?

Volume of NaOH = 25.00 cm3 ÷ 1000 = 0.025 dm3

Number of moles of NaOH = concentration × volume = 0.8 mol dm–3 × 0.025 dm3 = 0.02 mol

From the equation, 1 mol of H2SO4 reacts with 2 mol of NaOH, so there will be 0.01 mol of H2SO4

Volume of H2SO4 = number of moles ÷ concentration = 0.01 mol ÷ 1.0 mol dm–3 = 0.01 dm3 (or 10 cm3)

Notice that 1 mol of H2SO4 reacts with 2 mol

of NaOH. This is different than the situation

when hydrochloric acid is used, when 1 mol of

HCl reacts with just 1 mol of NaOH:

HCl + NaOH → NaCl + H2O

The ‘mole ratio’ between sulfuric acid and

sodium hydroxide solution is 1:2 – this is why

the number of moles of sulfuric acid is half

the number of moles of sodium hydroxide.

Divide by 1000 to convert

from cm3 to dm3.

(or )

Multiply by 1000 to convert from dm3 to cm3.

1. Calculate the volume of 0.5 mol dm–3 hydrochloric acid needed to neutralise 25.00 cm3 of 0.20 mol dm–3 sodium hydroxide solution. (3 marks)

2. Calculate the volume of 0.10 mol dm–3 sodium hydroxide solution needed to neutralise 20.00 cm3 of 0.25 mol dm–3 hydrochloric acid. (3 marks)

3. Calculate the volume of 0.25 mol dm–3 sulfuric acid needed to neutralise 20.00 cm3 of 0.10 mol dm–3 sodium hydroxide solution. (3 marks)

sulfuric acid needed to neutralise 20.00 cmsolution.

target

C-A*

1. Calculate the volume of 0.5 mol dm

2. Calculate the volume of 0.10 mol dm

target

C-A*

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CHEMISTRY

Chemistry extended writing 1

This is a basic answer. The � ame test is not correct – ammonium ions do not give a � ame colour. The test for the bromide ion is correct, although it does not give the other reagent added. However, it does identify the substance that makes the cream precipitate.

Remember that the method you need here is titration. You need to say how this method works. Remember that the � nal crystals made need to be dry and pure.

This is a good answer. The description of the bromide ion test is very good: the chemical substances suggested for the test are correct and ‘creamy’ is OK although ‘cream’ is more usual. The equation is also correct, including the state symbols. The test for the ammonium ion is mostly correct, but there are some important details missing. The solution usually needs warming, and there should be a test given for ammonia gas (using moist red litmus paper). These details would make the answer excellent – if you added the equation for the reaction of ammonium bromide with sodium hydroxide to make ammonia gas, the examiner would be very impressed indeed!

When answering extended writing questions you should try to:✓ apply your scienti� c knowledge✓ present your answer in a logical and organised way✓ write full sentences, and make sure that your spelling, punctuation and grammar are

as good as you can make them.

Ammonium bromide, NH 4Br, is a white solid.

Describe chemical tests that you could carry on an unknown solid to identify if it is ammonium bromide. You should include balanced chemical equations for any tests that you describe. (6 marks)

Sample answer 1You can do a � ame test for the ammonium ion and the � ame goes yellow. To test for bromide ions, you add silver nitrate. This makes a sort of creamy-white precipitate of silver bromide.

Sample answer 2Ammonium ion: add sodium hydroxide solution. This makes ammonia gas, which has a nasty smell. It’s also alkaline.Bromide ion: add nitric acid and then silver nitrate solution. This makes a creamy precipitate. The equation is NH4Br (aq) + AgNO3 (aq) → AgBr (s) + NH4NO3 (aq)

1. A chemist wants to make a sample of the soluble salt potassium chloride. He has been given a solution of potassium hydroxide and some hydrochloric acid.

Describe how he would make pure, dry crystals of potassium chloride. (6 marks)

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