Chemistry 3 & 4, Chapter 4 – Volumetric Analysis TEST SOLUTIONS
Multiple Choice
Identify the choice that best completes the statement or answers the question.
Chemistry 3 & 4, Chapter 2
Multiple Choice
Identify the choice that best completes the statement or answers the question.
1. For a substance to be suitable as a primary standard in a volumetric analysis, it should
have:
A a known formula with a low molar mass
B a very low melting point
C no tendency to absorb moisture from the air
D no acid–base or redox properties.
ANSWER: C
0/1
POINTS:
FEEDBACK: In most volumetric analyses, the primary standard must be an acid, a
base, an oxidant or a reductant. Since it must be a solid at room
temperature so that it can be accurately weighed out, it needs to have a
relatively high MP. Since it must be accurately weighed out, it should
not absorb any substance from the air. But a low molar mass means that
there will be a higher percentage of error should the measured mass
differ from the true mass by, for example, 0.005 g.
pp48–9
REF:
2. In a volumetric analysis of vinegar to determine the concentration of ethanoic acid
present, a pair of students obtained the following results:
Initial burette reading
0.10
0.15
0.00
0.25
0.30
(mL)
Final burette reading
22.70 22.60 22.50 22.73 22.60
(mL)
The mean titre is:
A 22.60 mL
B 22.48 mL
C 22.477 mL
D 22.47 mL
ANSWER: B
0/1
POINTS:
FEEDBACK: The titres are 22.60, 22.45, 22.50, 22.48 and 22.30 mL respectively. The
concordant results are 22.45, 22.48 and 22.50 mL. The mean of these is
22.48 mL, since a mean cannot have more decimal places than the
results used to calculate it.
pp45–6
REF:
3. Which one of the following substances is not suitable for use as a primary standard in
volumetric analysis?
A Sulfuric acid
B Potassium hydrogen phthalate
C Oxalic acid
D Anhydrous sodium carbonate
ANSWER: A
1/1
POINTS:
FEEDBACK: Answers B, C and D all obey the criteria for a primary standard, but
sulfuric acid does not, since it is a solution of a gas in water and its
concentration can only be precisely known by performing a set of
titrations.
pp49–50
REF:
4. What term is used to describe a substance that, when exposed to air, absorbs so much
water from the air that it eventually forms a solution?
A Hydrated
B Efflorescent
C Hygroscopic
D Deliquescent
ANSWER: D
0/1
POINTS:
FEEDBACK: A hydrated compound contains water molecules within the solid
structure; these can become part of the lattice, as the crystal is formed
when the solvent is evaporated from a solution of the compound. An
efflorescent substance releases water into the air. A hygroscopic
substance also absorbs moisture from the air but does not form a
solution.
p49
REF:
5. The uncertainty of the volume of a solution delivered by a 20.00 mL pipette would be:
A 2 mL
B 0.2 mL
C 0.02 mL
D 0.002 mL
ANSWER: C
0/1
POINTS:
FEEDBACK: As a general rule, the uncertainty of a pipette or burette is in the last
decimal place of the reading. In this case it means that all we can be sure
of is that the volume delivered by the pipette is between 19.98 and 20.02
mL.
pp44–5
REF:
6. To standardise a supplied sodium hydroxide solution, a pair of students placed the base
in a burette and titrated it against a standard potassium hydrogen phthalate solution,
KH(C8H4O4), which was previously prepared using a 250-mL volumetric flask. Which
items of equipment should the students have rinsed out with several amounts of de-
ionised water just before use, and with no other chemicals, during this procedure?
A The volumetric flask and reaction flask only
B The pipette and burette only
C The reaction flask only
D All of the equipment
ANSWER: A
1/1
POINTS:
FEEDBACK: All equipment should first be rinsed with water, to remove any
contaminants. But after that the pipette and burette must be rinsed with
the solution they will contain, so that it is not diluted. Otherwise we will
not know precisely what amount has been measured into the reaction
vessel and hence cannot determine the unknown concentration. If the
reaction flask and volumetric flask were rinsed with one of the reactants,
however, we will never know what amount of that reactant is present
and again we will not be able to determine the unknown concentration.
pp42–3
REF:
7. One reason that sodium hydroxide is not suitable as a primary standard is that:
A it is highly soluble in water.
B it is deliquescent.
C it is dangerous to handle.
D its solution is colourless.
ANSWER: B
0/1
POINTS:
FEEDBACK: Substances used in a volumetric analysis need to be soluble. The fact
that sodium hydroxide is dangerous does not prevent it from being used
as a primary standard, nor does the fact that it is colourless, since this
can be overcome by using an acid-base indicator or pH meter. But since
it is deliquescent (it absorbs a significant amount of water from the air),
it cannot be weighed out accurately. Hence we cannot know the precise
concentration of a solution of NaOH without performing a set of
titrations.
pp48–51
REF:
8. A sample of commercial brick cleaner was analysed against a 20.00 mL aliquot of a
solution of the primary standard anhydrous sodium carbonate, of concentration 0.05342
mol L-1. The equation for the titration reaction was:
2HCl(aq) + Na2CO3(aq)  2NaCl(aq) + H2O(l) + CO2(g)
The pH of the reaction mixture at the equivalence point should be:
A exactly 7, since this is a neutralisation reaction.
B less than 7, since some of the CO2 will remain in solution.
C greater than 7, since some of the CO2 will remain in solution.
D less than 7, since the amount of acid reacting is twice that of the base.
ANSWER: B
0/1
POINTS:
FEEDBACK: At the equivalence point, only the products are present, so the
stoichiometric ratio of the reacting acid and base has no effect on pH. At
REF:
the equivalence point in this case, the solution will contain NaCl, which
has no acid-base properties, and some dissolved CO2, which forms weak
carbonic acid. This acid will lower the pH of the solution.
p55
9. Following is the pH curve obtained for a particular volumetric analysis.
This curve shows that in all probability in this titration:
A a strong acid was being added to a solution of a weak base.
B a solution of a weak base was being added to a strong acid.
C a weak acid was being added to a solution of a strong base.
D a solution of a strong base was being added to a strong acid.
ANSWER: C
0/1
POINTS:
FEEDBACK: The initial pH is close to 14, indicating a solution of a strong base. The
final pH after the neutralisation reaction has occurred is approx 4,
showing an excess of a weak acid. A solution made up of about 20 mL
of a strong acid of approx. concentration 0.1 mol L-1 and 20 mL of a salt
solution (from the neutralisation reaction) would have a pH just greater
than 1.
pp54–6
REF:
10. Following is the pH curve obtained for another particular volumetric analysis.
Of the following list of acid-base indicators, which would be the most suitable indicator
to use for this titration?
A Alizarin yellow R, which changes colour over the pH range 10.1–12.0
B Bromothymol blue, which changes colour over the pH range 6.0–7.6
C Methyl red, which changes colour over the pH range 4.8–6.0
D Phenolphthalein, which changes colour over the pH range 8.2–10.0
ANSWER: D
0/1
POINTS:
FEEDBACK: The most suitable indicator must change colour as close as possible to
the equivalence point, which occurs when the pH of the reaction mixture
corresponds to the centre of the vertical section of the pH curve.
p57
REF:
11. To what volume of water must 5.0 mL of nitric acid of concentration 10 mol L-1 be
added to produce a solution of pH zero?
A 5.0 mL
B 45.0 mL
C 50.0 mL
D 95.0 mL
ANSWER: B
0/1
POINTS:
FEEDBACK: It must be remembered that pH itself is not a concentration. In this
dilution of conc. nitric acid,
c1 = 10 mol L-1, V1 = 5.0 mL,
c2 = 100 mol L-1 = 1 mol L-1, and V2 is unknown.
Substitution into the dilution formula gives
REF:
V2= 50.0 mL. But this includes the volume of the
acid, so the volume of water needed to dilute the
acid is (50.0 – 5.0) mL = 45.0 mL.
p59
12. A group of students was given the task of standardising solutions of ammonia (a weak
base), sodium hydroxide and calcium hydroxide (strong bases), using hydrochloric acid
(a strong acid). It was known that the concentration of each base was approximately 0.1
mol L-1. In each case the students pipetted 20.00 mL of the base into a reaction flask
and then delivered the acid by burette. The concentration of the acid used was the same
in each case. If the indicator used in each case was chosen so that it only changed
colour when the reaction was complete, which statement about the volume of acid
required for each set of titrations is CORRECT?
A All three bases should require the same amount of acid.
B NH3 should require less acid than NaOH or Ca(OH)2.
C NH3 and NaOH should require about half the amount of acid as Ca(OH)2.
D NH3 and NaOH should require about twice the amount of acid as Ca(OH)2.
ANSWER: C
0/1
POINTS:
FEEDBACK: Since HCl is a strong acid, it will react completely with both a strong
base and a weak base. In each case we have the same amount of base,
since their C and V is the same. When we write the balanced equation
for each titration, it can be seen that 1 mol of each of NH3 and NaOH
requires just 1 mol HCl for a complete reaction, so both should require
the same amount of HCl. But in the case of Ca(OH)2, twice as much
HCl should be required, since 1 mol of Ca(OH)2 requires 2 mol HCl.
p55
REF:
13. To determine the concentration of ethanoic acid, CH3COOH, in a sample of vinegar, a
10.00 mL aliquot of the vinegar was made up to 100.00 mL in a volumetric flask then
titrated against 20.00 mL aliquots of sodium hydroxide solution of concentration
0.09950 mol L-1. The equation for the titration is:
CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l)
If the mean titre was 19.45 mL, the concentration of ethanoic acid in the vinegar was:
A 0.09676 mol L-1
B 0.1023 mol L-1
C 0.9676 mol L-1
D 1.023 mol L-1
ANSWER: D
0/1
POINTS:
FEEDBACK: For each titration, n(NaOH) = 1.990 × 10-3 mol.
From the equation, n(CH3COOH) in 19.45 mL
of the diluted solution is 1.990 × 10-3 mol, which
means [CH3COOH] in the diluted solution is
0.1023 mol L-1. Since the dilution factor was
, which equals 10.00, [CH3COOH] in
the original vinegar solution is 1.023 mol L-1.
REF:
pp60–1
14. The active ingredient in a certain toilet cleaner that claims to ‘remove lime scale and
rust’ is hydrochloric acid of approximate concentration 3 mol L-1. To analyse this toilet
cleaner, a quality control chemist first delivered a 10.00 mL aliquot of the toilet cleaner
into a 250-mL volumetric flask, then made up the diluted solution. She then titrated a
20.00 mL aliquot of this diluted solution against standard sodium hydroxide solution of
concentration 0.1044 mol L-1. The expected titre would be approximately:
A 14.4 mL
B 17.4 mL
C 23.0 mL
D 28.7 mL
ANSWER: C
0/1
POINTS:
FEEDBACK:
The dilution factor is
, which equals 25.00,
so [HCl] in diluted solution is approx.
mol L-1,
or 0.12 mol L-1. For the titration,
n(HCl) reacting = 0.12 × 0.02000 mol = 2.4 × 10-3 mol.
From the equation for the reaction,
n(NaOH) reacting = n(HCl) = 2.4 × 10-3 mol.
REF:
Hence V(NaOH) in the titration =
= 0.0230 L = 23.0 mL. (Note all of these values are approximate.)
pp59–61
15. An analyst was commissioned to determine the concentration of substance X present in
a solution. He was provided with some data about X and as a result decided that a
simple volumetric analysis would not yield a very accurate result. Which one of the
following properties of X may have influenced this decision?
A Weak reductant
B Soluble in water
C Highly volatile
D Colourless
ANSWER: C
0/1
POINTS:
FEEDBACK: A substance must be soluble in water to be suitable for a volumetric
analysis. A colourless, weak reductant can be analysed by a simple
volumetric analysis – we would simply select a strong oxidant that
undergoes a colour change to react with it, such as acidified KMnO4.
But if it is highly volatile, a significant amount would escape from
solution and be lost, which would create an inaccurate result.
p50
REF:
16. Vitamin C, or ascorbic acid, C6H8O6, is classified as an antioxidant. To determine the
percentage by mass of Vitamin C in a tablet, a student weighed one tablet into a conical
flask and then crushed it in deionised water. The Vitamin C solution was then titrated
against iodine solution of concentration 0.1620 mol L-1. The equation for the reaction
is:
C6H8O6(aq) + I2(aq)  C6H6O6(aq) + 2H+(aq) + 2I- (aq)
Given the titre was 27.40 mL and the mass of the tablet was 1.335 g, what was the
percentage of Vitamin C in the tablet?
A 29.26 %
B 58.52 %
C 78.12 %
D 84.39 %
ANSWER: B
0/1
POINTS:
FEEDBACK: In the titration,
n(I2) reacting = 0.1620 × 0.02740 mol
= 4.439 × 10-3 mol.
From the equation,
n(C6H8O6) reacting = 4.439 × 10-3 mol.
Hence m(C6H8O6) in the tablet = 4.439 × 10-3 ×176.0 g
= 0.7812 g. Thus % Vitamin C in the tablet
REF:
=
pp64–6
= 58.52 %.
17. An experimental value for the concentration of hydrochloric acid in a sample of
commercial brick cleaner was obtained by titrating a diluted solution of the brick
cleaner against a standard solution of anhydrous sodium carbonate. This experimental
value was found to be lower than that claimed by the manufacturer. Which one of the
following explanations offered by different students could reasonably account for this
difference?
A Some HCl gas evaporated from solution during the analysis.
B The brick cleaner may have absorbed carbon dioxide out of the air.
C The Na2CO3 used contained some water of hydration.
D The pipette was rinsed out with water instead of Na2CO3 solution.
ANSWER: A
1/1
POINTS:
FEEDBACK: Hydrochloric acid is prepared by bubbling HCl gas into water, but some
will escape from open containers. This means a greater titre would be
required to neutralise the base, which could give an experimental value
of [HCl] that is less than the true value. In the case of B, C and D, each
aliquot will contain a lower amount of base than the amount calculated
to be present. Thus the amount of acid required to react will be less and
the mean titre will be lower, which would lead to an experimental value
of [HCl] that is higher than the true result.
Appendix 1, Student Activity Manual
REF:
18. The alcohol content of a certain low alcohol beer was determined using volumetric
analysis. In this analysis, a 10.00-mL sample of the beer was pipetted into a 100-mL
volumetric flask and the solution was made up to the mark. Then 20.00 mL aliquots of
this solution were titrated against acidified potassium dichromate solution, K2Cr2O7 of
concentration 0.0500 mol L-1. A mean titre of 16.35 mL was obtained. The equation for
the analysis is:
2Cr2O72-(aq) + 3C2H5OH(aq) + 16H+(aq)  4Cr3+(aq) + 3CH3COOH(aq) +
11H2O(l)
What was the molarity of the alcohol in the beer?
A 0.0273 mol L-1
B 0.0613 mol L-1
C 0.273 mol L-1
D 0.613 mol L-1
ANSWER: D
0/1
POINTS:
FEEDBACK: In the titration, n(Cr2O72-) reacting
= 0.0500 × 0.01635 mol = 8.18 × 10-4 mol.
From the equation,
n(C2H5OH) = × n(Cr2O72-)
= 1.23 × 10-3 mol.
Hence [C2H5OH] in the diluted beer
= 1.23 ×
mol L-1 = 0.0613 mol L-1.
Thus [C2H5OH] in the original beer
REF:
= 0.0613 ×
pp59–61
mol L-1 = 0.613 mol L-1.
19. Two pairs of VCE students were analysing the same sodium hydroxide solution by
titrating 20.00 mL aliquots against the same standard solution of potassium hydrogen
phthalate. The first pair obtained a mean titre of 21.45 mL while the second pair
obtained a mean titre of 21.80 mL. Which of the following might explain the difference
in the mean titres?
A The first pair rinsed their burette with only water and not acid.
B The second pair rinsed their burette with only water and not acid.
C The NaOH used by the second pair had absorbed more CO2 out of the air.
D The first pair blew the last drop of base out of their pipette.
ANSWER: B
0/1
POINTS:
FEEDBACK: If the burette is only rinsed with water, the acid added to it will be
diluted and so a greater volume of the acid will be required to react with
the base. If CO2, which is an acidic oxide, is absorbed by the base, it will
have partially reacted and less acid would be required to react with it,
not more. If extra base is blown out of the pipette, slightly more acid
would be required to react with it, not less.
Appendix 1, Student Activity Manual
REF:
20. An impure sample of anhydrous sodium carbonate was analysed as follows. Several
1.000 g samples of the impure mixture were weighed out into separate conical flasks
and about 20 mL of distilled water was added to each. The mixtures were stirred and
then 4 drops of an acid–base indicator were added to each flask. These were then
standardised against hydrochloric acid of concentration 0.1542 mol L-1, according to the
reaction:
Na2CO3(aq) + 2HCl(aq)  2NaCl(aq) + H2O(l)
Three concordant results were obtained for the titres: 21.25 mL, 21.29 mL and 21.30
mL. The percentage purity of the sodium carbonate was:
A 8.696%
B 17.39%
C 17.90%
D 34.78%
ANSWER: B
0/1
POINTS:
FEEDBACK: The mean titre is 21.28 mL.
Hence n(HCl) reacting in the titration
= 0.1542 × 0.02128 mol = 3.281 × 10-3 mol.
From the equation, n(Na2CO3) = × 3.281 × 10-3 mol
= 1.641 × 10-3 mol.
Thus m(Na2CO3) in each sample
= 1.641 × 10-3 mol. × 106.0 g = 0.1739 g.
REF:
Therefore the percentage purity is
= 17.39 %.
pp64–6