Oxford Revise AQA GCSE Chemistry | Chapter C9 answers

C9: Reactions of acids

Question

Answers

Extra information

Mark

AO / Specification reference

01.1

1 – strong acid

5 – weak acid

9 – weak alkali

1

1

1

AO1

4.4.2.4

01.2

1 mol/dm3 solution of hydrochloric acid

1

AO2

4.4.2.4

01.3

hydrogen

H+

‘+’ must be superscript and on the right of the ‘H’ symbol

1

1

AO1

4.4.2.4

01.4

water

1

AO1

4.4.2.4

02.1

the H+ ion concentration in solution A is 10 times the H+ ion concentration in solution B

1

AO2

4.4.2.6

02.2

copper chloride, carbon dioxide and water

1

AO2

4.4.2.2

02.3

CuCl2

1

AO2

4.4.2.2

03.1

pipette

1

AO1

4.4.2.5

03.2

two from:

  • swirl the conical flask to ensure good mixing
  • put it on a white tile so that you can see the colour
  • add the solution slowly so that you know exactly when the end point has been reached

one mark for each correct answer

2

AO3

4.4.2.5

03.3

when she has obtained three concordant readings

‘concordant’ must be present for the mark to be awarded

1

AO1

4.4.2.5

03.4

\(
\frac{{(21.20 + 21.30 + 21.25)}}{3} = 21.25
\)

1

AO2

4.4.2.5

03.5

number of moles of acid = concentration × volume in dm3 = 0.00 425 mol

one mole of NaOH reacts with one mole of HCl, so moles of NaOH = 0.00 425

\(
\begin{array}{l}
{\rm{concentration\ of\ NaOH\ }} = \frac{{{\rm{number\ of\ moles}}}}{{{\rm{volume}}}}\\
= \frac{{{\rm{0}}{\rm{.00425}}}}{{{\rm{0}}{\rm{.0250}}}} = 0.170{\rm{\ mol/d}}{{\rm{m}}^{\rm{3}}}
\end{array}
\)

allow error carried forward throughout

1

1

1

1

AO2

4.4.2.5

04.1

two from:

  • stir/swirl the conical flask to ensure good mixing
  • warm
  • use (finer) powder of zinc oxide

one mark for each correct answer

2

AO3

4.4.2.3

04.2

stop when they see unreacted zinc oxide

1

AO1

4.4.2.3

04.3

to remove unreacted solid/zinc oxide

1

AO1

4.4.2.3

04.4

water bath and Bunsen burner or electric heater

1

AO1

4.4.2.3

04.5

evaporate only some of the water (until crystals start to form)

then remove the heat source and allow to crystallise from the solution at room temperature

1

1

AO3

4.4.2.3

04.6

ZnO(s) + 2HCl(aq) → ZnCl2(aq) + H2O(g)

1 mark for formulae and state symbols of reactants

1 mark for formulae and state symbols of products

or

1 mark for correct formulae

1 mark for correct state symbols

1 mark for balancing

3

AO2

4.4.2.2

4.3.1.1

04.7

\({\rm{number\ of\ moles\ of\ HCl\ }} = \frac{{25.0}}{{1000}} \times 0.5 = 0.0125{\rm{\ mol}}\)

from balanced equation, two moles of acid make one mole of zinc chloride

0.0 125 mol of acid makes 0.5 × 0.0 125 mol = 0.00 625 mole of zinc chloride

mass of one mole of zinc chloride = 65 + (35.5 × 2) = 136 g

maximum mass of zinc chloride

= number of moles × mass of one mole

= 0.00 625 × 136 g

= 0.85 g

1 mark for correct answer, 1 mark for correct number of significant figures

1

1

1

1

2

AO2

4.3.2.2

4.3.4

05.1

magnesium

because it loses electrons

Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)

1 mark for formulae and state symbols of reactants

1 mark for formulae and state symbols of reactants

or

1 mark for correct formulae

1 mark for correct state symbols

1 mark for balancing

1

1

3

AO2

4.4.2.1

05.2

MgCl2

H2

1

1

AO2

4.4.2.1

06.1

weak – one from citric/ethanoic/carbonic

strong – one from hydrochloric/sulfuric/nitric

accept any correct acids

1

1

AO1

4.4.2.6

06.2

5

as pH decreases by one unit, H+ increases by a factor of 10

Here, H+ has decreased by a factor of 100, so pH increases by two units

AO1 × 1

AO2 × 2

4.4.2.6

06.3

\(
\begin{array}{l}
{{\rm{H}}^{\rm{\ +\ }}}{\rm{concentration\ in\ A\ }} = \frac{{{\rm{20}}}}{{{\rm{100}}}} \times 5{\rm{\ mol}}/{\rm{d}}{{\rm{m}}^3} = 1{\rm{\ mol}}/{\rm{d}}{{\rm{m}}^3}\\
{{\rm{H}}^{\rm{\ +\ }}}{\rm{concentration\ in\ B\ }} = \frac{{{\rm{100}}}}{{{\rm{100}}}} \times 2{\rm{\ mol}}/{\rm{d}}{{\rm{m}}^3} = 2{\rm{\ mol\ }}/{\rm{\ d}}{{\rm{m}}^3}
\end{array}
\)

H+ concentration in B is higher

B has the lower pH

1

1

1

1

AO2

4.4.2.6

07.1

2-

1

AO2

4.4.2.1

07.2

MgSO4

1

AO2

4.4.2.1

07.3

MnCl2

1

AO2

4.4.2.1

07.4

hydrogen

1

AO1

4.4.2.1

08.1

sulfuric acid

1

AO1

4.4.2.2

08.2

Level 3: The description of the method is detailed and accurate. Apparatus is named correctly, and the reasons given are clear and coherent.

5-6

AO1

4.4.2.3

Level 2: The descriptions of the method is correct, although lacks detail. Apparatus is named correctly, and reasons are given for some steps, although these may not be clearly explained.

3-4

Level 1: The method is outlined correctly. The names of one or two pieces of apparatus are given, as well as reasons for one or two steps only. The description overall lacks clarity and coherence.

1-2

No relevant content

0

Indicative content

  • use a spatula to add excess copper hydroxide to the acid in a conical flask/beaker
  • excess copper hydroxide is used so that all the acid reacts
  • filter using filter paper and funnel to remove excess/unreacted copper hydroxide
  • heat the filtrate in an evaporating basin over a water bath/with an electric heater until crystals begin to appear
  • remove the heat and allow the rest of the water to evaporate slowly, to allow big crystals to form

allow correct diagrams, which should use standard representations of equipment, to aid description

08.3

Cu(OH)2 + H2SO4 → CuSO4 + 2H2O

3

AO2

4.4.2.2

08.4

\(
\begin{array}{l}
32.5 \times \frac{{{\rm{30}}}}{{{\rm{1000}}}}\\
\; = 0.975{\rm{\ g}}\\
\frac{{75}}{{97.5}}\\
= 0.01{\rm{\ moles\ Cu}}{\left( {{\rm{OH}}} \right)_2}
\end{array}
\)

1

1

1

1

AO2

4.3.2.1

08.5

0.01 moles Cu(OH)2 = 0.01 moles CuSO4

Mr of CuSO4 = 63.5 + 32 + (16 × 4) = 159.5

0.01 × 159.5

= 1.595g

= 1.6g

1

1

1

1

1

AO2

4.3.2.1

09.1

Mr of zinc nitrate = 189 g

\(
{\rm{number\ of\ moles\ of\ zinc\ nitrate\ }} = \frac{{{\rm{mass}}}}{{{{\rm{M}}_{\rm{r}}}}} = \frac{{9.4}}{{189}} = 0.05{\rm{\ mol}}
\)

from equation, one mol of zinc carbonate makes one mole of zinc nitrate

Mr of zinc carbonate = 125 g

mass = number of moles × Mr of ZnCO3 = 0.05 × 125 g = 6.25 g

1

1

1

1

1

AO2

4.3.2.2

09.2

pH at start between one and three

pH increases as zinc carbonate added

because zinc carbonate neutralises the acid

1

1

1

AO1 × 1

AO2 × 2

4.4.2.4

10.1

pH probe

universal/broad range indicator

1

1

AO1

4.4.2.4

10.2

A

1

AO1

4.4.2.4

10.3

E

1

AO1

4.4.2.4

10.4

B

1

AO2

4.4.2.4

10.5

increases

1

AO2

4.4.2.4

11.1

2NaOH(aq) + H2SO4(aq) → Na2SO4 + 2H2O

if NaSO4 is used, but equation is still balanced, then maximum of 2 marks can be awarded

3

AO2

4.4.2.2

11.2

  • measure set volume/named volume of sulfuric acid and place in conical flask
  • using pipette
  • add a suitable indicator to the conical flask
  • put conical flask on a white tile
  • fill burette with sodium hydroxide
  • add sodium hydroxide 1cm3 at a time to the conical flask
  • swirl after each addition
  • when indicator changes colour, record the volume of sodium hydroxide (this is a rough titre)
  • refill burette with sodium hydroxide solution
  • set up new conical flask with same volume of sulfuric acid and indicator
  • run sodium hydroxide into conical flask until near the end point
  • swirling continuously
  • add sodium hydroxide dropwise swirling after each addition
  • until indicator changes colour
  • record volume of sodium hydroxide added
  • repeat until at least 3 concordant results are obtained

6

AO1

4.4.2.5

11.3

Mr of H2SO4 = 98

\(
\begin{array}{l}
\frac{29.4}{98}\\
= 0.3{\rm{\ mol/d}}{{\rm{m}}^3}\\
0.3 \times \frac{{25}}{{100}}\\
= 7.5 \times 10^{-3}{\rm{\ mol\ of\ }}{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}\\
7.5 \times {10^{-3}} \times 2 = 0.015{\rm{\ mol\ of\ NaOH}}
\end{array}
\)

1

1

1

1

1

1

AO2

4.4.2.5

11.4

Mr of NaOH = 40

\(
\begin{array}{l}
\frac{{{\rm{20}}}}{{{\rm{40}}}}\\
= 0.5{\rm{\ mol/d}}{{\rm{m}}^3}\\
\frac{{0.015}}{{0.5}}\\
= 0.03{\rm{\ d}}{{\rm{m}}^3}\\
= 30{\rm{\ c}}{{\rm{m}}^3}
\end{array}
\)

1

1

1

1

1

1

AO2

4.4.2.5

12.1

Z

1

AO3

4.4.2.6

12.2

W and Y

for a given concentration of solution, citric acid has the lower hydrogen ion, H+, concentration and higher pH

for the two solutions of concentration 0.1 mol/dm3, W has the higher pH

for the two solutions of concentration 1 mol/dm3, Y has the higher pH

1

1

1

1

AO3

4.4.2.6

12.3

a weak acid is partially dissociated in aqueous solution/a strong acid is fully dissociated in aqueous solution

1

AO1

4.1.2.6

12.4

strong – one from:

  • sulfuric
  • nitric

weak – one from:

  • ethanoic
  • carbonic

accept any correct acids

1

1

AO1

4.1.2.6

12.5

Level 3: Points that support and do not support the statement are made in detail, and a judgement made and justified. The answer is clearly and coherently written.

5-6

AO3

4.1.2.6

Level 2: Points that support and do not support the statement are made, but a judgement is not be included. The answer is reasonably clear, but not organised in a logical way.

3-4

Level 1: One or two relevant points are made. The answer is not clearly written nor is it logically organised.

1-2

No relevant content

0

Indicative content

  • pH is a measure of the hydrogen ion concentration
  • the greater the hydrogen ion concentration, the lower the pH
  • for weak and strong acids of the same concentration, the hydrogen ion concentration is always smaller in the weak acid, so the pH is always higher
  • so the statement is true for solutions of the same concentration
  • but a dilute solution of a strong acid could have a smaller hydrogen ion concentration than a more concentrated solution of a weak acid
  • so the statement is not always true for solutions of different concentrations

13.1

Noble Gases

1

AO1

4.1.2.4

13.2

2,8,8

1

AO2

4.1.1.7

13.3

they have stable arrangements of electrons/full outer shell

1

AO1

4.1.2.4

13.4

increases

1

AO2

4.1.1.5

14.1

31

1

AO2

4.1.1.5

14.2

69 – 31 = 38

1

AO2

4.1.1.5

14.3

31 – 3 = 28

1

AO2

4.1.1.5

14.4

one from:

  • boron
  • aluminium
  • indium
  • thallium

1

AO2

4.1.2.1

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