NCEA Level 3 Chemistry 91392 (3.6) — page 1 of 6
SAMPLE ASSESSMENT SCHEDULE
Chemistry 91392 (3.6): Demonstrate understanding of equilibrium principles in aqueous systems
Assessment Criteria
Achievement
Achievement with Merit
Demonstrate understanding typically involves
describing, identifying, and giving an account of
aqueous systems using equilibrium principles. This
requires the use of chemistry vocabulary, symbols and
conventions and may include related calculations.
Demonstrate in-depth understanding typically involves
using equilibrium principles to explain properties of
aqueous systems. This requires explanations that use
chemistry vocabulary, symbols and conventions and
may include related calculations.
Achievement with Excellence
Demonstrate comprehensive understanding typically
involves elaborating, justifying, relating, evaluating,
comparing and contrasting, or analysing properties
of aqueous systems in terms of equilibrium
principles. This requires the consistent use of
chemistry vocabulary, symbols, and conventions and
may include related calculations.
Evidence Statement
One
Expected coverage
Achievement
In (a) (i), equation correct
(a) (i)
HCOOH + H2O
« HCOO– + H3O+
OR
in (a) (ii), THREE species
correct.
Merit
In (a) (ii), reasons for
species concentrations
given.
HCOOH > H3O+  HCOO– > OH–
(a) (ii)
(a) (iii)
Methanoic acid is a weak acid so the equilibrium favours HCOOH.
Dissociation produces similar amounts of H3O+ and HCOO–. Dissociation
of HCOO– produces a small amount of OH–/ water dissociates to form a
small amount of H3O+ and OH- in equal amounts and after the
dissociation of HCOOH there will be slightly more H3O+ than HCOO-.
HCOOH + H2O
« HCOO– + H3O+
In (a) (iii), correct
calculation of [H3O+].
In (a) (iii), correct
calculation.
Excellence
NCEA Level 3 Chemistry 91392 (3.6) — page 2 of 6
éë HCOO – ùû éë H 3O+ ùû
Ka =
[ HCOOH ]
(10 )
–2.78 2
10
–3.74
=
[ HCOOH ]
[HCOOH] = 0.0151 mol L–1
(b)
HCl is a strong acid and completely dissociates into its ions. .It has a low
pH due to a high [H3O+]..
In (b), pH is due to H3O+
(or OH–) in solution
NH4Cl is an acidic salt that completely dissociates into its ions producing
NH4+. NH4+ is a weak acid so partially dissociates into H3O+, however,
although acidic its pH is not as low as HCl. The concentration of [H3O+] is
lower.
OR
in (b), conductivity is due
to ions in solution.
NH3 is a weak base, its reaction with water produces only a limited
amount of OH– compared to NaOH which is a strong base and fully
dissociates to produce the highest concentration of OH– and hence the
highest pH.
In (b), difference in pH of
two substances explained
in terms of H3O+ (or OH–)
OR
in (b), difference in
conductivity of two
substances explained in
terms of ions in solution.
Conductivity relates to the number of ions in solution.
Since HCl, NH4Cl and NaOH completely dissociate a large number of
ions are produced, hence high conductivity. However since NH 3 partially
dissociates there are fewer ions in solution hence low conductivity.
Not Achieved
NØ
No response; no relevant evidence.
N1
Candidate provides some accurate statements without answering any question completely.
N2
Candidate provides any ONE statement for Achievement.
A3
Candidate provides any TWO statements for Achievement.
A4
Candidate provides any THREE statements for Achievement.
Achievement
In (b), properties of
THREE substances
justified in terms of pH
related to H3O+ (or OH–)
AND conductivity related
to ions in solution of three
substances.
NCEA Level 3 Chemistry 91392 (3.6) — page 3 of 6
M5
Candidate provides any ONE statement for Merit.
M6
Candidate provides any TWO statements for Merit.
E7
Either pH or conductivity justified from the Excellence criteria.
E8
Candidate provides ALL the evidence from the Excellence criteria.
Merit
Excellence
Two
Expected coverage
FeS
(a) (ii)
Ks = [Fe2+][S2-]
Let solubility = s
(a) (iii)
Ks =
s2
4.9
´
Merit
In (a) (i) AND (ii), BOTH
correct.
« Fe2+ + S2-
(a) (i)
Achievement
In (a) (iii), correct answer.
10-18 = s2
s = 4.9 ´ 10 –18
= 2.21 ´ 10-9 mol L–1
[H3O+]2[S2–] = 1.1
[H3O+]
=
[H3O+]
(b) (i)
10-4.20
= 6.31
éëS2– ùû =
= 2.76
´
´
10-23
when pH = 4.20
´
(b) (ii)
´
In (b) (i), sulfide
concentration calculated.
In (b) (ii), common ion
used to calculate [Fe2+]
(error made).
In (b) (ii), [Fe2+] calculated
correctly.
10-5 mol L–1
1.10 ´ 10 –23
(10 )
–4.20 2
10-15 mol L–1
Ks = [Fe2+][S2-]
4.90
In (b) (i), [H3O+]
calculated correctly
10–18 = [Fe2+]
[Fe2+] = 1.78
´
´
2.76
´
10–15
10-3 mol L-1
i.e., solubility of FeS is 1.78
4.20.
´
10-3 mol L-1 when the pH of the solution is
Excellence
NCEA Level 3 Chemistry 91392 (3.6) — page 4 of 6
When hydrogen sulfide is bubbled through a solution containing Cu2+ and
Zn2+ the following equilibrium is established.
H2S
(c)
« 2H+ + S2-
In (c), states the
relationship between IP
and Ks for precipitation to
occur.
As the pH of a saturated solution (by the addition of HCl) decreases, the
equilibrium shifts in the reverse direction reducing [S2-].
In (c) links pH decrease
to decrease in [S2–] using
equilibrium principles.
In (c), links solubility to a
comparison of ionic
product and solubility
product for ZnS OR CuS.
For precipitation to occur IP>Ks. Hence only the metal sulfide with the
lowest Ks will precipitate. In this case CuS as IP(CuS)> Ks(CuS). The ZnS
will not precipitate. As the pH then decreases [S2–] will increase. This
enables metal sulfides with larger values of Ks in this case ZnS, to
precipitate as well as the CuS.
Not Achieved
NØ
No response; no relevant evidence.
N1
Candidate provides some accurate statements without answering any question completely.
N2
Candidate provides any ONE statement for Achievement.
A3
Candidate provides any TWO statements for Achievement.
A4
Candidate provides any THREE statements for Achievement.
M5
Candidate provides any TWO statements for Merit.
M6
Candidate provides any THREE statements for Merit.
E7
Minor errors (eg omission or inaccuracies) from the Excellence criteria.
E8
Candidate provides ALL the evidence from the Excellence criteria.
Achievement
Merit
Excellence
In (c), links pH decrease
to decrease in [S2–] using
equilibrium principles,
AND links solubility to a
comparison of ionic
product and solubility
product for BOTH ZnS
and CuS.
NCEA Level 3 Chemistry 91392 (3.6) — page 5 of 6
Three
(a) (i)
Expected coverage
In this region CH3COOH and CH3COO– are both present. CH3COOH /
CH3COO– is a conjugate acid/base pair. When base is added to the
buffer system it will react with CH3COOH, thus maintain the pH by
removing OH–.
Achievement
In (a) (i), recognises acid /
base pair.
Merit
Excellence
In (a) (i), links buffer to
addition of base and
includes equation
OR
CH3COOH + OH–  CH3COO– + H2O
X put on the graph at 10.00 mL
In (a) (ii), X at 10mL with
attempt at reason.
in (a) (ii), links X at 10 mL
to reason for most
efficient buffer action.
In b (i), [CH3COONa]
correct.
In (b) (i), method correct
using incorrect
[CH3COONa] (leading to
error in pH 8.92).
(a) (ii)
The most efficient buffering occurs when the pH of the solution is equal to
pKa ie [HA]=[A–].
Salt formed at equivalence point is CH3COONa.
[CH3COONa] = 0.125/2
= 6.25 ´ 10-2 mol L-1.
Ka = 10-4.76 = 1.74
´
10-5
CH3COOH + H2O  CH3COO– + H3O+
[ H3O+] =
(b) (i)
=
K a ´ Kw
–
[CH COO ]
3
1.74 ´10–5 ´1´10–14
6.26 ´10–2
= 1.67
´
pH = 8.78
10–9
In (b) (i), pH shown to be
8.78 using correct value
of salt concentration.
NCEA Level 3 Chemistry 91392 (3.6) — page 6 of 6
(b) (ii)
Phenolphthalein is a suitable indicator as its pKa is within 1 pH of
equivalence point. Hence it will change colour at the equivalence point of
the reaction in the steepest part of the graph.
Methyl orange will change colour in the buffer region as it’s between pH
2.7 and 4.7 which is in the buffer region making this indicator unsuitable.
Phenolphthalein is a weak acid and dissociates in water
HIn + H2O
(b) (iii)
« H3O+ + In–
In (b) (ii), limited reason
given for use of
phenolphthalein or nonuse of methyl orange.
In (b) (ii), both indicators
compared linked to
reasons
In (b) (iii), pH range of
indicator given.
In (b) (iii), link made
between addition of acid
or base to colour seen
and pH range.
When a base is added to the solution the equilibrium shifts in the forward
direction. Therefore In- is purple.
OR
When acid is added the equilibrium shifts in the reverse direction,
therefore HIn is colourless. Indicators are effective in the range pH = ± 1
pKa, ie between 8.60 and 10.6.
Not Achieved
NØ
No response; no relevant evidence.
N1
Candidate provides some accurate statements without answering any question completely.
N2
Candidate provides any ONE statement for Achievement.
A3
Candidate provides any TWO statements for Achievement.
A4
Candidate provides any THREE statements for Achievement.
M5
Candidate provides any TWO statements for Merit.
M6
Candidate provides any THREE statements for Merit.
E7
Candidate provides any ONE statement for Excellence.
E8
Candidate provides BOTH statements for Excellence.
Achievement
Merit
Excellence
In (b) (iii), equilibrium
principles used to explain
addition of acid and of
base linked to colour
change and correct
species over pH range.