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Edexcel IGCSE Chemistry

Topic questions Quantitative chemistry Concentration of solutions

Concentration of solutions

22 questions · 259 marks · about 4.5 hours

Concentration of solutions questions

Edexcel 4CH1 Concentration of solutions 22 questions

Chemistry Paper 2 June 2025 (R), Question 6 15 marks

Also involves Amounts of substance in equations

A student does a titration to find the volume of phosphoric acid solution (H3PO4) needed to neutralise a sample of potassium hydroxide solution (KOH).

One of the products of the reaction is potassium phosphate.

This is the student’s method.

• use a measuring cylinder to transfer 25 cm3 of potassium hydroxide solution to a conical flask

• add two drops of an indicator

• fill a burette with phosphoric acid solution and record the initial reading on the burette

• add the acid to the conical flask

• record the final burette reading when the indicator changes colour

This is the equation for the reaction.

Question 6, shown as an image
(a)

Explain two improvements to the student’s method that would give more accurate results.

[4]
Reveal mark scheme Mark scheme for question 6(a)
(b)

The student repeats the titration correctly.

The potassium hydroxide solution has a concentration of 0.150 mol / dm3.

The phosphoric acid solution has a concentration of 0.0400 mol / dm3.

Calculate the volume, in cm3, of phosphoric acid solution needed to neutralise 25.0 cm3 of 0.150 mol / dm3 potassium hydroxide solution.

Give your answer to three significant figures.

[4]
Reveal mark scheme Mark scheme for question 6(b)
(c)

Potassium phosphate solution reacts with calcium chloride solution.

One of the products is the insoluble solid, calcium phosphate.

(c)(i)

Complete the equation for the reaction.

Question 6, shown as an image
[1]
(c)(ii)

The student reacts a solution containing 0.0400 mol of potassium phosphate with an excess of calcium chloride solution.

The student obtains 5.70 g of calcium phosphate.

Show that the percentage yield for the reaction is about 90%

[for Ca3(PO4)2 Mr = 310]

[3]
Reveal mark scheme Mark scheme for question 6(c)(ii)
(c)(iii)

Describe how the student can obtain a pure, dry sample of calcium phosphate from the mixture at the end of the reaction.

[3]
Reveal mark scheme Mark scheme for question 6(c)(iii)

Chemistry Paper 2 November 2024, Question 7 12 marks

Also involves Titrations

This question is about alcohols and carboxylic acids.

(a)

These are the structural formulae of alcohol A and carboxylic acid B.

Alcohol A
CH3CH2CH2OH

Carboxylic acid B CH3(CH2)nCOOH

(a)(i)

Name alcohol A.

[1]
Reveal mark scheme Mark scheme for question 7(a)(i)
(a)(ii)

Draw the displayed formula of alcohol A.

[1]
Reveal mark scheme Mark scheme for question 7(a)(ii)
(a)(iii)

Carboxylic acid B has a chain of carbon atoms with no branches.

The number of CH2 units is represented by the letter n.

Calculate the value of n.

[for carboxylic acid B, Mr = 242]

[3]
Reveal mark scheme Mark scheme for question 7(a)(iii)
(a)(iv)

Alcohol A and carboxylic acid B react together to form an ester.

Give the other product of the reaction.

[1]
Reveal mark scheme Mark scheme for question 7(a)(iv)
(b)

These are the structural formulae of dicarboxylic acid C and diol D.

Dicarboxylic acid C HOOCCH2CH2COOH

Diol D
HOCH2CH2OH

(b)(i)

This is the equation for the reaction between dicarboxylic acid C and sodium hydroxide solution.

HOOCCH2CH2COOH + 2NaOH maths symbol NaOOCH2CH2COONa + 2H2O

25.0 cm3 of 0.150 mol/dm3 sodium hydroxide solution is completely neutralised by 17.5 cm3 of a solution of dicarboxylic acid C.

Calculate the concentration, in mol/dm3, of the solution of dicarboxylic acid C.

Give your answer to three significant figures.

concentration =........ mol/dm

[4]
Reveal mark scheme Mark scheme for question 7(b)(i)
(b)(ii)

Dicarboxylic acid C and diol D react to form a polyester.

Draw the displayed formula of the repeat unit of this polyester.

[2]
Reveal mark scheme Mark scheme for question 7(b)(ii)

Chemistry Paper 2 June 2024, Question 6 14 marks

A scientist reacts lead(II) nitrate solution with sodium bromide solution.

This is the equation for the reaction.

Pb(NO3)2(aq) + 2NaBr(aq) maths symbol PbBr2(s) + 2NaNO3(aq)

(a)

Describe how the scientist could obtain a pure dry sample of lead(II) bromide (PbBr2) from the reaction mixture.

[3]
Reveal mark scheme Mark scheme for question 6(a)
(b)

The scientist reacts an excess of lead(II) nitrate solution with 25 cm3 of sodium bromide solution of concentration 2.0 mol/dm3.

(b)(i)

Show that the amount of sodium bromide used is 0.050 mol.

[1]
Reveal mark scheme Mark scheme for question 6(b)(i)
(b)(ii)

Show that the maximum theoretical mass of lead(II) bromide is approximately 9 g.

[for PbBr2 Mr = 367]

[2]
Reveal mark scheme Mark scheme for question 6(b)(ii)
(c)

The scientist electrolyses molten lead(II) bromide using this apparatus.

Figure for question 6
(c)(i)

Explain why lead(II) bromide needs to be molten rather than solid for electrolysis to occur.

[3]
Reveal mark scheme Mark scheme for question 6(c)(i)
(c)(ii)

The electrolyte is at a temperature of 400maths symbolC.

Explain a suitable material for the electrodes.

[2]
Reveal mark scheme Mark scheme for question 6(c)(ii)
(c)(iii)

Give the half‑equation that occurs at the negative electrode.

[1]
Reveal mark scheme Mark scheme for question 6(c)(iii)
(d)

Bromine forms at the positive electrode.

This is the half‑equation for the reaction at the positive electrode.

Question 6, shown as an image
(d)(i)

State what is observed at the positive electrode.

[1]
Reveal mark scheme Mark scheme for question 6(d)(i)
(d)(ii)

State why the half‑equation represents an oxidation reaction.

[1]
Reveal mark scheme Mark scheme for question 6(d)(ii)

Chemistry Paper 2 November 2023, Question 6 11 marks

(a)

The diagram shows two pieces of apparatus used in a titration.

Figure for question 6

Give the names of these pieces of apparatus.

[2]
Reveal mark scheme Mark scheme for question 6(a)
(b)

Give the name of a suitable indicator that can be used in an acid-alkali titration.

[1]
Reveal mark scheme Mark scheme for question 6(b)
(c)

A student does a titration using sodium carbonate solution and dilute nitric acid.

This is the equation for the reaction.

Question 6, shown as an image

The table shows the concentrations of the two solutions and the volume of sodium carbonate used in the titration.

Figure for question 6

Use the equation and the data in the table to answer these questions.

(c)(i)

Calculate the volume of dilute nitric acid that the student would need to neutralise the sodium carbonate solution.

[3]
Reveal mark scheme Mark scheme for question 6(c)(i)
(c)(ii)

Calculate the volume, in cm3, of carbon dioxide gas at rtp that would be produced from the 25.0 cm3 of the sodium carbonate solution.

[at rtp, molar volume = 24 000 cm3]

[2]
Reveal mark scheme Mark scheme for question 6(c)(ii)
(d)

Describe a test to show that sodium carbonate solution contains carbonate ions.

[3]
Reveal mark scheme Mark scheme for question 6(d)

Chemistry Paper 2 June 2023, Question 4 11 marks

A student does a titration to find the concentration of potassium hydroxide solution.

This is the student’s method.

• add 25.0 cm3 of the potassium hydroxide solution to a conical flask

• add a few drops of methyl orange indicator to the conical flask

• fill a burette with dilute sulfuric acid and record the initial burette reading

• place the conical flask on a white tile

• add the acid from the burette to the mixture in the conical flask, swirling the flask continuously

• when the indicator changes colour at the end point, record the final burette reading

Repeat the titration to obtain concordant results.

(a)

Give the name of the most suitable piece of apparatus to measure out 25.0 cm3 of potassium hydroxide solution.

[1]
Reveal mark scheme Mark scheme for question 4(a)
(b)

Give the colour of methyl orange in potassium hydroxide solution and in dilute sulfuric acid.

colour in potassium hydroxide solution

colour in dilute sulfuric acid

[2]
Reveal mark scheme Mark scheme for question 4(b)
(c)

State why the student places the conical flask on a white tile.

[1]
Reveal mark scheme Mark scheme for question 4(c)
(d)

State why the student swirls the flask continuously.

[1]
Reveal mark scheme Mark scheme for question 4(d)
(e)

State what is meant by the term concordant results.

[1]
Reveal mark scheme Mark scheme for question 4(e)
(f)

The student finds that 15.00 cm3 of sulfuric acid of concentration 0.180 mol / dm3 neutralises 25.0 cm3 of potassium hydroxide solution.

This is the equation for the reaction.

Question 4, shown as an image

Calculate the concentration of the potassium hydroxide solution.

concentration =........ mol / dm

[3]
Reveal mark scheme Mark scheme for question 4(f)
(g)

This is the ionic equation for the reaction between an acid and an alkali.

Question 4, shown as an image

Explain why the OH− ion is a proton acceptor in this reaction.

[2]
Reveal mark scheme Mark scheme for question 4(g)

Chemistry Paper 2 June 2023 (R), Question 6 15 marks

Also involves Amounts of substance in equations · Titrations

(a)

Describe the forces of attraction in metallic bonding.

[2]
Reveal mark scheme Mark scheme for question 6(a)
(b)

When a small piece of potassium is added to water, bubbles of hydrogen gas are observed.

(b)(i)

Give the test for hydrogen gas.

[1]
Reveal mark scheme Mark scheme for question 6(b)(i)
(b)(ii)

Give two other observations that would be made.

[2]
Reveal mark scheme Mark scheme for question 6(b)(ii)
(c)

Explain why lithium is less reactive than potassium.

Refer to atomic structure in your answer.

[3]
Reveal mark scheme Mark scheme for question 6(c)
(d)

This is the equation for the reaction between sodium and water.

Question 6, shown as an image

A mass of 0.750 g of sodium is reacted with an excess of water.

Calculate the volume of hydrogen gas produced, in cm3, at room temperature.

[molar volume of hydrogen at rtp = 24 000 cm3]

[for Na,Ar = 23]

Give your answer to three significant figures.

[4]
Reveal mark scheme Mark scheme for question 6(d)
(e)

This is the equation for the reaction between sodium hydroxide and sulfuric acid.

Question 6, shown as an image

A volume of 25.0 cm3 of sodium hydroxide solution is completely neutralised by 16.3 cm3 of 0.0500 mol / dm3 sulfuric acid.

Calculate the concentration of the sodium hydroxide solution in mol / dm3.

concentration =........ mol / dm

[3]
Reveal mark scheme Mark scheme for question 6(e)

Chemistry Paper 2 January 2023 (R), Question 7 11 marks

Also involves Amounts of substance in equations

A student is asked to find the concentration of a solution of nitric acid by doing a titration.

The student is provided with sodium hydroxide solution with a concentration of 0.350 mol/dm3

These are the first four steps in the student’s method.

• use a pipette to add exactly 25.0 cm3 of the nitric acid solution to a conical flask placed on a white tile

• add a few drops of indicator to the flask

• use a burette to add sodium hydroxide solution to the flask until the indicator has changed colour

• record the volume needed for the indicator to change colour

(a)

Give the reason for using the white tile.

[1]
Reveal mark scheme Mark scheme for question 7(a)
(b)

Universal indicator is not a suitable indicator to use in a titration.

Complete the table to show the name of a suitable indicator and its final colour in this titration.

Figure for question 7
[2]
Reveal mark scheme Mark scheme for question 7(b)
(c)

The student wants to obtain an accurate value for the volume of sodium hydroxide solution needed to neutralise the 25.0 cm3 of nitric acid.

Describe all the further steps the student should take.

[5]
Reveal mark scheme Mark scheme for question 7(c)
(d)

This is a summary of the student’s results.

volume of HNO3 used = 25.0 cm3

concentration of NaOH solution = 0.350 mol/dm3

volume of NaOH solution needed for neutralisation = 18.80 cm3

The equation for the reaction is

HNO3 + NaOH maths symbol NaNO3 + H2O

(d)(i)

Calculate the amount, in moles, of NaOH that reacts.

[1]
Reveal mark scheme Mark scheme for question 7(d)(i)
(d)(ii)

Determine the amount, in moles, of HNO3 that reacts.

amount of HNO3 =........

[1]
(d)(iii)

Calculate the concentration, in mol/dm3, of the HNO3

concentration =........ mol/dm

[1]
Reveal mark scheme Mark scheme for question 7(d)(iii)

Chemistry Paper 2 January 2023, Question 7 13 marks

Also involves Amounts of substance in equations

This question is about hydrazine, N2H4

(a)

This is the displayed formula for a molecule of hydrazine.

Figure for question 7

Complete the dot‑and‑cross diagram for hydrazine.

Figure for question 7
[2]
Reveal mark scheme Mark scheme for question 7(a)
(b)
Figure for question 7
(b)(i)

Use the data in the table to calculate the total energy needed to break all the bonds in the reactants.

[2]
Reveal mark scheme Mark scheme for question 7(b)(i)
(b)(ii)

Use the data in the table to calculate the total energy released when all the bonds in the products are made.

[2]
Reveal mark scheme Mark scheme for question 7(b)(ii)
(b)(iii)

Calculate the enthalpy change, Δ H, in kJ/mol, for the reaction.

ΔH =........ kJ/mol

[1]
Reveal mark scheme Mark scheme for question 7(b)(iii)
(b)(iv)

Explain, in terms of bonds broken and bonds made, why this reaction is exothermic.

[2]
Reveal mark scheme Mark scheme for question 7(b)(iv)
(c)

A sample of hydrazine is completely decomposed.

This is the equation for the decomposition of hydrazine.

Question 7, shown as an image

The products of the decomposition are bubbled through 1100 cm3 of water.

The ammonia completely dissolves in the water, but nitrogen is insoluble in water.

The nitrogen has a volume of 1570 cm3 at room temperature and pressure (rtp).

Calculate the concentration, in mol/dm3, of the ammonia solution.

Give your answer to 3 significant figures.

[for a gas at rtp, molar volume = 24 000 cm3]

concentration =........ mol/dm

[4]
Reveal mark scheme Mark scheme for question 7(c)

Chemistry Paper 2 June 2022, Question 2 9 marks

Also involves Amounts of substance in equations

(a)

The diagram shows two pieces of apparatus, X and Y.

Figure for question 2
(a)(i)

Give the name of each piece of apparatus.

[2]
Reveal mark scheme Mark scheme for question 2(a)(i)
(a)(ii)

In a titration, a student adds 25.0 cm3 of barium hydroxide solution to a conical flask.

Give a reason why it is better to use Y rather than X.

[1]
Reveal mark scheme Mark scheme for question 2(a)(ii)
(b)

The student uses methyl orange indicator in the titration.

(b)(i)

State the colour of methyl orange in barium hydroxide solution.

[1]
Reveal mark scheme Mark scheme for question 2(b)(i)
(b)(ii)

Give a reason why universal indicator is not suitable for use in a titration.

[1]
Reveal mark scheme Mark scheme for question 2(b)(ii)
(c)

The student adds some dilute nitric acid to a burette and does the titration.

The equation for the reaction is

Question 2, shown as an image

The student finds that 21.50 cm3 of nitric acid of concentration 0.600 mol/dm3 neutralises 25.0 cm3 of barium hydroxide solution.

Calculate the concentration, in mol/dm3, of the barium hydroxide solution.

concentration =........ mol/dm

[3]
Reveal mark scheme Mark scheme for question 2(c)
(d)

State why sulfuric acid would not be a suitable acid to use in this titration.

[1]
Reveal mark scheme Mark scheme for question 2(d)

Chemistry Paper 2 June 2022 (R), Question 6 12 marks

A student uses this method to do a titration.

• use a measuring cylinder to obtain 25 cm3 of sodium hydroxide solution

• transfer the solution to a conical flask

• add a few drops of universal indicator to the flask

• fill a burette with dilute sulfuric acid and record the initial burette reading

• add the acid to the flask, swirling the flask continuously

• add the acid slowly near the end‑point

• record the final burette reading at the end‑point

The student repeats the titration until at least two concordant results are obtained.

(a)

State what is meant by concordant results.

[1]
Reveal mark scheme Mark scheme for question 6(a)
(b)

Explain two improvements to the student’s method so that more accurate results are obtained.

[4]
Reveal mark scheme Mark scheme for question 6(b)
(c)

The student makes the improvements and repeats the titration.

The sulfuric acid has a concentration of 0.600 mol/dm3.

The sodium hydroxide solution has a concentration of 1.50 mol/dm3.

This is the equation for the reaction.

Question 6, shown as an image

Calculate the volume, in cm3, of sulfuric acid that the student needs to completely react with 25.0 cm3 of the sodium hydroxide solution.

[3]
Reveal mark scheme Mark scheme for question 6(c)
(d)

The student plans to obtain pure dry crystals of hydrated sodium sulfate.

They add the calculated volume of sulfuric acid to 25.0 cm3 of the sodium hydroxide solution to form sodium sulfate solution.

Describe what the student should do to obtain pure dry crystals of hydrated sodium sulfate from the solution.

[4]
Reveal mark scheme Mark scheme for question 6(d)

Chemistry Paper 2 January 2022 (R), Question 3 11 marks

Also involves Amounts of substance in equations

A student does a titration to find the concentration of a solution of dilute sulfuric acid.

The student uses these solutions and this apparatus.

• dilute sulfuric acid

• potassium hydroxide solution of concentration 0.240 mol / dm3

• methyl orange indicator

Figure for question 3
(a)

The student wants to find the volume of sulfuric acid needed to neutralise 25.0 cm3 of the potassium hydroxide solution.

Describe how the student should do this titration.

Assume that all pieces of apparatus are clean and dry.

[6]
Reveal mark scheme Mark scheme for question 3(a)
(b)

The student needs 15.00 cm3 of sulfuric acid to neutralise 25.0 cm3 of the potassium hydroxide solution.

This is the equation for the reaction.

Question 3, shown as an image
(b)(i)

Calculate the amount, in moles, of KOH in 25.0 cm3 of potassium hydroxide solution of concentration 0.240 mol / dm3.

[2]
Reveal mark scheme Mark scheme for question 3(b)(i)
(b)(ii)

Calculate the amount, in moles, of H2SO4 in 15.00 cm3 of the sulfuric acid.

amount of H2SO4 =........

[1]
(b)(iii)

Calculate the concentration, in mol / dm3, of the sulfuric acid.

concentration of sulfuric acid =........ mol / dm

[2]
Reveal mark scheme Mark scheme for question 3(b)(iii)

Chemistry Paper 2 January 2022, Question 7 15 marks

Also involves Amounts of substance in equations

This question is about esters.

Ester A reacts with water to form ethanoic acid and ethanol.

The displayed formulae of the reactants and products are shown in this equation

Question 7, shown as an image

ester A

The molar enthalpy change (Δ H) for the reaction is 0 kJ/mol.

(a)(i)

Draw a ring around the functional group in ester A.

[1]
Reveal mark scheme Mark scheme for question 7(a)(i)
(a)(ii)

Give the name of ester A.

[1]
Reveal mark scheme Mark scheme for question 7(a)(ii)
(a)(iii)

Describe a chemical test, other than using an indicator, to show that the reaction mixture contains ethanoic acid.

[2]
Reveal mark scheme Mark scheme for question 7(a)(iii)
(b)

Explain why the molar enthalpy change (Δ H) for the reaction between ester A and water is 0 kJ/mol.

In your answer, refer to the bonds broken and the bonds formed.

[2]
Reveal mark scheme Mark scheme for question 7(b)
(c)

A mixture of ester A and water is left in a sealed container until the reaction mixture reaches dynamic equilibrium.

(c)(i)

Describe what is meant by dynamic equilibrium.

[2]
Reveal mark scheme Mark scheme for question 7(c)(i)
(c)(ii)

Explain why adding a catalyst does not change the position of equilibrium.

[2]
Reveal mark scheme Mark scheme for question 7(c)(ii)
(d)

The ethanoic acid produced in the reaction is completely neutralised by 22.75 cm3 of 0.150 mol/dm3 barium hydroxide solution.

The equation for the neutralisation reaction is

Question 7, shown as an image

Calculate the amount, in moles, of ethanoic acid neutralised.

Give your answer to 3 significant figures.

[3]
Reveal mark scheme Mark scheme for question 7(d)
(e)

The structures of two organic compounds are shown.

Question 7, shown as an image
Question 7, shown as an image
Question 7, shown as an image

These compounds react together to form a polymer.

Give the repeat unit of the polymer formed.

[2]
Reveal mark scheme Mark scheme for question 7(e)

Chemistry Paper 2 November 2021, Question 6 11 marks

Also involves Amounts of substance in equations · Titrations · Addition polymerisation

This question is about alcohols, carboxylic acids and esters.

(a)

Ethanol can be manufactured by reacting ethene with steam in the presence of a phosphoric acid catalyst.

Which row gives the correct conditions of temperature and pressure for this reaction?

Figure for question 6
[1]
Reveal mark scheme Mark scheme for question 6(a)
(b)

Give the displayed formula of butanol.

[1]
Reveal mark scheme Mark scheme for question 6(b)
(c)

Ethanoic acid (CH3COOH) is a carboxylic acid present in vinegar.

(c)(i)

The concentration of CH3COOH in vinegar can be found by titration with aqueous potassium hydroxide (KOH).

The equation for the reaction is

CH3COOH + KOH maths symbol CH3COOK + H2O

In a titration, a 25.0 cm3 sample of vinegar is neutralised by 45.00 cm3 of KOH solution of concentration 0.400 mol/dm3.

Calculate the concentration, in mol/dm3, of CH3COOH in this sample of vinegar.

concentration =........ mol/dm

[2]
Reveal mark scheme Mark scheme for question 6(c)(i)
(c)(ii)

A sample of vinegar containing 0.0030 mol of CH3COOH is poured into a flask.

Calculate the maximum volume, in cm3, of carbon dioxide gas formed at rtp when excess sodium carbonate is added to the flask.

The equation for the reaction is

Question 6, shown as an image

[Assume that the molar volume of carbon dioxide at rtp is 24 000 cm3]

[2]
Reveal mark scheme Mark scheme for question 6(c)(ii)
(d)

Alcohols react with carboxylic acids to form esters.

Which alcohol could react to form the ester ethyl propanoate?

Question 6, shown as an image
[1]
Reveal mark scheme Mark scheme for question 6(d)
(e)

Polyesters are formed in condensation polymerisation reactions between dicarboxylic acids and diols.

(e)(i)

State one difference between condensation polymerisation and addition polymerisation.

[1]
Reveal mark scheme Mark scheme for question 6(e)(i)
(e)(ii)

The repeat unit of a polyester is

O O H H

Question 6, shown as an image

H H

Give the displayed formula of each of the two monomers needed to form this polyester.

Figure for question 6
[2]
Reveal mark scheme Mark scheme for question 6(e)(ii)
(e)(iii)

Give one advantage of biopolyesters.

[1]
Reveal mark scheme Mark scheme for question 6(e)(iii)

Chemistry Paper 2 January 2021 (R), Question 6 10 marks

Also involves Amounts of substance in equations

A student does a titration using dilute sulfuric acid to find the concentration of a solution of potassium hydroxide.

The student adds 25.0 cm3 of the potassium hydroxide solution to a conical flask. He then adds a few drops of methyl orange indicator.

The student does the titration four times.

(a)(i)

Name the piece of apparatus the student should use to add the potassium hydroxide solution.

[1]
Reveal mark scheme Mark scheme for question 6(a)(i)
(a)(ii)

What is the colour of methyl orange in an alkaline solution?

Question 6, shown as an image
[1]
Reveal mark scheme Mark scheme for question 6(a)(ii)
(b)

The table shows the student’s results.

titration1234
volume of acid added in cm320.6520.6020.9020.55
concordant results

Concordant results are those within 0.20 cm3 of each other.

(b)(i)

Place ticks (✓) in the table to show which results are concordant.

[1]
Reveal mark scheme Mark scheme for question 6(b)(i)
(b)(ii)

Use the concordant results to calculate the mean (average) volume of acid added.

[2]
Reveal mark scheme Mark scheme for question 6(b)(ii)
(c)

This table shows the student’s results for another titration.

Figure for question 6

The equation for the reaction is

(c)(i)

Calculate the amount, in moles, of KOH in 25.0 cm3 of the potassium hydroxide solution.

[2]
Reveal mark scheme Mark scheme for question 6(c)(i)
(c)(ii)

Calculate the amount, in moles, of H2SO4 in 21.20 cm3 of sulfuric acid.

Question 6, shown as an image
[1]
Reveal mark scheme Mark scheme for question 6(c)(ii)
(c)(iii)

Calculate the concentration, in mol/dm3, of the sulfuric acid.

concentration of sulfuric acid =........ mol/dm

[2]
Reveal mark scheme Mark scheme for question 6(c)(iii)

Chemistry Paper 2 January 2021, Question 6 15 marks

Also involves Limiting reactants

This question is about the insoluble salt silver chloride (AgC l).

Silver chloride can be made by the reaction between copper(II) chloride and silver nitrate.

(a)

Describe how a student could prepare a pure, dry sample of silver chloride starting with copper(II) chloride solution and silver nitrate solution.

[4]
Reveal mark scheme Mark scheme for question 6(a)
(b)

A student investigates the quantity of silver chloride produced when different volumes of silver nitrate solution are added to copper(II) chloride solution.

This is the student’s method. maths symbol pour 5.0 cm3 of copper(II) chloride solution into a test tube maths symbol add 1.0 cm3 of silver nitrate solution to the test tube

maths symbol allow the silver chloride precipitate to settle

maths symbol measure the height of the precipitate

The student repeats the method using different volumes of silver nitrate solution.

The table shows the student’s results.

Volume of silver nitrate added in cm3Height of precipitate in cm
0.00.0
1.00.5
2.01.0
3.01.2
4.02.0
5.02.5
6.03.0
7.03.0
8.03.0
(b)(i)

Plot the student’s results.

[2]
Reveal mark scheme Mark scheme for question 6(b)(i)
(b)(ii)

Draw two straight lines of best fit, ignoring the anomalous result.

Figure for question 6

Volume of silver nitrate added in cm3

[1]
Reveal mark scheme Mark scheme for question 6(b)(ii)
(b)(iii)

Suggest a mistake the student made to cause the anomalous result.

[1]
Reveal mark scheme Mark scheme for question 6(b)(iii)
(b)(iv)

Give a reason why the last three heights are the same.

[1]
Reveal mark scheme Mark scheme for question 6(b)(iv)
(c)

The equation for the reaction between copper(II) chloride and silver nitrate is

CuCl2(aq) + 2AgNO3(aq) maths symbol 2AgCl(s) + Cu(NO3)2(aq)

A student measures 25.0 cm3 of 0.500 mol/dm3 copper(II) chloride solution and reacts it with silver nitrate solution.

(c)(i)

Name a piece of apparatus suitable for measuring 25.0 cm3 of copper(II) chloride solution.

[1]
Reveal mark scheme Mark scheme for question 6(c)(i)
(c)(ii)

Calculate the maximum mass, in grams, of silver chloride that could be produced.

[Mr of AgCl = 143.5]

[3]
Reveal mark scheme Mark scheme for question 6(c)(ii)
(c)(iii)

In an experiment using different solutions, the mass of silver chloride produced is 0.744 g.

The maximum mass of silver chloride that could be produced is 0.850 g.

Calculate the percentage yield.

[2]
Reveal mark scheme Mark scheme for question 6(c)(iii)

Chemistry Paper 2 November 2020, Question 6 11 marks

Also involves Amounts of substance in equations

A student is provided with a bottle containing a colourless solution X.

Solution X is thought to be dilute sulfuric acid of concentration 0.10 mol/dm3.

The student does some experiments on samples of solution X to try to show that it is dilute sulfuric acid.

The student adds a few drops of litmus to a sample of solution X.

The litmus turns red.

(a)

The student knows that the products of the electrolysis of dilute sulfuric acid are hydrogen and oxygen.

She carries out the electrolysis using this apparatus.

Figure for question 6
(a)(i)

Suggest why the student does not use zinc electrodes in her experiment.

[1]
Reveal mark scheme Mark scheme for question 6(a)(i)
(a)(ii)

State what is observed at both the anode and the cathode during the electrolysis.

[1]
Reveal mark scheme Mark scheme for question 6(a)(ii)
(a)(iii)

Which of these tests shows that the gas formed at the cathode is hydrogen?

Question 6, shown as an image
[1]
Reveal mark scheme Mark scheme for question 6(a)(iii)
(b)

Describe a test to show that solution X contains sulfate ions.

[2]
Reveal mark scheme Mark scheme for question 6(b)
(c)

The student then does a titration to see if the concentration of the dilute sulfuric acid is 0.10 mol/dm3.

She measures 25.0 cm3 of potassium hydroxide solution into a conical flask, and then adds a few drops of indicator solution.

(c)(i)

Name the piece of apparatus the student should use to measure 25.0 cm3 of the potassium hydroxide solution.

[1]
(c)(ii)

The concentration of potassium hydroxide in the solution is 0.125 mol/dm3.

Calculate the amount, in mol, of KOH in 25.0 cm3 of this solution.

[2]
Reveal mark scheme Mark scheme for question 6(c)(ii)
(c)(iii)

The equation for the reaction in the titration is

Question 6, shown as an image

Calculate the volume, in cm3, of 0.10 mol/dm3 sulfuric acid needed to neutralise 25.0 cm3 of the potassium hydroxide solution.

[3]
Reveal mark scheme Mark scheme for question 6(c)(iii)

Chemistry Paper 2 November 2020 (R), Question 6 13 marks

A student wants to prepare sodium chloride crystals from sodium hydroxide solution and dilute hydrochloric acid.

He does a titration to find the volume of dilute hydrochloric acid needed to neutralise the sodium hydroxide solution.

This is his method.

● add 25.0 cm3 of sodium hydroxide solution to a conical flask

● add a few drops of phenolphthalein indicator to the conical flask

● titrate the solution with the hydrochloric acid

(a)

Name a suitable piece of apparatus that the student should use to measure 25.0 cm3 of sodium hydroxide solution.

[1]
(b)(i)

Give the colour of the phenolphthalein indicator in sodium hydroxide solution and in hydrochloric acid.

colour in sodium hydroxide solution........

colour in hydrochloric acid........

[2]
Reveal mark scheme Mark scheme for question 6(b)(i)
(b)(ii)

Suggest why universal indicator is never used in a titration.

[1]
Reveal mark scheme Mark scheme for question 6(b)(ii)
(c)

The student finds that 21.50 cm3 of hydrochloric acid is needed to neutralise 25.0 cm3 of sodium hydroxide solution.

(c)(i)

Describe what the student should do next to prepare a pure solution of sodium chloride.

[2]
Reveal mark scheme Mark scheme for question 6(c)(i)
(c)(ii)

Describe how the student could obtain dry crystals of sodium chloride from the pure sodium chloride solution.

[4]
Reveal mark scheme Mark scheme for question 6(c)(ii)
(d)

The student needs 21.50 cm3 of hydrochloric acid to neutralise 25.0 cm3 of sodium hydroxide solution of concentration 0.800 mol/dm3.

The equation for the reaction is

Question 6, shown as an image

Calculate the concentration, in mol/dm3, of the hydrochloric acid.

concentration =........ mol/dm

[3]
Reveal mark scheme Mark scheme for question 6(d)

Chemistry Paper 2 January 2020 (R), Question 3 6 marks

Solutions of silver nitrate and potassium chloride react together to make the insoluble salt, silver chloride.

A student uses this method to prepare a sample of silver chloride.

Step 1 add 25 cm3 of silver nitrate solution to a conical flask

Step 2 add potassium chloride solution to the flask

Step 3 filter off the silver chloride

(a)

What term is used for this reaction?

Question 3, shown as an image
[1]
Reveal mark scheme Mark scheme for question 3(a)
(b)

Give two more steps that will produce a pure, dry sample of silver chloride.

[2]
Reveal mark scheme Mark scheme for question 3(b)
(c)

Acidified silver nitrate solution is used to test for chloride ions.

Give a reason why hydrochloric acid is not used to acidify silver nitrate solution.

[1]
Reveal mark scheme Mark scheme for question 3(c)
(d)

The chemical equation for the reaction between solutions of silver nitrate and potassium chloride is

AgNO3(aq) + KCl(aq) maths symbol AgCl(s) + KNO3(aq)

A student adds an excess of potassium chloride solution to 25.0 cm3 of 0.100 mol/dm3 silver nitrate solution.

Calculate the maximum mass of silver chloride, in grams, that can be produced.

[Mr of AgCl = 143.5]

[2]
Reveal mark scheme Mark scheme for question 3(d)

Chemistry Paper 2 January 2020, Question 4 9 marks

Also involves Amounts of substance in equations

A student investigates the reaction between sodium hydroxide solution and dilute sulfuric acid.

He does a titration to find the concentration of the sulfuric acid.

This is his plan for the titration. There are some mistakes and omissions in his plan.

● rinse a conical flask with the sodium hydroxide solution

● use a measuring cylinder to measure out 25 cm3 of the sodium hydroxide solution and add it to the conical flask

● add a few drops of methyl orange indicator to the conical flask

● rinse a burette with water and then fill it with the sulfuric acid

● add the acid from the burette to the conical flask until the indicator changes colour at the end‑point of the titration

● record the final burette reading

(a)

Give the colour change of the methyl orange indicator at the end‑point.

[2]
Reveal mark scheme Mark scheme for question 4(a)
(b)

Describe four changes that the student could make to improve his plan.

[4]
Reveal mark scheme Mark scheme for question 4(b)
(c)

The student then does the titration correctly.

He finds that 16.70 cm3 of the dilute sulfuric acid neutralises 25.0 cm3 of sodium hydroxide solution of concentration 0.200 mol/dm3

The equation for the reaction is

Question 4, shown as an image

Calculate the concentration, in mol/dm3, of the sulfuric acid.

concentration of sulfuric acid =........ mol/dm

[3]
Reveal mark scheme Mark scheme for question 4(c)

Chemistry Paper 2 January 2020 (R), Question 5 13 marks

Also involves Amounts of substance in equations · Titrations

This question is about Group 1 metals and their reactions.

(a)

When lithium is added to water, bubbles of hydrogen gas are observed.

(a)(i)

Give two other observations that could be made.

[2]
Reveal mark scheme Mark scheme for question 5(a)(i)
(a)(ii)

Give the test for hydrogen gas.

[1]
Reveal mark scheme Mark scheme for question 5(a)(ii)
(b)(i)

Give one observation that would be different if potassium is used instead of lithium.

[1]
Reveal mark scheme Mark scheme for question 5(b)(i)
(b)(ii)

The diagram represents an atom of lithium and an atom of potassium.

Figure for question 5

Explain why potassium is more reactive than lithium.

[3]
Reveal mark scheme Mark scheme for question 5(b)(ii)
(c)

The equation for the reaction between lithium and water is

Question 5, shown as an image
(c)(i)

A mass of 0.500 g of lithium reacts with an excess of water.

Calculate the volume, in cm3, of hydrogen gas produced at rtp.

[molar volume of a gas at rtp = 24 000 cm3]

Give your answer to three significant figures.

[3]
Reveal mark scheme Mark scheme for question 5(c)(i)
(c)(ii)

In a reaction between lithium and water, 150 cm3 of lithium hydroxide solution is formed.

The lithium hydroxide solution is then completely neutralised by 24.85 cm3 of 0.100 mol/dm3 sulfuric acid.

The equation for the neutralisation is

Question 5, shown as an image

Calculate the concentration, in mol/dm3, of the lithium hydroxide solution.

concentration =........ mol/dm

[3]
Reveal mark scheme Mark scheme for question 5(c)(ii)

Chemistry Paper 2 June 2019 (R), Question 6 15 marks

Also involves Amounts of substance in equations

When a bottle of wine is left open for several days, some of the ethanol in the wine turns to ethanoic acid, CH3COOH

(a)

A scientist uses a titration method to investigate how much ethanoic acid is formed if a bottle of white wine is left open for one week.

She uses this method.

● fill a burette with the white wine and record the reading

● add 25.0 cm3 of sodium hydroxide solution to a conical flask

● add a few drops of phenolphthalein indicator to the flask

● swirl the flask continuously while adding wine from the burette

● add the wine drop by drop near the end point

● record the reading at the end point

(a)(i)

Name the piece of apparatus that would be most suitable for measuring the 25.0 cm3 of sodium hydroxide solution.

[1]
Reveal mark scheme Mark scheme for question 6(a)(i)
(a)(ii)

Suggest why red wine would not be suitable to use for this investigation.

[1]
Reveal mark scheme Mark scheme for question 6(a)(ii)
(a)(iii)

State why she swirls the flask continuously.

[1]
Reveal mark scheme Mark scheme for question 6(a)(iii)
(a)(iv)

State why she adds the wine drop by drop near the end point.

[1]
Reveal mark scheme Mark scheme for question 6(a)(iv)
(b)

The diagram shows the burette readings at the start and end of one of the titrations.

Figure for question 6
Question 6, shown as an image

Use the readings to complete the table.

Give your values to the nearest 0.05 cm3.

Figure for question 6
[3]
Reveal mark scheme Mark scheme for question 6(b)
(c)

The scientist repeats the titration four more times.

The table shows her results for these four titrations.

titration number1234
volume of wine added in cm320.4020.1020.3520.45
concordant results

Concordant results are those within 0.20 cm3 of each other.

(c)(i)

Add ticks (maths symbol) to the table to show the concordant results.

[1]
Reveal mark scheme Mark scheme for question 6(c)(i)
(c)(ii)

Use your ticked results to calculate the mean (average) volume of wine added.

[2]
Reveal mark scheme Mark scheme for question 6(c)(ii)
(d)

Another scientist repeats the titration with a different bottle of white wine that has been left open for a week.

The equation for the reaction that occurs in this titration is

CH3COOH + NaOH maths symbol CH3COONa + H2O

The mean volume of wine added is 19.50 cm3.

(d)(i)

The concentration of the sodium hydroxide solution is 0.0500 mol/dm3.

Calculate the amount, in moles, of NaOH in 25.0 cm3 of sodium hydroxide solution.

[2]
Reveal mark scheme Mark scheme for question 6(d)(i)
(d)(ii)

Deduce the amount, in moles, of CH3COOH in 19.50 cm3 of the wine.

[1]
Reveal mark scheme Mark scheme for question 6(d)(ii)
(d)(iii)

Calculate the concentration, in mol/dm3, of CH3COOH in the wine.

concentration of CH3COOH =........ mol/dm3

[2]
Reveal mark scheme Mark scheme for question 6(d)(iii)

Chemistry Paper 2 June 2019, Question 8 7 marks

Also involves Amounts of substance in equations · Hazard symbols and risk assessment

The concentration of NaC l O(aq) in a solution of bleach is found by reacting it with hydrochloric acid.

The equation for the reaction is

Question 8, shown as an image

An excess of dilute hydrochloric acid is added to 4.00 cm3 of bleach solution.

60.0 cm3 of chlorine gas is produced.

(a)

Explain a safety precaution that should be taken when doing this experiment.

[2]
Reveal mark scheme Mark scheme for question 8(a)
(b)(i)

Calculate the amount, in moles, of chlorine gas produced. Assume one mole of chlorine gas occupies 24 000 cm3.

[2]
Reveal mark scheme Mark scheme for question 8(b)(i)
(b)(ii)

Determine the amount, in moles, of NaClO in 4.00 cm3 of bleach.

[1]
Reveal mark scheme Mark scheme for question 8(b)(ii)
(b)(iii)

Calculate the concentration, in mol/dm3, of the bleach solution.

concentration =........ mol/dm3

[2]
Reveal mark scheme Mark scheme for question 8(b)(iii)

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