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1. NCEA Level 3 Chemistry (90700) 2011 — page 1 of 4
Assessment Schedule – 2011
Chemistry: Describe properties of aqueous systems (90700)
Evidence Statement
Question Evidence Achievement Merit Excellence
ONE NH3 weak base TWO of: THREE from part (a) correct.
(a) NaCl neutral
NH4Cl weak acid • THREE from part (a) correct. AND
HF weak acid
(b)(i) NH3 + H2O  NH4+ + OH– ONE explanation to Merit level.
Equilibrium is to the left, so the greatest concentration
of a species is NH3. For each NH3 that reacts equal AND
amounts of NH4+ and OH– are formed and are greater
than the OH– and H3O+ formed by the dissociation of
water. ONE full explanation. (all 4
• Correct equation AND correct
NH3 > OH– ≥ NH4+ > H3O+ • Correct equation. species).
order of species for BOTH
OR (b)(i) and (b)(ii).
(ii) HF + H2O  F– + H3O+ • Correct rank for b(i) or b(ii) OR
Equilibrium is to the left, so the greatest concentration • Correct equation, order of AND
of a species is HF. For each HF that reacts equal species AND full explanation
amounts of F– and H3O+ are formed and are greater than (all 4 species) for EITHER
the OH– and H3O+ formed by the dissociation of water. (b)(i) or (b)(ii). Correct answer with units, and
HF > H3O+ ≥ F– > OH– appropriate number of sig. fig.
EITHER
(c) Ka = 6.76 × 10–4
HF + H2O  H3O+ + F– • Ka correct.
AND
Assume [H3O+] = [F–] OR
• Correct [H3O+]. • Correct answer with minor
[H 3O + ]2 [H 3O + ]2 error (incorrect sig. fig. or
Ka = ! [HF] =
[HF] Ka units).
[H 3O + ] = 4.57 " 10#3 mol L#1
[HF] = 0.0309 mol L#1

2. NCEA Level 3 Chemistry (90700) 2011 — page 2 of 4
TWO Zn(OH)2(s)  Zn2+(aq) + 2OH–(aq) TWO of:
(a)(i)
• Part (a) correct.
(ii) Ks = [Zn2+][OH–]2
(b) Let s be solubility Solubility calculated correctly, Solubility calculated correctly,
Ks = 4s 3 (incorrect sig. fig.). 3 sig. fig. and s is defined.
3 • Method correct, but error in
Ks calculation. (Allow s2 follow on
s= from part (a) or 2s3error but if so, AND AND
4
must have calculated s value
s = 1.96 ! 10 –6 mol L"1 correctly according to the candidates ONE of:
follow on.) • Recognises that a complex ion Complex ion forms, precipitate
(c) Raising the pH will increase the concentration of OH– will form and links this to re-dissolves, as equilibrium
ions. either less solid remaining or shifts in the forwards direction /
This will initially cause additional precipitate to form. equilibrium shifting to the to RHS. This shift to the right
• Recognises that [OH–] has increased.
Once the pH has been increased sufficiently (enough right. will occur so more Zn2+ and OH–
OH- has been added) the formation of a complex ion • Identifies equilibrium shifting will dissolve into solution so that
with Zn2+ will occur, lowering OH– ion concentration in • Recognises equilibrium will shift to to the left due to additional the solution becomes saturated
solution. the left. OH–. again.
Thus the precipitate will redissolve as a complex ion • Explains equilibrium shifting to
and less precipitate will be at the bottom of the test tube. the left in terms of the I.P. now
exceeding Ks.

3. NCEA Level 3 Chemistry (90700) 2011 — page 3 of 4
THREE TWO of:
(a) HG + H2O  G– + H3O+
OR • Part (a) and (b) correct.
HOCH2COOH + H2O  HOCH2COO – + H3O+
(b) [G ! ][H 3O + ]
Ka = (must have equilibrium arrow)
[HG] • EITHER
Correct value for Ka
(c) Correct answer with minor error. Correct answer with
[H 3O + ] = Ka ! [HG] OR appropriate number of sig. fig.
Ka = 1.50 ! 10"4 Correct rearrangement of Ka
expression to make [H30+] subject.
[H 3O + ] = 9.99 ! 10"3 mol L"1 AND
pH = 2.00
AND
(d) [H 3O + ] = 1.00 ! 10"4 mol L"1 Correct [G–].
Ka ! [HG] • EITHER
[G " ] = +
= 1.48 mol L"1 OR
[H 3O ] Correct [H3O+].
Thus in 200 mL = 0.2 ! 1.48 = 0.296 mol OR Correct method for [G–] and n(G–)
Alternative method Ka expression rearranged for [G–] or calculation but incorrect answer. Correct n(G–) to 3 sig. fig.
other appropriate method for [G–]
[weak base]
pH = pK a + log10 stated and rearranged for [G–].
[weak acid]
[base]
4.00 = 3.83+ log10
[acid]
log10 [base] = 0.17
[base] = 1.48 mol L–1
Thus, in 200 mL = 00.2 ! 1.48 = 0.296 mol

4. NCEA Level 3 Chemistry (90700) 2011 — page 4 of 4
FOUR A ONE of:
(a) At point A, there is an equi-molar mixture of HEt and • Recognises that at point A there is a
Et–. On addition of OH– ions, the acid part of the buffer buffer solution. Describes how a buffer works Shows recognition of equimolar
neutralises the OH– ions, by donating a proton. The acid (for when both acid AND base HEt and Et– thus pKa = pH
reacts with the base: • States that equimolar amounts of are added) by: and discusses how the buffer
HEt + OH– → Et– + H2O acid / base conjugate are present at EITHER solution works and links to
A. • Giving equations for the equations.
On addition of H3O+, the ethanoate will accept a proton specific buffer
from the hydronium ion: OR
• States that pH will not change when
Et– + H3O+ → HEt + H2O small amounts of acid or base are • Writing about how a buffer
added. works in general terms
Candidate may discuss equilibrium shift.
pKa = pH = 4.76 (accept 4.5 – 4.9) OR
• Correct pKa / Ka • Links that due to equimolar
(b) B HEt and Et– thus pKa = pH
At the equivalence point all the HEt has been AND
neutralised by NaOH. AND AND
ONE of:
HEt + NaOH → EtNa + H2O • Recognises that all the HEt has been • Recognises that none of the
used up at B. original HEt remains as ithas
The Et– reacts further to a small extent with water. all reacted with NaOH
• That the pH of equivalence point is OR
Et– + H2O  HEt + OH– greater than 7. (must have clearly Uses two equations to explain
• That the pH of equivalence why the pH is above 7. (One
indicated that point B is the
point is greater than 7 with a equation may be implied in the
Thus the pH of the equivalence point is above 7 due to equivalence point)
valid reason. candidate’s written answer.)
presence of OH–.
Judgement Statement
Achievement Achievement with Merit Achievement with Excellence
3A 2M+1A 2E+1A