Chemistry 2007
Annotated indicative response and judgments within criteria
Supervised assessment: Acid and bases,
and equilibrium
This sample demonstrates on-balance judgments within criteria. It provides information about
student achievement where the indicative response matches the qualities of the A standards.
Criteria assessed
• Knowledge and conceptual understanding
• Investigative processes
• Evaluating and concluding
Assessment instrument
The response presented in this sample is in response to assessment items.
The task is a supervised assessment on the topic of acids and bases, and equilibrium.
Students have had the learning experiences of conducting titrations.
14364
The school considers the objectives of the syllabus when designing the assessment items.
The objectives, described as syllabus standards A–E, inform the design of the items and the
allocation of marks.
The matching of the standards with the items is demonstrated on the instrument-specific criteria and
standards on pages 2 and 3, and throughout the indicative response starting on page 4.
The item-specific marking scheme shows expected Standard A responses, and the marking scheme
on pages 20 and 21 shows the match of marks to standard descriptors.
Knowledge and conceptual understanding
Instrument-specific criteria and standards
Standard A
Standard B
Standard C
Standard D
Standard E
The student work has the
following characteristics:
The student work has the
following characteristics:
The student work has the
following characteristics:
The student work has the
following characteristics:
The student work has the
following characteristics:
•
•
•
•
reproduction of simple
ideas and concepts
•
reproduction of isolated
facts
•
description of simple
processes and
phenomena
•
recognition of isolated
simple phenomena
•
application of algorithms,
principles, theories and
schema
•
application of simple
given algorithms
reproduction and
interpretation of complex
and challenging
concepts, theories and
principles
reproduction and
interpretation of complex
or challenging concepts,
theories and principles
reproduction of concepts,
theories and principles
Q4
•
comparison and
explanation of complex
concepts, processes and
phenomena
•
comparison and
explanation of concepts,
processes and
phenomena
•
explanation of simple
processes and phenomena
Q6
•
linking and application of
algorithms, concepts,
principles, theories and
schema to find solutions
in complex and
challenging situations
•
linking and application of
algorithms, concepts,
principles, theories and
schema to find solutions
in complex or challenging
situations
Q9
Preliminary
grade
boundaries
39–24
Q3
•
application of algorithms,
principles, theories and
schema to find solutions in
simple situations
Q2, 8, 9
24–16
Q1, 5, 7, 9
16–13
13–9
8–1
Note: Preliminary grade boundaries are determined based on the school’s experience with similar assessment instruments and the relative number of A, B and C marks
available (see above).
All grade boundaries must be confirmed once they have been applied to student responses and compared to syllabus standards.
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Queensland Curriculum & Assessment Authority
July 2014
Page 2 of 22
Standard A
Standard B
Standard C
Standard D
Standard E
The student work has the
following characteristics:
The student work has the
following characteristics:
The student work has the
following characteristics:
The student work has the
following characteristics:
The student work has the
following characteristics:
•
•
•
•
•
Investigative
processes
systematic analysis of
secondary data to
identify relationships
between patterns and
trends
analysis of secondary
data to identify patterns
and trends
Q13, 14
Preliminary
grade
boundaries
Evaluating
and
concluding
Q13, 14
12–11
identification of obvious
patterns
recording of data
Q11, 12, 13, 14
10–9
8–7
The student work has the
following characteristics:
The student work has the
following characteristics:
The student work has the
following characteristics:
The student work has the
following characteristics:
The student work has the
following characteristics:
•
•
•
•
•
exploration of
scenarios and possible
outcomes with
justification of
conclusions
explanation of
scenarios and possible
outcomes with
discussion of
conclusions
Q15
Preliminary
grade
boundaries
analysis of secondary
data to identify obvious
patterns and trends
10–9
description of
scenarios and possible
outcomes with
statements of
conclusion
Q15
8–7
identification of
scenarios or possible
outcomes
statements about
outcomes
Q15
6–5
Note: Preliminary grade boundaries are based on the school’s experience with similar assessment instruments and on the relative number of marks available based on the
task-specific descriptors (drawn from the syllabus exit standards). All grade boundaries must be confirmed once they have been applied to student responses and matched
to syllabus standards.
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Queensland Curriculum & Assessment Authority
July 2014
Page 3 of 22
Indicative response — Using marks
Relevant exit
standard
descriptor
Question 1 (KCU C: 2 marks)
+
-6
Calculate the pH of the following solution [H ] = 1.0 x 10 M.
State whether it is acidic, basic or neutral.
Expected response
pH = log H 3 O +
[ ]
= log [1 × 10 ]
−6
application of
algorithms and
theories to find
solutions in
simple situations
Marking
scheme
Question 1 KCU:
2 marks C

=6
The solution is below 7 which is neutral, so it is slightly acidic. 
Question 2 (KCU B: 3 marks)
+
A solution has a pH of 12.68. Calculate the [H ] and [OH ].
Expected Response
Question 2 KCU:
pH = 12.68
[H ] = 10
[H ] = 10
reproduction of
a concept
+
− pH
+
−12.68
2B marks
[H ][OH ] = 1 ×10
2.089 × 10 × [OH ] = 1 × 10
1 × 10
[OH ] = 2.089
× 10
−13
−
−14
−
−14
1 mark for the
definition of pH

1 mark for the
calculation of
[H+]
−14
−
application of
algorithms and
theories to find
solutions in a
complex
situation
2C marks

= 2.089 × 10 −13
+
−13
= 0.0487
2 marks for the
calculation of
[OH-]

= 4.807 × 10 − 2 M
Question 3 (KCU C: 1 mark)
Which of the following lists includes all species present in a
-1
1.0 mol L solution of acetic acid CH 3 COOH ?
(a)
CH 3 COOH , H2 O
(b)
H 3 O + , CH3 COO −
CH 3 COOH , CH 3 COO − , H + , H 2 O, OH −
CH 3 COOH , CH 3 COO − , H + , H 2 O
(c)
(d)
1 mark for the
algorithm and
correct solution
1 mark for the
statement
regarding acidity
interpretation of
a complex
concept
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 5 of 22

Question 3 KCU:
B 1 mark
1 mark for the
correct response
Queensland Curriculum & Assessment Authority
July 2014
Relevant
exit
standard
descriptor
reproduction
of a concept
Question 4 (KCU C: 1 mark)
Which of the following will determine whether an acid will be strong or
weak?
(a)
its concentration
(b)
its ability to donate protons

(c)
its solubility in water
(d)
its pH
Question 5 (KCU C: 4 marks)
Based on the pH values shown in the figure, which of the following
statements about the concentration of hydrogen ions is correct?
Marking
scheme
Question 4
KCU:1
mark C
1 mark for
the correct
response
Question 5
KCU: 4
marks C
(a) It is twice as great in milk as it is in lemon juice.
(b) It is 1 000 000 times greater in soap than in wine.
(c) It is three times greater in wine than in bleach solution.
(d) It is 1 000 times greater in distilled water than in soap.
application
of algorithms
and theories
to find
solutions in
a simple
situation
Show reasoning that applies the concept of the pH scale to support
your answer:
Expected response
The pH scale is a logarithmic scale. The hydrogen ion concentration at
the positions indicated is:
+
-6
Milk [H ] = 10
+
-3
Lemon juice [H ] = 10
+
The [H ] in lemon juice is 1000 times greater than in milk, so the
statement is incorrect.
+
-10

Soap [H ] = 10
+
-4
Wine [H ] = 10
+
6
The [H ] in wine is 10 times greater than in soap, so the statement is
incorrect.
+
-12
Bleach solution [H ] = 10

+
-4
Wine [H ] = 10
+
8
The [H ] in wine is 10 times greater than that in soap, so the statement
is incorrect.
+
-7
Distilled water [H ] = 10

+
-10

Soap [H ] = 10
+
4
The [H ] in distilled water is 10 times greater than that in soap, so this
so the statement in correct.
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 6 of 22
1 mark for
each
calculation
and
statement
Queensland Curriculum & Assessment Authority
July 2014
Relevant exit
standard
descriptor
Question 6 (KCU B 3 marks)
Write TWO chemical equations to show that the dihydrogen
phosphate ion (H2PO4 ) is amphiprotic.
Compare the
equations to explain the term amphiprotic.
Expected response

A substance is amphiprotic if it acts like both an acid and a
base.
In the following equation a hydrogen ion is being donated,
showing that the dihydrogen phosphate ion is acting as an
acid.
H 2 PO 4 − ( aq ) → H + ( aq ) + HPO4
_ −
(aq)

In the following equation a hydrogen ion is being accepted,
showing that the dihydrogen phosphate ion is acting as a base.
H 2 PO 4 −
( aq )
+ H + ( aq ) → H 3 PO 4
(aq)

Marking
scheme
Question 6 KCU: 3
marks B
1 mark for
explanation of the
term amphiprotic
1 mark for writing
each equation and
explaining whether
the H+ ion id
donated or
accepted
comparison and
explanation of
concepts
Question 7 (KCU C: 3 marks)
-1
A titration is carried out in which 0.40 mol L potassium
hydroxide (KOH) solution in a burette is titrated against
10.00mL of hydrochloric acid (HCl) solution. The volume of
base required to reach the end point was 12.5mL.
Write a balanced equation for this reaction then calculate the
concentration of the hydrochloric acid solution. Show all
working.
Expected response
KOH ( aq ) + HCl ( aq ) → KCl ( aq ) + H 2 O( aq )

Question 7 KCU: 3
marks C
conc1 × vol1 = conc 2 × vol 2
0.4 × 0.0125 = conc 2 × 0.01

0.4 × 0.0125
conc 2 =
0.01
= 0.5M

application of
algorithms and
theories to find
solutions in a
simple situation
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 7 of 22
1 mark for writing
the correct
chemical equation
1 mark for
substitution into
the equation
1 mark for the
correct response
Queensland Curriculum & Assessment Authority
July 2014
Relevant exit
standard
descriptor
Question 8 (KCU A: 4 marks)
Marking scheme
Borax or sodium tetraborate decahydrate,
Na2B4O7.10H2O, is a base which can easily be obtained
in very pure form. (This salt dissociates to form anion,
cation and 10H2O) It reacts with strong acids such as
hydrochloric to form the extremely weak boric acid.
2-
B4O7
(aq)
+ 2H
+
(aq)
+ 5 H2O(l) → 4 H3BO3 (aq)
Question 8 KCU: 4
marks A
This is a multi-step
(i.e. complex) and
unfamiliar (i.e.
challenging)
situation.
Borax is frequently used as a primary standard in acidbase titrations.
In one particular experiment 7.738 g of borax was
dissolved in water and made accurately to 250mL.
25 mL of this solution required 27.65 mL of a solution of
hydrochloric acid for exact neutralization.
Calculate the concentrations of the borax and
hydrochloric acid solutions.
Expected response
Molecular weight of borax = (10.81 × 4 ) + (15.99 × 7 )
= 43.24 + 111.93
= 155.17
linking and
application of
concepts, principles,
theories and schema
to find solutions in a
complex and
challenging situation

7.738
155.17
= 0.0498 moles

number of moles
volume in litres
0.0498
=
0.25
= 0.19947 moles / L
= Conc acid × vol acid

Moles of borax =
1 mark for the
calculation of the
molecular weight
1 mark for the
calculation of the
number of moles of
borax
1 mark for the
calculation of the
concentration of
borax
1 mark for the
calculation of the
concentration of the
acid
Concentration of borax =
Conc borax × vol borax
0.19947 × 0.025 = Conc acid × 0.02765
Conc acid = 0.18035 mol / L
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 8 of 22

Queensland Curriculum & Assessment Authority
July 2014
Question 9 (KCU A: 10 marks)
Most swimming pools use 'chlorine' as the sanitizing agent to
kill bacteria and viruses. There are a variety of ways of
generating the 'chlorine' in pool water. The most widely used
method is the addition of hypochlorite compounds. The active
chemical produced in each method is HOCl or hypochlorous
acid, sometimes written as HClO.
Once in the water, an equilibrium is established between the
strong oxidant HOCl and the weaker OCl ion.
−
HOCl ( aq ) + H 2 O( aq ) ⇔ H 3 O( aq ) + OCl ( aq ) K a = 3.0 x 10 −8 Eqn 1
This equilibrium system is pH dependent and the following
graph shows how the concentrations of HOCl and OCl change
with the pH.
Marking
scheme
Question 9 KCU
10 marks A
This is a multi-step
(i.e. complex) and
unfamiliar (i.e.
challenging)
situation.
4 mark C
3 marks B
3 marks A
-
The level of 'free chlorine' (ie; HOCl + OCl ) present in a pool is
most commonly measured using a pool testing kit. This involves
adding a special chemical to a water sample and checking if the
colour produced shows that the amount of chlorine is in or
outside of the accepted range.
The amount of free chlorine can be determined more accurately
by carrying out a redox titration reaction on a sample of pool
water. The titration involves two simultaneous redox reactions
as shown below:
OCl − ( aq ) + 2 I − ( aq ) + 2 H + ( aq ) ⇒ Cl − ( aq ) + I 2( aq ) + H 2 O(l ) Eqn 2
2 S 2 O3 − − ( aq ) + I 2( aq ) ⇒ S 4 O6 − − ( aq ) + 2 I − ( aq )
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 9 of 22
Eqn 3
Queensland Curriculum & Assessment Authority
July 2014
Relevant
exit
standard
descriptor
2-
A solution of known concentration of thiosulphate ions (S2O3 ) is
reacted with a sample of pool water containing OCl ions and when
the reaction is complete the number of moles of OCl can be
calculated.
Refer to the information on the previous page in answering the
following questions:
A Year 12 student invites some of her friends over for a pool party
during schoolies week. The pool is regularly maintained by the
student’s parents and so the pH level was at 7.5 and the 'free
chlorine' at the recommended level of 2ppm (ie; [HOCl] + [OCl ] =
-5
3.84 x 10 M) before the students used the pool.
After the party was over the student tested the water and found that
the pH was 8.5 and the 'free chlorine' level was 0 – 0.5ppm.
The student also collected a 100mL sample of the pool water to
carry out a titration to check if the chlorine level was as low as
-6
indicated by the test kit. A volume of 38mL of 5.0 x 10 M
thiosulphate solution was found to react with the OCl ions in the
pool water sample.
(a) Only OCl is shown as reacting with thiosulphate ions in
equations (1) & (2) for the titration. Explain how titrating for the OCl
ion only, enables the amount of 'free chlorine' to be determined and
then show that the concentration of ‘free chlorine’ in the pool water
-6
is 9.5 x 10 mol/L.
(b) Show that the ratio of [HOCl]: [OCl ] agrees with what is
indicated in the graph when the pH is 8.5, and then determine the
concentration of both HOCl and OCl that are present.
Marking
scheme
Expected response
(a) The overall equation that occurs during the titration is
1 mark for the
combination of
OCl − ( aq ) + 2 H + ( aq ) + 2 S 2 O3 − − ( aq ) ⇒ Cl − ( aq ) + H 2 O(l ) + S 4 O6 − − ( aq ) Eqn 4 equation 2 and
3 into an
overall
2where OCl (aq) reacts with 2S2O3 (aq) in the ratio 1:2.

equation 4 and
As the OCl (aq) is used up in the reaction its concentration decreases stating the ratio
-
in the pool water sample. However, OCl (aq) is part of the equilibrium
system
HOCl ( aq ) + H 2 O( aq ) ⇔ H 3 O( aq ) + OCl − ( aq ) K a = 3.0 x 10 −8

-
reproduction
and
interpretation
of complex
and
challenging
concepts
Decreasing the concentration of OCl (aq) causes the system to
oppose this change and so the reaction moves to the right
producing more OCl (aq) but decreasing the concentration of
HOCl(aq). This continues until all of the HOCl(aq) is changed into OCl
(aq). Thus the titration enables the determination of the concentration
of OCl (aq), which will be equal to the combined concentrations of
OCl (aq) and HOCl(aq) that were present in the pool water i.e. the ‘free’
chlorine concentration.
1 mark for the
reference to
equation 1
1 mark for the
explanation of
the free
chlorine in the
system

Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 10 of 22
Queensland Curriculum & Assessment Authority
July 2014
Relevant exit
standard
descriptor
In the actual titration,
Volume of S 2O3
−−
Marking
scheme
= 38mL = 0.38 L
−−
and [ S 2O3 ] = 5.0 × 10 − 6 M
linking and
application of
algorithms and
concepts to find
solutions in a
complex and
challenging
situation
therefore, number of moles S 2O3
−−
1 mark for the
calculation of
the free
chlorine
= Conc x Vol
= 5.0 × 10 − 6 x 0.38
= 1.9 × 10 − 6 moles
From the equation for the titration
Number of moles S 2O3
−−

= 2 × number of moles OCl
−
and the free chlorine concentration at pH = 8.5 is 9.5 × 10 − 6 M
Now [ HOCl ] + [OCl − ] = 9.5 × 10 − 6
[ HOCl ] 1
= = 0.11 ( from the graph )
[OCl − ] 9
1 mark for
substituting in
equations

therefore [ HOCl ] = 0.11 × [OCl − ]
Substituting
0.11 [OCl − ] + [OCl − ] = 9.5 × 10 − 6
1.11 [OCl − ] = 9.5 × 10 − 6
9.5 × 10 − 6
therefore [OCl ] =
1.11
−6
= 8.6 × 10 M
−
And [ HOCl ] = 9.5 × 10
= 9.5 × 10
−6
−6
1 mark for the
calculation of
the [OCl-]

1 mark for the
calculation of
[HOCl]
−
− [OCl ]

− 8.6 × 10 − 6
= 9.0 × 10 − 7 M
number of moles S 2O3
Therefore, number of moles OCl =
2
−6
1.9 × 10
=
2
9
.
5
=
÷ 10 − 6
Volume of pool water = 100mL = 0.100mL
9.5 × 10 − 7 
Therefore, [OCl − ] =
0.100
−6
= 9.5 × 10 M
−−
−
linking and
application of
concepts,
principles,
theories and
schema to find
solutions in a
complex and
challenging
situation
(b) When pH = 8.5, the graph indicates that [ HOCl ] : [OCl − ] = 1 : 9
At pH = 8.5, [ H 3O + ] = 10 − 8.5 = 3.16 × 10 − 9 M
K a = 3.0 × 10 − 8 for HOCl
HOCl
( aq )
+ H 2O(l ) ⇔ H 3O
Therefore , K a =
Therefore,
+
( aq )
+ OCl

−
1 mark for the
calculation of
the free
chlorine
2 marks for the
steps in
showing that
the graph ratio
agrees with the
calculations
( aq )
[ H 3O + ] [OCl − ]
[ HOCl ]

[ HOCl ] [ H 3O + ] 3.16 × 10 − 9
=
=
=
0.11
[OCl ]
Ka
3.0 × 10 − 8
This is equivalent to 1 : 9.
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 11 of 22
Queensland Curriculum & Assessment Authority
July 2014
Relevant exit
standard
descriptor
Question 10 (KCU A: 5 marks)
In red wine production the grape skins are left in the crushing. The
skin contains compounds called anthocyanins which have
complex molecular shapes. They react with water as follows:
A+
( aq ) +
H 2 O(l ) ⇒ AOH ( aq ) + H + ( aq )
Equation 1
A + ( aq ) form is red , AOH form is colourless
Sulphur dioxide is added to the process to kill any bacteria and
wild yeast and SO2 enters an equilibrium with water as follows:
SO 2( g ) + H 2 O(l ) ⇔ H + ( aq ) + HSO3 − ( aq )
Marking
scheme
Question 10
KCU
5 marks A
Equation 2
−
The HSO3 ( aq ) also reacts with anthocyanins to form colourless
compounds.
A + ( aq ) + HSO3 − ( aq ) ⇔ ASO3 H ( aq )
Equation 3
reproduction
and
interpretation of
complex and
challenging
concepts,
theories and
principles
A + ( aq ) form is red , ASO3 H form is colourless
By analysing and evaluating the information given above, explain
how the addition of SO2(g) can alter the ‘taste’ of a wine, by
referring to pH changes that might occur.
Expected response

+
The red A (aq) form when added to water will eventually become
+
more acidic with the production of H (aq) ions. pH would
decrease. This may make the wine taste sour.
When the sulphur dioxide is added, the solution becomes even
+
+
more acidic with the production of H (aq) ions. As more H (aq)
ions are produced Equations 1 and 2 will move in the reverse
+
direction. Moving to the left will produce A (aq) which is red and
+
use up the H ions. Some SO2(g) may stay in solution.

+
−
Some A (aq) will react with HSO3 ( aq ) in Equation 3 causing
colourless ASO3H to be produced. Red colour will lessen. 
The amount of ASO3H produced will be small as only a certain
amount of SO2(g) would be added and once it is used up in the
+
forward direction, the eventual outcome will be more A (aq)

produced.
The pH would initially be lowered but would be raised as both the
+
concentration of H (aq) ions and HSO3 (aq) ions is lowered. The
overall outcome is that the solution becomes redder in colour and
less acidic.
1 mark for
reference to
Equation 1
producing H+
ions
1 mark for
reference to
Equation 2
1 mark for
reference to
the interaction
of both
equations
2 marks for
discussion of
the pH of the
solution

Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 12 of 22
Queensland Curriculum & Assessment Authority
July 2014
Relevant exit
standard
descriptor
analysis of
secondary data to
identify obvious
patterns and trends
The following table shows the colours of some indicators in
acidic, alkaline and neutral solutions.
Indicator
Colour in
acidic
solution
Colour in
alkaline
solution
Colour in
neutral
solution
RB
red
blue
purple
VR
violet
red
red
RY
red
yellow
orange
YB
yellow
blue
blue
CR
colourless
red
colourless
CB
colourless
blue
colourless
Question 12 (IP C: 1 mark)
Equal amounts of an unknown liquid are poured into each of
three test tubes. To the first test tube two drops of neutral
indicator CR are added and the resulting solution is found to
be colourless. To the second test tube two drops of neutral
indicator CB are added and the resulting solution is found to
be colourless. To the third test tube two drips of neutral
indicator YB are added and the resulting solution is found to
be blue. Which one to the following conclusions can be
justified?
analysis of
secondary data to
identify obvious
patterns and trends
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 13 of 22
Question 10
IP C 1 mark
1 mark for
the correct
response
Question 11 (IP C: 1 mark)
The leaves of certain cabbage plants are purple when boiled in
pure water, red when “pickled” in vinegar and blue or green
when boiled in water containing baking soda (an alkaline
solution). Which one of the following statements could explain
the observed effects?
The cabbage leaves contain
A.
RB
B.
VR
C.
RY

D.
YB
The unknown liquid is
A.
acidic
B.
alkaline
C.
neutral
D.
water
Marking
scheme

Question 11
IP C 1 mark
1 mark for
the correct
response
Question 12
IP C 1 mark
1 mark for
the correct
response
Queensland Curriculum & Assessment Authority
July 2014
Relevant exit
standard
descriptor
Question 13 (IP A: 5 marks)
Marking
scheme
The following figure shows the curves for weak acid-strong base
titrations for the titration of 50.0mL of 0.1.M HA of varying
concentrations with 0.100M NaOH. The pKa values for HA are (a) 1,
(b) 3, (c) 5, (d) 7, (e) 9, and (f) 11 respectively.
Question 13
IP A 5 marks
Systematically analyse the data to identify the relationships and/or
trends of the curves and relate this to the pKa values of the acids.
Relate acid (f) to another acid in your response.
Expected response
Line (a) has the lowest pH and pKa value of 1 and so is the
strongest acid. Similarly, line (f) is the weakest acid as the pH and
pKa value of 11 is highest.
systematic
analysis of
secondary data
to identify
relationships
between
patterns and
trends
All lines show a similar trend. As more NaOH is added the curve
trend is upward towards the equivalence point. This occurs at the
point where 50mL of NaOH has been added. The length of the
curve at the equivalence point differs depending on the strength of
the acid. As the acid becomes weaker this part of the line becomes
shorter because there is not so much change in pH at this point.
As the acid becomes weaker, the initial part of the curve takes
longer to show a change in pH. For example, there is difference in
the first part of the curve in going from acid (c) to acid (d). This
trend continues as the acids become weaker.
As the acids become weaker i.e. in going from (a) to (f) the curve
straightens out and the equivalence point is missing when pKa is 11
–11
or the Ka is 10 . This would not be seen by an indicator.
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 14 of 22
Queensland Curriculum & Assessment Authority
July 2014
Note: This shows an expected A standard response. Marks will also be awarded for alternative responses
using valid analytical approaches. Responses will be marked using the table below. It is assumed that any
response will contain information recorded in the table for the standard below it.
Item-specific marking scheme
Standard
Marks
Description of response
A
5
Systematic analysis of secondary data to identify relationship between
patterns, e.g.
• The equivalence point occurs at different pHs for the various strength
acids.
• The weaker the acid the higher the pH at the equivalence point.
• The equivalence point for the acid solution with pKa=11 is not detectable
and would not be seen by an indicator.
B
4
Analysing data to identify pattern, e.g.
• The length of the lines varies at the equivalence point i.e. at the point
when 50mL of NaOH is added.
• This is related to the strength of the acid.
C
3
Analysing data to identify an obvious pattern, e.g.
• Graphs have similar slopes.
• Equivalence point occurs when 50 mL of NaOH has been added..
D
2
Identifying obvious patterns, e.g.
• Graphs have the same general shape.
E
1
Recording data, e.g.
• Records the fact that each line represents acids with different pKa values.
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 15 of 22
Queensland Curriculum & Assessment Authority
July 2014
Question 14 (IP A: 5 marks)
Marking
scheme
The following diagrams and information relate to the titration of
phosphoric acid and sodium hydroxide.
Question 14
IP: 5 marks
This question
involves
several items
of data and
requires a
system by the
student to
analyse
Figure 1: Relative fractions against pH
3rd equivalence point
2nd equivalence point
1st equivalence point
Figure 2: Titration of 50mL of 0.1M H3PO4 with 0.1M NaOH
Phosphoric acid, H3PO4(aq), is a polyprotic acid where the
dissociation occurs stepwise as shown below.
H 3 PO 4 ( aq ) ⇔ H 2 PO 4 − ( aq ) + H + ( aq )
H 2 PO 4 − ( aq ) ⇔ HPO4 − − ( aq ) + H + ( aq )
HPO 4
−
( aq )
⇔ PO 4
−−−
( aq )
+ H + ( aq )
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 16 of 22
Equation
1
Equation 2
Equation 3
Queensland Curriculum & Assessment Authority
July 2014
Relevant exit
standard
descriptor
Marking
scheme
Figure 3: Acid base indicator colours
Systematically analyse the data to identify a combination of
indicators that could be used to find the first two equivalence
points. Discuss the possibility of using an indicator to find the
third equivalence point.
systematic
analysis of
secondary data to
identify
relationships
between patterns
and trends
This question also
provides
opportunities for
students to link
and apply
algorithms to find
solutions (KCU).
The school has
chosen not to
assess this, but
considered it part
of the analysis
required for the
demonstration of
the IP standards.
Expected response
The equivalence points occur at pHs equal to 4.67, 9.46 and
11.93. Methyl orange shows a distinct colour change from red to
yellow-orange at approximately 4.7. At pHs below 4.67 this
indicator would be red and changing to yellow orange at
approximately 4.7. Bromothymol blue would not be suitable
because it is already into its violet blue tones. Bromocresol green
is just at the start of any change phase. So, methyl orange would
be suitable to show the first equivalence point.
At the second equivalence point at pH =9.46 thymophthalein
would change from colourless to blue. Thymol blue would be in
its darkest colour phase. With phenolphthalein it would be hard to
distinguish any change in the red colour.
Methyl orange and thymophthalein blue could be combined to
show the equivalence points. The two indicators could be
combined as the thymophthalein is colourless at pH less than
9.5.
The third equivalence point does not show sufficient change in
pH to allow an indicator to be used (curve is almost horizontal).
So a third indicator is unnecessary and if used would not show a
colour change.
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 17 of 22
Queensland Curriculum & Assessment Authority
July 2014
Note: This shows an expected A standard response. Marks will also be awarded for alternative responses
using valid analytical approaches. Responses will be marked using the table below. It is assumed that any
response will contain information recorded in the table for the standard below it.
Item-specific marking scheme
Standard
Marks
Description of response
A
5
Systematic analysis of secondary data to identify relationship between patterns,
e.g.
• Correctly identifying the two indicators and the pH range.
• Identifies that only two indicators are required.
• Identifies the relationship of the pH and Pka at the half equivalence point.
• Identifies the relationship of the pH and Pka at the three equivalence points.
B
4
Analysing data to identify pattern, e.g.
• Correctly identifying the two indicators and the pH range.
• Identifies that at the half equivalence point the pH equals the pKa value.
C
3
Analysing data to identify an obvious pattern, e.g.
• Identifies that two indicators are needed but may not be correct.
• Identifies that at any half equivalence point half of the acid has dissociated.
D
2
Identifying obvious patterns, e.g.
• Comments about the shape of the curve.
• Identifies one indicator correctly.
E
1
Recording data, e.g.
• Records that the equivalence points occur at pHs on the graph.
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 18 of 22
Queensland Curriculum & Assessment Authority
July 2014
Relevant exit
standard
descriptor
Question 15 (EC A: 5 marks)
The legal limit of tin contamination in canned fruit is 250 ppm
(parts per million). A pawpaw canning plant in northern
Queensland encountered a sudden, unexpected tin
contamination problem. The reaction involving the
contamination has been suggested to be:
Marking
scheme
Question 15 EC
5 marks A
NO3 − ( aq ) + 4Sn ( s ) + 10 H + ( aq ) ⇒ NH 4 + ( aq ) + 4 Sn + + ( aq ) + 3H 2 O(l )
A comprehensive investigation was conducted on different
batches of pawpaw of varying acidity. Pawpaws vary in
acidity due to the conditions when picked. Nitrate
concentrations may vary due to the amount of nitrate in the
soil as the plant grows.
Nitrate ion concentrations in the investigation were 10ppm
and 40ppm. The results are summarised in the graphs
below:
Could the contamination problem be rectified by adjusting
the pH? Explore different possibilities and justify a
recommendation.
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 19 of 22
Queensland Curriculum & Assessment Authority
July 2014
Expected response
Nitrates are present in fruit due to the natural uptake by plants of nitrates from the soil.
All nitrate compounds are soluble. Nitrates in fertilisers are absorbed by plants and so
if a plant has been subjected to excessive fertilisers the concentration in the fruit will
be higher than normal. The nitrate ion is a strong oxidising agent and when present in
sufficient quantities will cause the corrosion of the tin can.
The equation suggests that any nitrate (NO3 -) present combines with the tin Sn(s) to
form tin ions Sn2+(aq) in acidic solution i.e. when H+ ions are present. Even after one
week of storage the cans at 40ppm nitrate are starting to show an increase in tin
contamination.
At the lower level of nitrates at 10 ppm, the limit of 250ppm of Sn2+(aq) is not reached
for any of the pH readings. At the lower nitrate concentration of 10ppm, the graph
shows there is an initial increase in Sn++ ions for pHs of 3.5, 3.8 and 4.2. The pH level
remains the same for pH of 4.8.
If the nitrate concentration is at 40ppm then the contamination occurs i.e. Sn2+(aq)
ions are formed. This means that there are sufficient nitrate ions and hydrogen ions to
cause the reaction to move to the right. At the higher level of nitrate at 40 ppm the
limit is reached only for pH levels at 3.8 or lower.
The equation shows that as the nitrate concentration is raised the reaction is moved
to the right producing more Sn2+(aq) ions. As the reaction moves to the right, more
H+(aq) ions are used up raising the pH. The pH during the time of testing was not
recorded.
Once the can is sealed, it is imposible to adjust the pH inside the can. The pH and the
nitrate level of the actual fruit needs to be taken into account before canning. Fruit of a
very low initial pH e.g.3.8 or lower should not be used or could have sugar added to
lessen the acidity. It is unclear whether or not the curves will continue to increase for
longer storage periods.It would seem unlikely as the curves seem to be levelling off
as the H+ ions are used up.
Further investigations could be carried out to see if the pH of the fruit changes over
the same or longer storage periods. The pH is that of the fruit when picked.
Relevant exit
standard
descriptor
exploration of
scenarios and
possible
recommendations
justification of
recommendations
Note: This shows an expected A standard response. Marks will also be awarded for alternative responses
using valid analytical approaches. Responses will be marked using the table below. It is assumed that any
response will contain information recorded in the table for the standard below it.
Item-specific marking scheme
Standard
A
Mark
5
Description of response
Exploration of the scenario and justification of a recommendation
• Explores the necessary conditions for the reaction to proceed.
•
Explores the two graphs.
B
4
Explanation of the scenario
• Explains the differences in the two sets of graphs.
C
3
Description of a scenario and statements made
• Describes the differences in the two sets of graphs.
D
2
Identification of a scenario
• Compares the two sets of graphs.
E
1
Statements about outcomes
• Statements about the graphs.
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Page 20 of 22
Queensland Curriculum & Assessment Authority
July 2014
Marking scheme
The Marking scheme on pages 21 and 22 shows the match of marks to standard descriptors.
Knowledge and conceptual understanding
A
Question
reproduction
and
interpretation
of complex and
challenging
concepts,
theories and
principles
comparison
and
explanation of
complex
concepts,
processes and
phenomena
B
linking and
application of
algorithms,
concepts,
principles,
theories and
schema to find
solutions in
complex and
challenging
situations
reproduction
and
interpretation
of complex or
challenging
concepts,
theories and
principles
comparison
and
explanation of
concepts,
processes and
phenomena
C
linking and
application of
algorithms,
concepts,
principles,
theories and
schema to find
solutions in
complex or
challenging
situations
reproduction of
concepts,
theories and
principles
explanation of
simple
processes and
phenomena
1
2
3
1
4
1
5
4
6
3
7
3
8
4
9
10
Totals
application of
algorithms,
principles,
theories and
schema to find
solutions in
simple
situations
2
2
10
Total
4
5
19
9
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
11
39
Queensland Curriculum & Assessment Authority
July 2014
Page 21 of 22
Investigative processes
Question
A
B
C
systematic analysis of secondary data to
identify relationships between patterns and
trends
analysis of secondary data to identify patterns and trends
analysis of secondary data to identify obvious
patterns and trends
11
1
12
1
13
5
14
5
IP totals
10
2
12
Evaluating and concluding
A
B
C
exploration of scenarios and possible
outcomes with justification of conclusions
explanation of scenarios and possible outcomes with
discussion of conclusions
description of scenarios and possible outcomes with
statements of conclusion
15
5
EC totals
5
5
Chemistry 2007
Annotated indicative response and judgments using marks within criteria
Queensland Curriculum & Assessment Authority
July 2014
Page 22 of 22