21st Century Chemistry
Structured Question in Topic 4
Acids and Alkalis
Unit 13-17
1.
For each of the following experiments, state ONE observable change and write a chemical
equation for the reaction involved.
(a) Magnesium strip is added into dilute hydrochloric acid.
Magnesium dissolves./ Gas bubbles are given off.
[1]
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
[1]
(b) Iron(II) hydroxide is added into dilute sulphuric acid.
Iron(II) hydroxide dissolves./ A pale green solution is formed.
[1]
[1]
Fe(OH)2(s) + H2SO4(aq) → FeSO4(aq) + 2H2O(l)
(c) Sodium carbonate solution is added into concentrated hydrochloric acid.
Effervescence occurs. / Gas bubbles are given off.
[1]
Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
[1]
(d) Sodium hydroxide solution is added into copper(II) sulphate solution.
A pale blue precipitate is formed. / The blue copper(II) sulphate solution becomes
paler.
[1]
2NaOH(aq) + CuSO4(aq) → Cu(OH)2(s) + Na2SO4(aq)
[1]
(e) Calcium chloride solution is added into sodium sulphate solution.
A white precipitate is formed.
[1]
CaCl2(aq) + Na2SO4(aq) → CaSO4(s) + 2NaCl(aq)
[1]
(10 marks)
2.
Consider the following chemicals:
sodium hydrogencarbonate, dilute sulphuric acid, sodium chloride solution and sodium
hydroxide solution
(a) Which chemical can be used as a constituent in baking powder? Explain your answer
with an appropriate equation.
Sodium hydrogencarbonate
[1]
When sodium hydrogencarbonate is heated, it is decomposed and carbon dioxide is
given off which makes the cake rise.
[1]
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
[1]
(b) Which chemical can remove rust? Explain your answer with an appropriate equation.
Dilute sulphuric acid
[1]
Rust(iron(III) oxide) dissolves in dilute sulphuric acid.
[1]
Fe2O3(s) + 3H2SO4(aq) → Fe2(SO4)3(aq) + 3H2O(l)
[1]
(c) Which chemical can be used to remove silver ions from a solution of silver nitrate and
zinc sulphate? Explain your answer with an appropriate equation.
Sodium chloride solution
[1]
Chloride ions react with silver ions to form insoluble silver chloride which can be
removed by filtration.
[1]
－
+
Ag (aq) + Cl (aq) → AgCl(s)
[1]
(d) Which chemical can be used to distinguish between iron(II) sulphate solution and
iron(III) sulphate solution? Explain your answer with appropriate equations.
Sodium hydroxide solution
[1]
Sodium hydroxide solution reacts with iron(II) sulphate and iron(III) sulphate solutions
to form green and brown precipitates respectively.
[1]
2+
−
Fe (aq) + 2OH (aq) → Fe(OH)2(s) (dirty green precipitate)
[1]
3+
−
Fe (aq) + 3OH (aq) → Fe(OH)3(s) (brown precipitate)
[1]
(13 marks)
3.
Suppose there are four unlabelled bottles of chemicals. They are either one of the following:
iron(II) nitrate, copper(II) nitrate, zinc nitrate and lead(II) nitrate. Suggest how you can
distinguish all of them with aqueous ammonia only. Illustrate your answer with appropriate
equations.
When aqueous ammonia is added to each of them separately, coloured precipitate or white
precipitate will be formed. Some of them can be identified by their characteristic colours.
Iron(II) nitrate will form a dirty green precipitate
[1]
and copper(II) nitrate will form a pale blue precipitate with aqueous ammonia.
[1]
2+
−
Fe (aq) + 2OH (aq) → Fe(OH)2(s) (dirty green precipitate)
[1]
2+
−
Cu (aq) + 2OH (aq) → Cu(OH)2(s) (pale blue precipitate)
[1]
Both zinc nitrate and lead(II) nitrate will form white precipitate with aqueous ammonia.
[1]
2+
[1]
Zn (aq) + 2OH−(aq) → Zn(OH)2(s) (white precipitate)
2+
−
Pb (aq) + 2OH (aq) → Pb(OH)2(s) (white precipitate)
[1]
However, zinc hydroxide will redissolve in excess ammonia solution to give a colourless
solution while lead(II) hydroxide does not.
[1]
(8 marks)
4.
Six compounds are classified into two groups as shown in the table below:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Gas
Solid
ammonia
copper(II) hydroxide
carbon dioxide
zinc hydroxide
nitrogen dioxide
potassium hydroxide
Which is/are coloured compound(s)? State its/ their colour(s).
Nitrogen dioxide : brown
[1]
Copper(II) hydroxide : blue
[1]
Which solid will give a solution with the highest pH value when 0.1 mole of each solid
is added to 500 cm3 of water respectively? Explain your answer.
Potassium hydroxide
[1]
Potassium hydroxide is soluble in water while the other two are not, so the
concentration of hydroxide ions in the solution of potassium hydroxide is the highest.
[1]
Which compound(s) is/are acidic in nature?
Carbon dioxide
[1]
Nitrogen dioxide
[1]
Suggest simple methods to prepare carbon dioxide and ammonia in laboratory.
Carbon dioxide:
Adding calcium carbonate into dilute hydrochloric acid / Heating calcium carbonate
[1]
Ammonia:
Heating an ammonium compound (e.g. NH4Cl) with an alkali (e.g. NaOH)
[1]
Suggest a test to identify ammonia.
Ammonia gas turns moist pH paper (or moist red litmus paper) blue.
[1]
Suggest a test to distinguish between 1 M potassium hydroxide solution and 1 M
ammonia solution. State the observable change and explain your answer briefly.
Add a piece of pH paper (or a few drops of universal indicator) to each solution. [1]
1 M potassium hydroxide solution has a higher pH value than 1 M ammonia solution.
[1]
It is because ammonia solution is a weak alkali while potassium hydroxide solution is a
strong alkali. / Ammonia only partially ionizes while potassium hydroxide completely
ionizes to give hydroxide ions in water.
[1]
or,Measure the electrical conductivity of the solutions.
1 M potassium hydroxide solution has a higher electrical conductivity than 1 M
ammonia solution.
It is because ammonia solution is a weak alkali while potassium hydroxide solution is a
strong alkali. / Ammonia only partially ionizes while potassium hydroxide completely
ionizes to give hydroxide ions in water.
A student suggested that if some potassium hydroxide solution was split onto the hand,
vinegar should be used to neutralize the potassium hydroxide solution left on the hand.
Explain why the action is inappropriate and suggest a proper action.
A lot of heat energy is released when vinegar reacts with potassium hydroxide solution.
[1]
This will cause skin burn.
[1]
Proper action:Wash the hand with large amount of water.
[1]
(15 marks)
5.
To determine the effect of concentration of hydrogen ions on reaction rate, same mass of
magnesium ribbon was put into 0.1 M hydrochloric acid (pH = 1), 0.001 M hydrochloric acid
(pH = 3) and 0.1 M ethanoic acid (pH = 3) respectively.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Write an ionic equation for the reaction between magnesium and hydrochloric acid.
Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)
[1]
Describe, giving the names of the apparatus used, how 0.001 M hydrochloric acid can
be prepared from 0.1 M hydrochloric acid.
Use a pipette to transfer 25 cm3 of 0.1 M hydrochloric acid into a 250 cm3 volumetric
flask.
[1]
Then add distilled water to the graduation mark.
[1]
The resulting solution is 0.01 M hydrochloric acid.
Repeat the above procedures for 0.01 M hydrochloric acid. 0.001 M hydrochloric acid
will be obtained.
[1]
Suggest a method which allows you to collect the gas given off.
Using syringe / By displacement of water
[1]
How can you compare the rate of the above reactions?
By comparing the volume of gas given off per unit time
[1]
Why was same volume of acid used in each conical flask?
To ensure a fair comparison as same amount of solution was used.
[1]
Explain why it is not appropriate to use sodium instead of magnesium.
Sodium reacts explosively with acid.
[1]
(i)
Arrange the acids in order of increasing initial rate of reaction with magnesium.
Explain your answer.
0.001 M HCl = 0.1 M ethanoic acid < 0.1 M HCl
[1]
The concentration of hydrogen ions is related to the pH value of an acid. The
lower the pH value, the higher the concentration of hydrogen ions and the faster
the acid reacts.
[1]
(ii) From the experiment, suggest two factors that affect the pH value of an acid.
Strength of acid and
[1]
concentration of acid used
[1]
(12 marks)
6.
Sally is a F.5 student and she likes chemistry very much. Last week, she tried an experiment
at home. The details are as follows:
She got some baking powder and vinegar from the kitchen. From a dictionary, she knew
that the chemical names of vinegar and baking powder are ethanoic acid and sodium
hydrogencarbonate respectively. She made a dilute solution of the vinegar and added it to
the baking powder.
(a)
(b)
(c)
(d)
(e)
(f)
What would happen when a piece of pH paper is dipped into vinegar?
The pH paper would turn red.
[1]
Apart from using pH paper, suggest another method to determine the pH of vinegar.
Using pH meter / universal indicator / data-logger with pH sensor.
[1]
(i)
What is the meaning of the term "weak acid"?
The acid which partially ionizes
[1]
to give hydrogen ions in water.
[1]
(ii) Do you think ethanoic acid is a strong acid or a weak acid?
It is a weak acid.
[1]
What would you expect to see when baking powder is added to the dilute solution of
ethanoic acid?
Baking powder dissolves.
[1]
Effervescence occurs. / Gas bubbles are given off.
[1]
Write an equation for the reaction between vinegar and baking powder.
[1]
CH3COOH(aq) + NaHCO3(s) → CH3COONa(aq) + CO2(g) + H2O(l)
To investigate the electrical conductivity of vinegar, Sally tried to construct an
experimental set-up.
(i)
Draw a diagram to show the experimental set-up.
carbon
electrodes
vinegar
[2]
(ii)
Can vinegar conduct electricity? Explain your answer.
Yes. Vinegar contains mobile ions.
[1]
(11 marks)
7.
Study the following reaction scheme of metal A.
burns with sparks;
a black solid B is produced
heat in air
metal A
dilute sulphuric acid
a pale green
solution C +
a colourless
gas D
dilute sodium
hydroxide solution
a green
precipitate E
(a)
(b)
(c)
(d)
Give the names of A, B, C, D and E.
A: iron
[1]
B: iron(II,III) oxide
[1]
C: iron(II) sulphate solution
[1]
D: hydrogen
[1]
E: iron(II) hydroxide
[1]
Write an equation for each of the following reactions:
(i)
A is heated in air.
3Fe(s) + 2O2(g) → Fe3O4(s)
[1]
(ii) A is added to dilute sulphuric acid.
Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)/ Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g)
[1]
(iii) Solution C is added to dilute sodium hydroxide solution.
FeSO4(aq) + 2NaOH(aq) → Fe(OH)2(s) + Na2SO4(aq)
－
/ Fe2+(aq) + 2OH (aq) → Fe(OH)2(s)
[1]
Suggest an industrial method from which metal A can be obtained from solid B.
Carbon reduction / heating with carbon
[1]
Suggest a chemical test to identify gas D.
Test the gas with a burning splint.
[1]
A "pop" sound will be heard.
[1]
(11 marks)
8.
Consider the following reaction scheme:
Blue solid A
dilute
hydrochloric acid
Blue solution B
sodium
hydroxide
solution
sodium carbonate
solution
Greenish blue
solid C
dilute
sulphuric acid
Blue solution D
limewater
Colourless gas E
Milky solution
heat
Black solid F
(a)
(b)
(c)
(i)
Name solid A.
Copper(II) hydroxide
[1]
(ii) Write an ionic equation for the reaction between solid A and dilute hydrochloric
acid.
[1]
Cu(OH)2(s) + 2H+(aq) → Cu2+(aq) + 2H2O(l)
(i)
What is solid C?
Copper(II) carbonate / CuCO3
[1]
(ii) Write an ionic equation for the formation of solid C from solution B and sodium
carbonate solution.
Cu2+(aq) + CO32-(aq) → CuCO3(s)
[1]
(iii) Write a chemical equation for the formation of gas E and solid F from the
thermal
decomposition of solid C.
CuCO3(s) → CuO(s) + CO2(g)
[1]
(i)
Name solution D.
Copper(II) sulphate
[1]
(ii) Write an equation for the reaction between solution D and sodium hydroxide
solution.
－
Cu2+(aq) + 2OH (aq) → Cu(OH)2(s) / CuSO4(aq) + 2NaOH(aq) → Cu(OH)2(s) +
(iii)
Na2SO4(aq)
(1) Name a reagent that can convert solid F to solution D.
Dilute sulphuric acid
(2) Write a chemical equation for the conversion.
CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l)
[1]
[1]
[1]
(d)
Describe briefly how large crystals of solid D can be obtained from solution D.
Warm copper(II) sulphate solution (solution D) to obtain a concentrated (or saturated)
solution.
[1]
Cool the solution slowly, large crystals of copper(II) sulphate are obtained.
[1]
Filter the mixture.
[1]
Wash the crystals with a small amount of cold distilled water and dry them with filter
paper.
[1]
(13 marks)
9.
The following tests were conducted to find out the identity of cations and anions of two
unknown ionic compounds.
Test
Compound X
Compound Y
(A) Flame test
Bluish-green flame
Brick-red flame
(B) Adding sodium hydroxide
Blue gelatinous
No observable change
solution
precipitate is formed
(C) Adding acidified silver
White precipitate is
No observable change
nitrate solution
formed
(D) Adding dilute hydrochloric No observable change Gas bubbles are given off
acid
which turn limewater milky
(a) From the above tests, suggest the possible cations that are present in unknown
compounds X and Y.
Cation in compound X: copper(II) ion
[1]
Cation in compound Y: calcium ion
[1]
(b) Name the acid that should be used to acidify silver nitrate solution.
Dilute nitric acid
[1]
(c) Draw one hazard warning label which should be displayed on a bottle of concentrated
hydrochloric acid.
Corrosive:
[1]
(d)
(i)
(ii)
(e)
(i)
(ii)
Consider compound X, write ionic equations to account for
(1) the formation of blue gelatinous precipitate in test B.
[1]
Cu2+(aq) + 2OH−(aq) → Cu(OH)2(s)
(2) the formation of white precipitate in test C.
Ag+(aq) + Cl−(aq) → AgCl(s)
[1]
Hence give a possible formula of compound X.
[1]
Compound X is CuCl2.
Draw the experimental set-up for testing the gas given off from the reaction
between compound Y and dilute hydrochloric acid.
[2]
Suggest what the gas evolved should be and write the equation for the reaction
between the gas and limewater.
The gas given off is carbon dioxide.
[1]
[1]
CO2(g) + Ca(OH)2(aq) → CaCO3(s) + H2O(l)
(iii)
10.
(a)
(b)
(c)
(d)
Hence give a possible formula of compound Y and write a chemical equation for
the reaction between compound Y and dilute hydrochloric acid.
[1]
CaCO3 / Ca(HCO3)2
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
/ Ca(HCO3)2(s) + 2HCl(aq) → CaCl2(aq) + 2CO2(g) + 2H2O(l)
[1]
(13 marks)
(i) Write an equation for the reaction between sodium hydroxide solution and dilute
nitric acid.
NaOH(aq) + HNO3(aq) → NaNO3(aq) + H2O(l)
[1]
(ii) What volume of 0.2 M sodium hydroxide solution is required to neutralize 150
cm3 of 0.12 M nitric acid?
[1]
No. of moles of HNO3 = 0.12 × 0.15 = 0.018 mol
According to the equation, 1 mole of NaOH reacts with 1 mole of HNO3.
Volume of NaOH used = 0.018 ÷ 0.2 = 0.09 dm3 (90 cm3)
[1]
0.2 M ammonia solution is used to neutralize 150 cm3 of 0.12 M nitric acid.
(i)
Compare the strength of ammonia solution and sodium hydroxide solution.
Ammonia solution is a weak alkali while sodium hydroxide solution is a strong
alkali.
[1]
(ii) Write an equation for the neutralization reaction between ammonia solution and
dilute nitric acid.
NH3(aq) + HNO3(aq) → NH4NO3(aq)
[1]
(iii) Is there any difference between the volume of 0.2 M sodium hydroxide solution
used and that of 0.2 M ammonia solution used to neutralize 150 cm3 of 0.12 M
nitric acid?
No. The volume of ammonia solution used is the same as the volume of sodium
hydroxide solution used.
[1]
Compare the heat released in the neutralizations in (a) (ii) and (b).
The heat released in the neutralization between ammonia solution and nitric acid is
less than that between sodium hydroxide solution and nitric acid.
[1]
Suggest an indicator that should be used to detect the end point in both
neutralizations in (a) (ii) and (b).
Methyl orange
[1]
(8 marks)
11.
A standard solution of oxalic acid with molarity of 0.050 M was prepared.
(a) What is the meaning of “standard solution”?
A solution of known concentration.
[1]
(b) You are provided with pure oxalic acid crystals (H2C2O4•2H2O), distilled water, a
beaker, a 250 cm3 volumetric flask and an electronic balance. Describe with steps how
you can prepare a standard solution of oxalic acid with molarity of 0.050 M.
First, weigh 1.575 g of oxalic acid with the electronic balance.
[1]
3
[1]
Dissolve the acid into 100 cm of distilled water in the beaker.
3
Transfer it into the 250 cm volumetric flask.
[1]
Add distilled water into the volumetric flask to the graduation mark and then shake it.
[1]
(c) (i)
Name the apparatus used to transfer 25 cm3 of the standard solution of oxalic
acid accurately.
Pipette
[1]
(ii) Suggest how the apparatus in (c)(i) should be cleaned before it is used.
It should be washed with distilled water first and then with the standard solution
of oxalic acid.
[1]
3
(d) 25 cm of 0.05 M oxalic acid was titrated against potassium hydroxide solution. 22.4
cm3 of 0.112 M potassium hydroxide solution was required for complete neutralization.
(i)
Suggest an indicator that should be used for the titration and the colour change
at
the end point.
Phenolphthalein
[1]
The colour changed from colourless to pink.
[1]
(ii) Why was a white tile placed under the conical flask?
For better observation of the colour change when end point was reached. [1]
(iii) Determine the basicity of oxalic acid.
Number of moles of oxalic acid in 25 cm3 solution
= 0.05 x 0.025
－
= 1.25 x 10 3 mol
[1]
Number of mole of KOH
= 0.112 x 0.0224
－
= 2.5 x 10 3 mol
Basicity of oxalic acid =
(iv)
[1]
2.5x10 −3
=2
1.25x10 −3
Write the equation for the neutralization.
2KOH(aq) + H2C2O4(aq) → K2C2O4(aq) + 2H2O(l)
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)
[1]
[1]
(14 marks)
12.
(a)
(b)
Antacid tablets are used to relieve stomach ache. They usually contain magnesium
hydroxide as the active ingredient.
(i)
Explain how antacid tablets can relieve stomach ache with the help of an
equation.
Magnesium hydroxide in antacid neutralizes the excess hydrochloric acid in
stomach.
[1]
Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l)
[1]
(ii) Explain why magnesium hydroxide instead of calcium carbonate is used as the
active ingredient in antacid nowadays.
Calcium carbonate reacts with hydrochloric acid to give out carbon dioxide. [1]
It builds up pressure in the stomach and causes discomfort.
[1]
(iii) Explain why the antacid tablets should be chewed before swallowing.
Chewing breaks down the tablets into smaller pieces. This increases the surface
area of the tablets and thus increases the reaction rate (brings faster relief of
pain).
[1]
Waste water discharged from some factories is usually acidic and contains metal ions
such as copper(II) ions.
(i)
How do metal ions lead to pollution?
The metal ions may poison the aquatic lives.
[1]
(ii) With the help of an equation, suggest how metal ions can be removed from the
waste water. Use copper(II) ions as an example.
Metal ions can be removed by adding sodium hydroxide solution into the liquid
waste to form insoluble metal hydroxide.
[1]
The precipitate is then removed from the liquid waste by filtration.
[1]
－
2+
Cu (aq) + 2OH (aq) → Cu(OH)2(s)
[1]
(iii) Suggest another reason why copper(II) ions should be removed from the waste
water before discharge.
To recover copper metal.
[1]
(iv) The liquid waste contains 0.3 M hydrochloric acid. To remove the acidity of the
liquid waste, slaked lime (calcium hydroxide) is used. 333 g of slaked lime is
required per minute to completely neutralize the hydrochloric acid present in the
liquid waste. Calculate the rate of liquid waste discharged in dm3 per minute.
2HCl(aq) + Ca(OH)2(s) → CaCl2(aq) + 2H2O(l)
No. of moles of Ca(OH)2 used = 333 ÷ [40.0 + 2 x (16.0 + 1.0)] = 4.5 mol
[1]
No. of moles of HCl(aq) used = 4.5 x 2 = 9 mol
[1]
3
Rate. of liquid waste discharged per minute = 9 ÷ 0.3 = 30 dm
[1]
(Relative atomic masses: H = 1.0, O = 16.0, Ca = 40.0)
(13 marks)
13.
A student was given 14.04 g of a dibasic acid (H2X). The dibasic acid was then dissolved in
water and made up to a 250 cm3 solution. 25.0 cm3 of the solution with indicator was then
titrated against 0.45 M sodium hydroxide solution until the end point was reached. The
results are shown in the following table.
Titration
1
2
3
4
Burette reading
Final (cm3)
50.00
32.70
40.80
45.40
Initial (cm3)
17.20
0.50
8.80
13.60
(a) What is the meaning of the term "dibasic acid"? Give an example of dibasic acid.
Dibasic acid is an acid with two ionizable hydrogen ions in each molecule.
[1]
Example: sulphuric acid.
[1]
(b) Write the chemical equation for the reaction.
[1]
H2X(aq) + 2NaOH(aq) → Na2X(aq) + 2H2O(l)
(c) If H2X is a strong acid, suggest an indicator that should be used in the titration. State
the colour change at the end point.
Methyl orange: from red to yellow
/ Phenolphthalein: from colourless to pink
[2]
(d) Draw the experimental set-up for the titration. Label all the apparatus used.
(e)
(f)
(1+1/2 mark for titration set-up ; 1/2 mark for each label for apparatus)
Find the reasonable average volume of sodium hydroxide solution used.
Volume: [(32.7 − 0.5) + (40.8−8.8) + (45.4 − 13.6)] ÷3 = 32.0 cm3
Calculate the molar mass of the acid.
－
No. of moles of NaOH(aq) used in titration = 0.45 x 0.032 = 1.44 x 10 2 mol
According to the equation, no. of moles of NaOH : no. of moles of H2A = 2 : 1
－
－
No. of moles of H2X(aq) in 25 cm3 solution = 1.44 x 10 2 ÷ 2 = 7.2 x 10 3 mol
3
－3
No. of moles of H2X in 250 cm solution = 7.2 x 10
－2
－2
x 10 = 7.2 x 10
－1
Molar mass of H2X = 14.04 ÷ (7.2 x 10 ) = 195 g mol
(Relative atomic masses: H = 1.0, O = 16.0, Na = 23.0)
mol
[3]
[1]
[1]
[1]
[1]
(12 marks)
14.
(a)
(b)
7.50 g of an impure sample of sodium carbonate required 31.2 cm3 of 2.0 M nitric acid
for complete reaction.
(i)
Write the chemical equation for the reaction.
[1]
Na2CO3(s) + 2HNO3(aq) → 2NaNO3(aq) + CO2(g) + H2O(l)
(ii) What is the percentage purity of the sample?
No. of moles of HNO3 = 2 x 0.0312 = 0.0624 mol
According to the equation, 1 mole of Na2CO3 requires 2 moles of HNO3 for
complete reaction.
No. of moles of Na2CO3 = 0.0624 ÷ 2 = 0.0312 mol
[1]
Mass of Na2CO3 in the sample = 0.0312 x (23.0 x 2 + 12.0 + 16.0 x 3)
= 3.307 g
[1]
Percentage purity of the sample = (3.307 ÷ 7.50) × 100% = 44.1 %
[1]
(iii) What assumption has been made on your calculation?
The impurities are insoluble in water. / The impurities do not react with nitric
acid.
[1]
3.35 g of a sample of hydrated sodium carbonate crystals, Na2CO3‧nH2O, was
dissolved in water and the solution was made up to 250 cm3. Using a suitable indicator
in titration, 25.0 cm3 of this solution required 27.0 cm3 of 0.10 M sulphuric acid for
complete reaction.
(i)
Write the chemical equation for the reaction.
Na2CO3(s) + H2SO4(aq) → Na2SO4(aq) + H2O(l) + CO2(g)
[1]
(ii) Suggest a suitable indicator for the titration and state the colour change at the
end point.
Methyl orange
[1]
The colour changed from yellow to red.
[1]
(iii) Calculate the value of n.
No. of moles of H2SO4 = 0.10 x 0.027 = 2.7 x 10−3 mol
According to the equation, 1 mole of Na2CO3 requires 1 mole of H2SO4 for
complete reaction.
No. of moles of Na2CO3 in 25.0 cm3 of solution = 2.7 x 10−3 mol
No. of moles of Na2CO3‧nH2O in the sample = 2.7 x 10−3 x 10 = 2.7 x 10−2 mol
[1]
−2
Formula mass of Na2CO3‧nH2O = 3.35 ÷ 2.7 x 10 = 124
23.0 x 2 + 12.0 + 16.0 x 3 + n (16.0 + 1.0 x 2) = 124
∴n = 1
[1]
[1]
(iv)
A student sets up the following apparatus to carry out the above titration:
funnel
0.1M sulphuric acid
sodium carbonate
solution + indicator
Point out any three errors in the experimental set-up.
Any three of the followings:
[1 x 3]
The funnel has not been removed.
The burette is not set up vertically.
A white tile should be placed under the conical flask to show the colour change at
the end point.
The tip of the burette should be placed below the mouth of the conical flask.
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0, Na = 23.0)
(14 marks)
15.
The sour taste of citrus fruits is mainly due to citric acid which is a solid weak acid at room
temperature. Its structure is shown in the diagram below:
CH2COOH
HO
C
COOH
CH2COOH
(a)
(b)
(c)
(d)
(e)
Explain the term "weak acid".
Weak acid is an acid which partially ionizes in water
[1]
to give hydrogen ions.
[1]
If you are provided with 0.1 M citric acid solution and 0.1 M hydrochloric acid, suggest
a test to identify these two acids.
Measure the pH value of these two acids.
[1]
Citric acid has a higher pH value than hydrochloric acid.
[1]
or,Dip two carbon electrodes into the acid and connect them to a d.c. supply with a
light bulb in the circuit. Compare the brightness of the light bulbs.
Hydrochloric acid has better electrical conductivity than citric acid and gives brighter
light bulb.
or,Compare the degree of vigor in their reactions with magnesium.
Hydrochloric acid is a strong acid which has a higher concentration of hydrogen ions
than citric acid, so magnesium reacts with it faster than with citric acid.
What will be observed when a magnesium ribbon is added to citric acid in
methylbenzene? Explain the observation.
There will be no observable change.
[1]
Citric acid does not ionize to give hydrogen ions in methylbenzene.
[1]
(i)
What is meant by the term “basicity” of an acid?
It refers to the maximum number of hydrogen ions produced by an acid
molecule.
[1]
(ii) What is the basicity of citric acid?
3
[1]
3
(i)
Calculate the mass of citric acid required to prepare 100 cm of 0.2 M citric acid
solution.
No. of moles of citric acid = 0.2 x 0.1 = 0.02 mol
－
Molar mass of citric acid = 12.0 x 6 + 1.0 x 8 + 16.0 x 7 = 192 g mol 1
[1]
Mass of citric acid required = 0.02 x 192 = 3.84 g
[1]
(ii) What is the volume of 0.46 M potassium hydroxide solution required to
completely neutralize 25.0 cm3 of the solution prepared in (i)?
Let the formula of citric acid be H3A.
H3A(aq) + 3KOH(aq) → K3A(aq) + 3H2O(l)
No. of moles of H3A = 0.2 x 0.025 = 5 x 10−3 mol
[1]
−3
No. of moles of KOH = 3 x 5 x 10 = 0.015 mol
[1]
－2
3
3
Volume of KOH used = 0.015 ÷ 0.46 = 3.26 x 10 dm (32.6 cm )
[1]
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)
(13 marks)
16.
Commercial glass cleaners contain ammonia. A student carried out an experiment to
determine which brand of glass cleaner (X or Y) is best to buy. The table below shows some
of the information about these two brands.
Concentration of
Brand
Price
Volume
ammonia
X
$4.00
200 cm3
3.96 g /dm3
Y
$8.00
500 cm3
?
To determine the concentration of ammonia in brand Y, the student titrated 25.0 cm3 of
brand Y against 0.22 M hydrochloric acid. 24.4 cm3 of the acid was required for complete
neutralization.
(a) Write the equation for the reaction between ammonia solution and hydrochloric acid.
(DO NOT accept ionic equation)
1]
NH3(aq) + HCl(aq) → NH4Cl(aq)
(b) (i)
Name the apparatus containing brand Y in titration.
Conical flask
[1]
(ii) How should the apparatus in (i) be washed before titration?
It should be washed with distilled water only.
[1]
−3
(c) Calculate the concentration of ammonia in brand Y in g dm .
No. of moles of hydrochloric acid = 0.22 x 0.0244 = 5.368 x 10−3 mol
Molarity of ammonia = 5.368 x 10−3 ÷ 0.025 = 0.215 mol dm−3
[1]
−3
Concentration of ammonia = 0.215 x (14.0 + 1.0 x 3) = 3.65 g dm
[1]
(d) Determine which brand of glass cleaner is better buy. Show your reasoning.
Mass of NH3 in brand X = 3.96 x 0.2 = 0.792 g
[1]
Price per gram of NH3 = 4 ÷ 0.792 = $5.05
Mass of NH3 in brand Y = 3.65 x 0.5 = 1.825 g
Price per gram of NH3 = 8 ÷ 1.825 = $4.38
[1]
∴ Brand Y is better buy.
[1]
(e) State one reason for using ammonia solution instead of sodium hydroxide solution in
window cleaners.
Ammonia solution is volatile (will not leave stain on glass). / Sodium hydroxide
solution is corrosive to skin (can attack glass).
[1]
(Relative atomic masses: H = 1.0, N = 14.0)
(9 marks)
17.
(a)
(b)
Sodium nitrate is a salt which can be prepared by reacting an acid with an alkali using
titration method.
(i)
Name an acid and an alkali which react to give sodium nitrate.
Nitric acid
[1]
Sodium hydroxide solution
[1]
(ii) Explain why titration method is suitable for preparation of sodium nitrate.
It is because sodium hydroxide solution, nitric acid and sodium nitrate are
soluble in water.
[1]
Lead(II) bromide is a salt which can be prepared by precipitation method.
(i)
Name suitable reagents for the preparation of lead(II) bromide.
Lead(II) nitrate solution
[1]
Potassium bromide solution
[1]
(Any soluble lead(II) compounds and bromide compounds are acceptable)
(ii) Explain why the precipitation method is suitable for the preparation of lead(II)
bromide.
It is because the reactants (PbNO3 and KBr) are soluble in water and the product
(PbBr2) is an insoluble salt which can be easily separated from the mixture. [1]
(iii) What are the steps required to obtain pure dry sample of lead(II) bromide from
the reaction mixture?
Filter the product from the reaction mixture.
[1]
Wash the residue with distilled water.
[1]
Dry the residue with filter paper, under lamp or use air-dry.
[1]
(9 marks)
18.
A description of the preparation of a pure crystalline sample of hydrated copper(II) sulphate
is given below.
A 25 cm3 sample of dilute sulphuric acid is heated gently. A small quantity of copper(II)
oxide is added to the hot acid with stirring. More copper(II) oxide is added until it is in
excess. The excess copper(II) oxide is filtered off from the solution. The filtrate is heated
until crystals just started to form and then left to cool. The crystals are collected and
washed with small amount of distilled water. Then the crystals are dried without heating
with filter paper.
(a) (i)
Explain the importance of each of the five underlined phrases in the preparation
of pure crystalline hydrated copper(II) sulphate.
“heated”: The acid is heated so that copper(II) oxide can react with the acid
readily / to speed up the reaction.
[1]
“with stirring”: The copper(II) oxide is uniformly distributed throughout the acid
for a fast and complete reaction.
[1]
“in excess”: To make sure that all acid has been reacted.
[1]
“cool”: This allows the concentrated filtrate to lose heat and hence copper(II)
sulphate crystals form.
[1]
“without heating”: To prevent the water of crystallization of hydrated copper(II)
sulphate crystals to be lost.
[1]
(ii) Write the equation and state the observation(s) for the reaction between
copper(II) oxide and dilute sulphuric acid.
[1]
CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l)
Black solid dissolved.
[1]
The solution changed from colourless to blue.
[1]
(iii) Draw a labelled diagram for the experimental set-up used in filtering excess
copper(II) oxide from copper(II) sulphate solution.
(b)
(1 mark for the diagram; 1 mark for labelling the filter funnel and filter paper)
[2]
(iv) A student followed the above description and tried to prepare crystalline sample
of hydrated copper(II) sulphate. However, he could not obtain any solid after
one day. Suggest an explanation.
The filtrate was not saturated (concentrated) enough.
[1]
Describe what will be observed when ammonia solution is added drop by drop with
stirring to aqueous copper(II) sulphate until it is in excess.
A pale blue precipitate is initially formed when ammonia solution is added.
[1]
When excess ammonia solution is added, the pale blue precipitate dissolves to give a
deep blue solution.
[1]
(13 marks)
19.
A domestic drain cleaner contains concentrated sulphuric acid as the active ingredient. An
titration experiment was carried out to determine the concentration of sulphuric acid in the
domestic drain cleaner.
25.0 cm3 of the drain cleaner was diluted to 1 000.0 cm3 with distilled water. 25.0 cm3 of the
diluted solution were then titrated against 1.0 M sodium hydroxide solution. 20.6 cm3 of 1.0
M sodium hydroxide solution were required for complete neutralization.
(a) Describe how the end point in this titration can be determined.
Adding a few drops of phenolphthalein (or methyl orange) to the conical flask. [1]
The colour changes from colourless to pink (or from red to yellow)
[1]
(b) Suggest one reason for diluting the drain cleaner before titration.
To save chemicals.
[1]
－3
(c) Calculate the concentration, in g dm , of sulphuric acid in the drain cleaner.
H2SO4(aq) + 2NaOH(aq) Æ Na2SO4(aq) + 2H2O(l)
No. of moles of NaOH used = 1 x 0.0206 mol
= 0.0206 mol
According to the equation, 1 mole of H2SO4 requires 2 moles of NaOH for complete
neutralization.
∴ No. of moles of H2SO4 in 25.0 cm3 of diluted drain cleaner
= 1/2 x 0.0206
= 0.0103 mol
[1]
Molarity of H2SO4 in the drain cleaner
1000
x 0.0103
25
=
25
1000
(d)
(e)
(f)
= 16.48 M
[1]
Concentration of sulphuric acid in the drain cleaner
= 16.48 x (1.0 x 2 + 32.0 + 16.0 x 4)
[1]
= 16.48 x 98.0
－
= 1 615.0 g dm 3
[1]
Suggest ONE disadvantage of using the drain cleaner for cleaning.
It may cause corrosion to the iron drainage pipes.
/ A large amount of heat will be released in dilution of concentrated sulphuric acid.
This may cause deforming of plastic pipes.
[1]
State ONE safety precaution needed when using the drain cleaner. Explain your
answer.
Wear plastic gloves. / Wear safety goggles.
[1]
It is because concentrated sulphuric acid is highly corrosive.
[1]
Describe briefly the procedure to prepare a burette containing the sodium hydroxide
solution for titration.
Wash the burette with distilled water,
[1]
and then with 1.0 M sodium hydroxide solution.
[1]
Fill the burette with the sodium hydroxide solution, making sure that the space
between the tap and the tip of the burette is filled with the alkali.
[1]
(Relative atomic masses: H = 1.0, O = 16.0, S = 32.0)
(13 marks)
20.
A student tried to make a hydrated salt CaCl2‧6H2O by adding excess calcium oxide to
dilute hydrochloric acid. After filtering excess calcium oxide, a saturated solution was
prepared. It was then left in air. Some crystals were formed after standing overnight. The
crystals were then collected and washed with distilled water.
(a) Explain the term "hydrated salt".
It refers to the salt containing water of crystallization.
[1]
(b) Write an ionic equation for the reaction between calcium oxide and dilute hydrochloric
acid.
CaO(s) + 2H+(aq) → Ca2+(aq) + H2O(l)
[1]
(c) Suggest the reason for adding excess calcium oxide.
To make sure all the hydrochloric acid has been reacted.
[1]
(d) Draw the experimental set-up for preparing saturated solution from the reaction
mixture.
(e)
(f)
(g)
[2]
Explain why it is unwise to evaporate the saturated solution to dryness in order to
obtain the salt.
It will boil away all the water, so hydrated salt cannot be prepared.
[1]
Moreover, some impurities may be trapped in the salt.
[1]
Explain why it is necessary to wash the crystals with distilled water.
To remove any soluble impurities.
[1]
3
If there is 100 cm of 2 M HCl, calculate the maximum mass of hydrated salt that can
be obtained.
No. of moles of HCl = 2 x 0.1 = 0.2 mol
According to the equation, 2 moles of HCl gives 1 mole of CaCl2.
No. of moles of CaCl2‧6H2O = 0.2 ÷ 2 = 0.1 mol
[1]
Molar mass of hydrated calcium chloride = 40.0 + 35.5 x 2 + 6 x (16.0 + 2 x 1.0)
－
= 219 g mol 1
1]
Mass of hydrated calcium chloride obtained = 0.1 × 219 = 21.9 g
(Relative atomic masses: H = 1.0, O = 16.0, Cl = 35.5, Ca = 40.0)
[1]
(11 marks)
21.
Decide whether it is appropriate to prepare the following substances by mixing the reagents
listed on the right. If it is appropriate, just write the equation for the reaction involved. If it
is inappropriate, state the reason and suggest other suitable reagents for preparing the
substance. Write a chemical equation for the reaction involved.
Substance to be prepared
Reagents used
(a) copper(II) sulphate
copper, dilute sulphuric acid
(b) calcium sulphate
calcium carbonate, dilute sulphuric acid
(c) hydrogen
magnesium, dilute hydrochloric acid
(d) zinc chloride
zinc nitrate and sodium chloride
(10 marks)
(a)
There is no reaction between copper and dilute sulphuric acid.
Suitable reagents:
copper(II) oxide (or copper(II) hydroxide) + dilute sulphuric acid
Equation:
CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l)
/ Cu(OH)2(s) + H2SO4(aq) → CuSO4(aq) + 2H2O(l)
[1]
[1]
[1]
(b)
Calcium carbonate reacts with dilute sulphuric acid to form insoluble calcium sulphate
on the surface which prevents further reaction.
[1]
Suitable reagents:
calcium nitrate solution + sodium sulphate solution
[1]
(Any soluble calcium salt + any soluble sulphate compound are acceptable)
Equation:
Ca(NO3)2(aq) + Na2SO4(aq) → CaSO4(s) + 2NaNO3(aq)
[1]
(c)
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
[1]
There is no reaction between zinc nitrate and sodium chloride.
Suitable reagents:
zinc + dilute hydrochloric acid
Equation:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
[1]
(d)
[1]
[1]
22.
Consider the following titration.
Step I
Place 25.0 cm3 of 0.600 M sodium hydroxide solution in a clean conical flask.
Step II Add a few drops of phenolphthalein to the sodium hydroxide solution.
Step III Fill a burette with dilute sulphuric acid.
Step IV Run the acid into the sodium hydroxide solution until the indicator just
changes colour.
Result: 15.0 cm3 of sulphuric acid is used.
(a) What apparatus is usually used to transfer 25.0 cm3 of sodium hydroxide solution into
the conical flask?
Pipette
[1]
(b) Suggest how the apparatus in (a) should be cleaned before the delivery of sodium
hydroxide solution.
Wash the pipette with distilled water
[1]
and then with sodium hydroxide solution.
[1]
(c) Draw a labelled diagram for the experimental set-up for this titration experiment.
(d)
(e)
(f)
(1 mark for the titration set-up; 2 marks for all four labels)
What is the colour change of the indicator at the end point?
From pink to colourless
(i)
Write the chemical equation for the neutralization.
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
(ii) Calculate the molarity of the sulphuric acid.
Number of moles of NaOH in 25.0 cm3 solution
= 0.6 x 0.025 = 0.015 mol
Number of moles of H2SO4 in 15.0 cm3 solution
= 0.015 ÷ 2 = 0.0075 mol
Molarity of H2SO4
= 0.0075 ÷ 0.015
= 0.5 M
Name the salt formed in this experiment.
Sodium sulphate
[3]
[1]
[1]
[1]
[1]
[1]
(g)
Describe briefly how you can obtain the crystals of the salt from the sodium
hydroxide and sulphuric acid above.
Mix 25 cm3 of the sodium hydroxide solution and 15 cm3 of the sulphuric acid. [1]
Heat the mixture to obtain a saturated solution.
[1]
Allow the saturated solution to cool in air until crystals are obtained.
[1]
Wash the crystals with small amount of cold water and dry them with filter paper.
[1]
(15 marks)
25 cm3 of 0.15 M sulphuric acid was added to excess freshly cleaned magnesium ribbons in
a reaction vessel. The vessel was connected to a gas syringe so that the gas given off by the
reaction can be collected and measured at regular time interval. The following table shows
the results obtained.
Time (min)
0
1
2
3
4
5
6
7
8
9
10
Volume of
0
40
60
70
77
82
86
88
89
90
90
gas collected
(cm3)
(a) Draw a labelled diagram for a possible experimental set-up for this experiment.
[2]
(d)
3
(c)
Suggest how you can prepare freshly cleaned magnesium ribbons.
Clean the magnesium ribbons with sand paper before experiment.
[1]
Write an ionic equation for the reaction.
Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)
[1]
(i)
Plot a graph to show the relationship between the volume of gas collected and
time.
(cm )
(b)
Volume of hydrogen collected
23.
100
90
80
70
60
50
40
30
20
10
0
0
1
2
3
4
5
6
7
8
9
10
Time (min)
[2]
(ii)
(e)
Explain the shape of the curve plotted.
The reaction rate can be reflected by the slope of the curve. The faster the
reaction rate, the steeper the curve.
[1]
The initial rate is the fastest since the concentration of acid is the highest initially.
[1]
Reaction rate gradually decreases as part of the acid has been reacted and the
concentration of acid decreases.
[1]
Reaction stopped at 9th min as all acid has been reacted and hence the curve
flattens.
[1]
Explain how the following changes will affect the experimental results.
(i)
Use excess calcium granules to replace excess magnesium ribbons.
Calcium reacts faster than magnesium initially as calcium is more reactive than
magnesium.
[1]
The reaction stops quickly as insoluble calcium sulphate is formed on the calcium
granules which prevents further reaction.
[1]
The volume of hydrogen collected is less than the theoretical value.
[1]
(ii) Use magnesium powder of the same mass to replace magnesium ribbons.
Magnesium powder reacts at a faster rate as it has a larger surface area. [1]
The volume of hydrogen collected remains unchanged.
[1]
3
3
(iii) Use 50 cm of 0.15 M ethanoic acid to replace 25 cm of 0.15 M sulphuric acid.
The reaction will be slower as the concentration of hydrogen ions is lower. [1]
The volume of hydrogen collected remains unchanged.
[1]
(17 marks)
A student performs two experiments to study the reactions between (1) calcium and excess
water and (2) calcium and excess hydrochloric acid. He uses 0.4 g of calcium in both
experiments and the volumes of hydrogen liberated at 10-second intervals are tabulated as
shown below (all gases are measured under the same conditions).
Time (s)
11
0 10 20 30 40 50 60 70 80 90 100
0
Volume of H2 given
off in experiment (1) 0 17 34 76 134 184 216 220 222 224 224 224
(cm3)
Volume of H2 given
off in experiment (2) 0 88 144 182 207 222 224 224 224 224 224 224
(cm3)
(a) (i)
Write the equations for the reactions between calcium and (1) water and (2)
dilute hydrochloric acid respectively.
[1]
(1)
Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g)
[1]
(2)
Ca(s) + 2HCl(aq) → CaCl2(aq) + H2(g)
(ii) Suggest an observable change when calcium reacts with water.
Calcium dissolves./ Gas bubbles are given off.
[1]
(b) Using "time" as the x-axis and "volume of hydrogen given off" as the y-axis, plot the
results for both experiments on the same graph.
240
3
Volume of hydrogen given off (cm )
24.
220
200
180
160
140
Experiment 1
120
Experiment 2
100
80
60
40
20
0
0
10
20
30
40
50
60
70
80
90
100
110
Time (s)
[3]
(c)
(d)
(e)
(f)
Which reaction, (1) or (2), reacts faster initially? Why?
(2)
[1]
Hydrochloric acid has a higher concentration of hydrogen ions than water.
[1]
Which reaction, (1) or (2), has a higher reaction rate at 50th second? Explain your
answer with reference to the shape of the curve.
(1)
[1]
The slope of curve (1) is steeper than that of curve (2) at 50th second.
[1]
When is calcium completely used up in each experiment?
Reaction (1) stops at 90th second.
[1]
Reaction (2) stops at 60th second.
[1]
Why are the final volumes of hydrogen collected the same in experiments (1) and (2)?
Same mass of calcium (limiting reactant) is used in each case.
[1]
(13 marks)
25.
Magnesium ribbons react with 1 M hydrochloric acid to liberate a colourless gas.
(a) Write a balanced equation, including state symbols, for the reaction.
Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) / Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) [1]
(b) Draw the electronic diagrams of the products formed (showing electrons in the
outermost shells only).
Magnesium chloride:
[1]
Hydrogen gas:
[1]
(c)
Suggest an experimental method to monitor the progress of the reaction. Draw a
labelled diagram to show the experimental set−up.
[1]
Measure the volume of H2(g) given off at regular time intervals.
Diagram of set−up
[2]
or Place the vessel on a top pan balance and record the mass of the conical flask plus
its content at regular time intervals.
Diagram of set−up
(d)
(e)
(f)
(g)
How do you know the reaction is completed?
The volume of hydrogen collected remains unchanged./ The mass of the conical flask
plus its content remains unchanged.
[1]
How will the reaction rate and the amount of products formed be affected when the
reaction mixture is put into a hot water bath?
The reaction rate will increase.
[1]
There will be no change in the amount of products formed.
[1]
How will the reaction rate be changed if magnesium powder of the same mass is used
instead of magnesium ribbons? Explain your answer.
The reaction rate will increase.
[1]
Magnesium powder has a larger surface area than magnesium ribbons.
[1]
How will the reaction rate be changed if 1 M sulphuric acid is used instead of 1 M
hydrochloric acid? Explain your answer.
The reaction rate will increase.
[1]
Sulphuric acid is a dibasic acid while hydrochloric is a monobasic acid.
[1]
Hence, sulphuric acid of the same molarity has a higher concentration of hydrogen
ions than hydrochloric acid.
[1]
(14 marks)
26.
A student carried out the following experiments at room temperature and pressure to study
the rate of reaction between magnesium and sulphuric acid:
Experiment
Reaction
1
0.7 g magnesium + 100 cm3 of 1 M H2SO4
2
0.7 g magnesium + 200 cm3 of 1 M H2SO4
3
0.7 g magnesium + 100 cm3 of 2 M H2SO4
For experiment 2, a plot of the volume of gas given off against time was obtained:
(a)
(b)
(c)
(d)
Write an equation for the reaction between magnesium and sulphuric acid.
Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g) / Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)
[1]
Suggest a method for measuring the volume of the gas liberated.
By connecting the reaction vessel to a gas syringe.
[1]
Explain why the slope of the curve is steeper at X than at Y in experiment 2.
The reaction rate at X is higher than that at Y. / The concentration of hydrogen ions at
X is higher than that at Y.
[1]
When was the reaction completed in experiment 2?
34th second
[1]
(e)
Copy the graph of experiment 2 in your answer book and draw on the same graph the
expected curves for experiments 1 and 3. Explain your answer.
[2]
The reaction rate of experiment 1 is the same as that of experiment 2. This is
because the concentration of hydrogen ions in both experiments is the same.
[1]
The reaction rate of experiment 3 is faster than that of experiment 2. This is because
the concentration of hydrogen ions in experiment 3 is higher than that in experiment 2.
[1]
The final volume of hydrogen given off in experiments 1, 2 and 3 are the same
because magnesium is the limiting reactant in the three experiments.
[1]
(9 marks)
The reaction between sodium thiosulphate solution and hydrochloric acid can be represented
by the following equation:
－
S2O32 (aq) + 2H+(aq) Æ H2O(l) + SO2(g) + S(s)
To investigate the effect of the concentration of sodium thiosulphate solution on the rate of
the above reaction at 25 oC, 30 cm3 of 0.4 M sodium thiosulphate solution was prepared in a
beaker. The beaker was then placed over a black cross mark. 20 cm3 of 2.5 M hydrochloric
acid was added to the beaker. The time required for the disappearance of the cross was
recorded.
The experiment was repeated with different volumes of 0.4 M sodium thiosulphate solution
and water according to the table below.
Volume of 0.4 M sodium
1/t
Volume of
Time
Experiment thiosulphate solution
－1
3
(s
)
)
water
(cm
required
(s)
(cm3)
1
30
0
21.6
0.046
2
20
10
30.2
0.033
3
15
15
38.3
0.026
4
10
20
55
0.018
5
5
25
100.1
0.010
(a) Explain why
(i)
the cross was placed under the beaker.
To find out the time for the disappearance of the cross.
/ To find out the time that a certain amount of sulphur is produced.
[1]
(ii) water was added in the experiments 2, 3, 4 and 5.
To vary the concentration of sodium thiosulphate solution while keeping the
volume of the solution constant at the same time.
[1]
(iii) the same volume of 2.5 M hydrochloric acid was used in each experiment.
To ensure that only the concentration of sodium thiosulphate solution is varied.
[1]
(b) Plot a graph of 1/t against the volume of 0.4 M sodium thiosulphate solution used.
0.05
0.04
-1
1/t (s )
27.
0.03
0.02
0.01
0
0
5
10
15
20
25
30
35
Volume of 0.4 M sodium thiosulphate solution used (cm3)
(1 mark for correctly labelled axis; 1 mark for straight line through the origin)
[2]
(c)
Deduce the relationship between the rate of reaction and the concentration with
reference to the graph.
As the volume of sodium thiosulphate solution used is directly proportional to its
concentration
[1]
and 1/t is directly proportional to the rate of reaction,
[1]
the concentration of sodium thiosulphate solution used is directly proportional to the
rate of reaction.
[1]
(d) How will the graph change if the temperature for the experiment is increased to 35 oC?
The slope of the curve will increase.
[1]
(9 marks)
28.
Briefly describe how you would test for the presence of ammonium ions (NH4+), iron(III)
ions (Fe3+) and water (H2O) in a sample of iron alum ((NH4)2SO4‧Fe2(SO4)3‧20H2O).
(You are required to give paragraph-length answer to this question. In this question, 6
marks
will be awarded for chemical knowledge and 3 marks for effective communication.)
(9 marks)
Add sodium hydroxide solution to the sample and heat the mixture.
[1]
A gas is given off which turns moist red litmus paper blue. This indicates the presence of
NH4+.
[1]
Add sodium hydroxide solution to the sample.
[1]
A brown precipitate (Fe(OH)3) is formed. This indicates the presence of Fe3+.
[1]
Heat a sample of the salt.
[1]
Water vapour will turn dry cobalt(II) chloride paper from blue to pink / anhydrous copper(II)
sulphate from white to blue. This indicates the presence of H2O.
[1]
(3 marks for effective communication)
29.
Describe the procedures for determining the exact concentration of a sample of sodium
carbonate solution (known to be approximately 0.1 M) using 0.20 M nitric acid. (No diagram
is required.)
(You are required to give paragraph-length answer to this question. In this question, 8
marks will be awarded for chemical knowledge and 3 marks for effective communication.)
(11 marks)
3
First wash a 25.0 cm pipette with distilled water and then with sodium carbonate solution.
[1]
Transfer 25.0 cm3 of sodium carbonate solution to a clean conical flask using the pipette and
pipette filler.
[1]
Add a few drops of methyl orange to the solution as an indicator.
[1]
Wash the burette first with distilled water and then with 0.20 M nitric acid. Fill the burette
with the acid.
[1]
Place the conical flask containing the sodium carbonate solution on a white tile. Run the acid
from the burette until the indicator in the conical flask just changes colour.
[1]
Note the volume of nitric acid added. This is the titre reading of the first trial.
[1]
Repeat the titration 2 or 3 times to check if the titre readings are coincident. Record the
results.
[1]
The exact concentration of the sample of sodium carbonate solution can be calculated from
the volume and concentration of nitric acid used according to the following equation:
2HNO3(aq) + Na2CO3(aq) Æ 2NaNO3(aq) + CO2(g) + H2O(l)
Number of moles of HNO3(aq)
= average volume of HNO3(aq) used x concentration of HNO3(aq)
Number of moles of Na2CO3(aq) = number of moles of HNO3(aq) ÷ 2
Molarity of Na2CO3(aq) =
no. of mole of Na 2 CO 3
25
1000
[1]
(3 marks for effective communication)
30.
A sample of sodium carbonate is contaminated with sodium nitrate. Describe how you would
conduct an experiment to determine the percentage by mass of sodium carbonate powder in
the sample.
(You are required to give paragraph-length answer to this question. In this question, 7
marks will be awarded for chemical knowledge and 3 marks for effective communication.)
(10 marks)
[1]
Weigh the sample with an electronic balance to obtain the mass (m1).
3
3
Prepare a 250 cm solution of the sample with a volumetric flask. Transfer 25 cm of the
solution into a conical flask with addition of methyl orange indicator.
[1]
Deliver known concentration (MA) of sulphuric acid (or other suitable acid) from burette.
Record the volume of acid (VA) needed to change the colour of indicator from yellow to red.
[1]
Determine the number of moles of sulphuric acid used from the calculation of MA x VA. [1]
The number of moles of sodium carbonate in the sample = (MA x VA) x 10
[1]
The mass of sodium carbonate (m2) in the sample = the number of moles of sodium
carbonate in the sample x the molar mass of sodium carbonate.
[1]
Calculate the percentage by mass of sodium carbonate in the sample by (m2 ÷ m1) x 100%.
[1]
(3 marks for effective communication)
31.
Describe how large crystals of sodium chloride can be prepared from sodium hydroxide
solution and dilute hydrochloric acid.
(You are required to give paragraph-length answer to this question. In this question, 6
marks will be awarded for chemical knowledge and 3 marks for effective communication.)
(9 marks)
Titrate sodium hydroxide solution with dilute hydrochloric acid
[1]
until the end point is reached.
[1]
(or, add dilute hydrochloric acid to sodium hydroxide solution in the mole ratio of 1 : 1)
Heat the solution until a saturated (more concentrated) solution is obtained.
[1]
Cool slowly to obtain large crystals of sodium chloride.
[1]
Filter the crystals.
[1]
Wash the crystals with distilled water and then dry with filter paper.
[1]
(3 marks for effective communication)
32.
You are provided with the following chemicals and materials:
calcium carbonate granules, 1 M hydrochloric acid, distilled water, test tubes, measuring
cylinder, balloons, stop-watches
Use the above materials to design an experiment which can demonstrate two factors
affecting the rate of a reaction. You should give brief explanation for the factors.
(You are required to give paragraph-length answer to this question. In this question, 7
marks will be awarded for chemical knowledge and 3 marks for effective communication.)
(10 marks)
The rate of a reaction can be measured by the inflation rate of a balloon using a stop-watch
when calcium carbonate reacts with an acid to give out carbon dioxide.
[1]
The first factor to be investigated is the surface area of the reactant.
First, add certain amount of calcium carbonate granules to a test tube. Then, certain volume
of 1 M hydrochloric acid is added. Immediately fill the mouth of the test tube with a balloon.
[1]
Repeat the experiment by using same volume of acid and same amount of calcium
carbonate granules which are crushed into powder.
[1]
The inflation rate of the balloon increases when calcium carbonate powder is used because
the contact surface area is greater.
[1]
The second factor to be investigated is the concentration of the reactant.
First, add certain amount of calcium carbonate to a test tube. Then, certain volume of 1 M
hydrochloric acid is added. Immediately fill the mouth of the test tube with a balloon. [1]
Repeat the experiment by using 0.1 M hydrochloric acid of the same volume (prepared by
diluting the 1 M hydrochloric acid) and the same amount of calcium carbonate granules.
[1]
The inflation rate of the balloon decreases because the concentration of hydrogen ions in
the acid is lower.
[1]
(3 marks for effective communication)