MS Acids Bases Buffers ALL PPQ
MS Acids Bases Buffers ALL PPQ
MS Acids Bases Buffers ALL PPQ
com
Questi Mark
Answer/Indicative content Guidance
on s
ALLOW ECF
FIRST CHECK THE ANSWER ON ANSWER LINE
If answer = 2.4 × 10–2 (mol dm–3) award 4 marks
ALLOW [HA] and [A–] in working
-----------------------------------------------------------------
4
(AO
3.3×3
Calculation of H+ in buffer
)
[H+] buffer = 10–4.00 OR 1 × 10–4 (mol dm–3) ✓
Hasselbalch equation (ALLOW –logKa for pKa) e.g. No volumes used = 3.6 x 10-2 2 marks
Total 8
ALLOW → for ⇌
(AO
3.2)
Examiner’s Comments
5.1.3 Acids, Bases and Buffers PhysicsAndMathsTutor.com
Total 4
cm3 ) ✓ (mol)
and
n(NaOH) = 0.02(00) x 0.1 = 0.002(00) (mol)
Examiner’s Comments
n(OH–)excess = 0.100 ×
FIRST CHECK THE ANSWER ON ANSWER LINE
If answer = 12.55 award 4 marks
Common errors
------------------------------------------
If initial V(NaOH) = 45 cm3
Excess mol of NaOH:
[OH–] = 0.0643 (mol)
[H+] = 1.56 × 10–13 (mol dm–3)
n(OH–)excess = n(OH–) – n(C2H5COOH)
pH = 12.81 award three marks (no 1st mark)
4
If n(OH–)excess is used in [H+] calculation
= (0.100 × – (0.0800 ×
n(OH–)excess = 0.0025 (mol)
Examiner’s Comments
AND 1
i Examiner’s Comments
pH range matches vertical section/rapid pH change (AO
v
OR 3.3)
Some candidates realised that the most
end point/colour change matches vertical section/rapid pH
suitable indicator for a weak acid / strong
change ✓
alkali titration would be cresol purple. For
incorrect responses, other indicators
appeared to be selected at random,
suggesting that candidates were unclear on
the criteria for selected a suitable indicator.
ALLOW use of HA
Ignore [HIO3]equilibrium < [HIO3 ]initial/undissociated
ALLOW
[HIO3]equilibrium ~ [HIO3]undissociated is no longer
a valid assumption
ALLOW
HIO3 dissociation is not negligible / dissociates to a significant [HIO3] has a larger Ka so the assumption
extent that [HIO3] at equilibrium = [HIO3] initially so
1
OR assumption is not valid
c (AO
Large Ka and HIO3 is ‘stronger’ (weak) acid
3.3)
OR Examiner’s Comments
[HIO3 ]eqm is significantly lower than [HIO3 ]initial/undissociated ✓
Very few candidates scored the mark for this
question.
Total 14
Action of buffer
Level 2 (3–4 marks)
Detailed explanation of equilibrium and the action of the buffer.
• Increase in H+ / addition of acid
OR
leads to:
Detailed explanation of equilibrium and correct calculation of
H+(aq) + HCO3–(aq) → H2CO3(aq)
[HCO3–] : [H2CO3] ratio.
OR HCO3– reacts with added acid
OR
• Increase in OH– / addition of alkali
Detailed explanation of the action of the buffer and correct
leads to:
calculation of [HCO3–] : [H2CO3] ratio. 6
H+(aq) + OH–(aq) → H2O(l)
(AO1.
OR
OR 1 ×2)
H2CO3(aq) + OH–(aq) → HCO3–
Partial explanations of equilibrium, and the action of the buffer (AO1.
(aq) + H2O(l)
4 and attempt calculation of [HCO3–] : [H2CO3] ratio. 2 ×2)
OR
(AO3.
H2CO3 reacts with added alkali
There is a line of reasoning presented with some structure. 1 ×1)
The information presented is relevant and supported by some (AO3.
Calculation of [HCO3–] : [H2CO3] ratio
evidence. 2 ×1)
0 marks ALLOW OR
No response or no response worthy of credit.
Examiner’s Comments
This Level of Response question was
generally well answered with many
candidates achieving maximum marks by
simply considering what was required in the
question.
Exemplar 5
Total 6
Examiner’s Comments
Total 1
ALLOW
CO32– + 2H2O → 2OH– + H2CO3
IGNORE
Na2CO3 + H2O → 2NaOH + CO2
Ionic equation required
Examiner’s Comments
Examiner’s Comments
Acid/H+/HCl reacts with OR protonates
Calculator = 0.03851851852
Calculator = 0.01303278689
------------------------------------------------
OR 0.0385….. (mol) ✓
----------------------------------------------
Common errors
Actual moles AO2. 35.2% → 2 marks
8×1
d
n(C6H5COOH) OR 0.013(0)…. AO2.
•
8×1
(mol) ✓
(3 sig fig) ✓
Examiner’s Comments
IGNORE
• Initial filtering
• hot filtration to remove insoluble
impurities
Examiner’s Comments
Dissolve in the minimum quantity of hot water/solvent ✓
Many candidates produced thorough
responses, showing that they had
Cool 2 encountered recrystallisation as a technique
e AND AO in their practical work.
Filter 3.3 x2
AND Most candidates were aware that the impure
(leave to) dry ✓ product is dissolved in a minimum volume of
All three needed hot solvent, although ‘minimum’ was
sometimes omitted.
Total 8
ALLOW 3-hydroxybutan-1-al
DO NOT ALLOW
• 3-hydroxybutal
• 3-hydroxylbutanal
7 i 3-hydroxybutanal ✓ 1
Examiner’s Comments
Exemplar 15
For connectivity,
ALLO
W
i
1
v (Connectivity not being assessed)
Examiner’s Comments
Total 6
8 a i 1
Examiner’s Comments
Examiner’s Comments
= 1.74 × 10−5 (mol dm−3) ✓
pKa
This three-step calculation was successfully
= −log Ka = −log 1.74 × 10−5 = 4.76 ✓
completed by almost all candidates.
3 SF required
Examiner’s Comments
Examiner’s Comments
Total 9
..............................................................
[H+] m= 10−pH = 10−2.440 = 3.63 × 10−3 ALLOW use of HA and A−
(mol dm−3) ✓ ALLOW 3 SF up to calculator value of
3.630780548 × 10−3 correctly rounded
Examiner’s Comments
Most candidates coped with this commonly
seen type of calculation and were able to
correctly calculate the concentration of the
weak acid
FCH2COO− + CH3COOH2+
Mark by ECF
.....................................................
[CH3COO-]
.........
Alternative method
(If both methods are attempted, mark the
method which produces the higher mark)
[H+]
[H+] = 10−pH = 10−4.50
= 3.16 × 10−5 (mol dm−3) ✓
[CH3COO−]
mass of CH3COONa
c i 5
(must come from calculated [H+])
pH
pH = −log (3.16 × 10−5) = 4.50 ✓
..............................................................
LAST 3 marks are NOT available using .....................................................
.........
• Ka square root approach (weak acid pH)
• Kw/10−14 approach (strong base pH)
Common errors
Examiner’s Comments
This question caused difficulty for all but the
more able. For many weaker candidates
getting beyond a concentration of
CH3COONa was a problem. Once again,
candidates should be advised to show every
step in their calculation. This would allow
method marks to be applied in the absence
of a correct final answer.
M2 is dependent upon M1
Examiner’s Comments
5.1.3 Acids, Bases and Buffers PhysicsAndMathsTutor.com
Total 12
Examiner’s Comment:
1 This part was very well answered.
i 1
0 Square brackets required Candidates responded with either near
molecular formulae, such as C4H9SH,
structural formulae or with skeletal formulae.
Some candidates made careless errors
such as omitting the negative charge or
showing [H+]2 as numerator rather than
[C4H9S–] [H+].
ALLOW C4H9SH
ALLOW CH3COOH
ii 2
Examiner’s Comment:
In this part, candidates were expected to
apply their knowledge and understanding of
Structure of thioester ✓ esterification to thiols and thioesters. Over
half the candidates obtained a correct
Complete equation ✓ structure of the thioester. Most of these
candidates constructed a balanced equation
although some omitted the water product.
Common errors included formation of a
conventional ester and H2S, and retaining
the O atom from the OH in the carboxyl
group to form –COOS–. As with 4(b)(i),
structural and skeletal formulae were used.
Candidates are less likely to omit H atoms if
the skeletal formula is used.
Examiner’s Comment:
Just over half the candidates drew the
correct structure, displaying a good
understanding of interpreting organic
nomenclature when drawing a structure.
Examiner’s Comment:
In this part, candidates were expected to
i apply their knowledge and understanding of
2
v Reactants ✓ condensation to an entirely new context.
One mark was allocated for the reactants
Products AND balanced equation ✓ and this was usually scored. The second
mark for the novel cyclic compound and
water was much more difficult, aimed at
stretch and challenge. A significant number
of candidates interpreted the information to
obtain a correct cyclic structure but this
mark was the domain of the most able
candidates.
Total 6
Observation: fizzing ✓
Total 5
1 ALL species MUST have square brackets State symbols not IGNORE state symbols, even if wrong
a i 1
2 required
TAKE CARE that ‘H’ is different on top and bottom of
IGNORE
expression
ii [vitamin C]
4
i
= 0.0341 (mol dm−3) ✓
= ✓
mass of C6H7O6Na =
Total 12
1 Examiner's Comments
a 1
3
IGNORE state symbols Almost all candidates successfully wrote the
expression for Ka. Responses using
[H+(aq)]2 were not credited. Rarely, the
expression was shown inverted or square
brackets were omitted from one or more of
the terms. For most candidates, this was an
easy mark.
Examiner's Comments
Calculation of [H+]
[H+] = 3.752 × 10−4 (mol dm−3) ✔ Mark by ECF from incorrect [HNO2] and
[NO2−]
pH to 2 DP (From 3.42573717) ONLY award marks for a pH calculation via
c i 4
pH = −log 3.752 × 10−4 = 3.43 ✔ Ka AND using concentrations / mol derived
from the question
NO marks are available using
Ka square root approach (weak acid pH) DO NOT ALLOW final pH mark by ECF if
Kw/10−14 approach (strong base pH) pH > 7
............................................... ...............................................
ALLOW alternative approach based on Henderson– COMMON ERRORS BUT CHECK
Hasselbalch equation (ALLOW −logKa for pKa) WORKING
pH = 2.82 3 marks
initial concs: 0.200 and 0.0625
pH = 3.23 3 marks
0.0400 and 0.0500 acid / base ratio inverted
pH = 3.83 2 marks
pH = 3.329 + 0.097 = 3.43 ✔
initial concs: 0.200 and 0.0625 and ratio
inverted
pH = 2.73 3 marks
Incorrect [NO2−] = 0.01 and correct [HNO2] =
5.1.3 Acids, Bases and Buffers PhysicsAndMathsTutor.com
0.04
pH = 4.03 3 marks
correct [NO2−] = 0.05 and incorrect [HNO2] =
0.01
Examiner's Comments
Control of pH: 2 marks (QWC) IGNORE just acid reacts with added alkali
Added HCl
NO2− reacts with added acid / HCl / H+ IGNORE just conjugate base / salt / base
OR NO2− + H+ → reacts with added acid
ii 4
OR more HNO2 forms ✔ DO NOT ALLOW salt / base reacts with
added acid
Added NaOH
HNO2 reacts with added alkali / NaOH / OH−
OR HNO2 + OH− →
OR more NO2− forms
OR H+ reacts with added alkali / NaOH AWARD ‘shift mark’ ONLY if correct
OR H+ + OH− → ✔ equilibrium equation has been given
IGNORE any other equilibria in response
Equilibrium shift:
1 mark for shifts in HNO2 ⇌ H+ + NO2− (See 1st mark) Examiner's Comments
Equilibrium for added acid → left
AND Equilibrium for added alkali → right ✔ (QWC) The role of buffers in controlling pH is a
common question and most candidates had
prepared their rehearsed answers.
Consequently these candidates could obtain
the four marks easily. As always, candidates
who had not learnt the work produced
5.1.3 Acids, Bases and Buffers PhysicsAndMathsTutor.com
Examiner's Comments
ii
pKw = 13.03 ✔ 1
i
Despite the novel context, almost all
candidates obtained the correct pKw value of
13.03.
FIRST, CHECK THE ANSWER ON ANSWER LINE FULL ANNOTATIONS MUST BE USED
i
IF answer = 10.76, award 3 marks 3 ...............................................
v
...............................................
5.1.3 Acids, Bases and Buffers PhysicsAndMathsTutor.com
Examiner's Comments
Total 18
around.
ALLOW ‘just acid’ and ‘base’ labels
throughout if linked by lines so that it is
clear what the acid-base pairs are.
Examiner's Comments
Total 2
Examiner's Comments
Proton / H+ donor
1
a AND 1
5 For most candidates, this was an easy
Partially dissociates / ionises ✔
mark, although some only responded for a
weak acid (partial dissociation) or for a
Brønsted–Lowry acid (proton donor).
Examiner's Comments
Answer: pH = 13.70
Examiner's Comments
ii 3
NOTE: The final two marks are ONLY available from ALLOW use of quadratic equation which
attempted use of Ka AND [C2H5COOH] gives same answer of 2.90 from 0.120 mol
dm−3
5.1.3 Acids, Bases and Buffers PhysicsAndMathsTutor.com
....................................................
.......
pH = 5.79 2 marks
−log(1.35 × 10−5 × 0.120) No √
pH = 5.19 1 mark
−log (1.35 × 10−5 × 0.480) Original conc,
no √
pH = 4.87 0 marks
−log(1.35 × 10−5) = 4.87 −log Ka
Examiner's Comments
Answer: pH = 2.90
Examiner's Comments
d i pH = −log 1.35 × 10−5 = 4.87 ✔ 1
Answer: pH = 4.87
Added ammonia
C2H5COOH removes added NH3 / alkali / base
ALLOW use of HA / weak acid / acid for
OR C2H5COOH + NH3 / OH− →
ii C2H5COOH;
OR NH3 / alkali reacts with / accepts H+
OR H+ + NH3 →
OR H+ + OH− → ✔
ALLOW use of NH4OH for NH3
Examiner's Comments
i.e.
For moles / concentration 1 mark (1 mark
pH = −log 4.5 × 10−6 = 5.35 2 dp required ✔ lost)
5.1.3 Acids, Bases and Buffers PhysicsAndMathsTutor.com
Examiner's Comments
Answer: pH = 5.35
Total 14
Examiner's Comments
Examiner's Comments
FIRST, CHECK THE ANSWER ON ANSWER LINE IF there is an alternative answer, check to
see if there is any
IF answer = 1.15 × 10−11, award 2 marks ECF credit possible using working below.
ii 2
.......................................... ..........................................
5.1.3 Acids, Bases and Buffers PhysicsAndMathsTutor.com
[H+] = 10−3.06 = 8.71 × 10−4 (mol dm−3) ✔ ALLOW 2 SF: 8.7 × 10−4 up to calculator
value of
8.7096359 × 10−4 correctly rounded
✔
Examiner's Comments
ALLOW Ca(CH3COO)2
ALLOW (CH3COO−)2Ca2+
BUT DO NOT ALLOW if either charge is
missing or incorrect
Examiner's Comments
c i 1
2CH3COOH + CaCO3 → (CH3COO)2Ca + CO2 + H2O ✔ The equations seen were certainly better
than in previous sessions, perhaps as
candidates will have practised similar
questions from past papers. Ionic signs
within the formula of calcium ethanoate
were allowed but both were then needed.
Common errors included an incorrect
formula of calcium ethanoate with one
ethanoate group only and an unbalanced
ethanoic acid on the left-hand side of the
equation.
Examiner's Comments
..........................................
IF more than one equilibrium shown, it must
be clear which one is being referred to by
labeling the equilibria.
ii
CH3COOH reacts with added alkali 5
i
OR CH3COOH + OH− →
ALLOW weak acid reacts with added alkali
OR added alkali reacts with H+
DO NOT ALLOW acid reacts with added
OR H+ + OH− → ✔
alkali
Examiner's Comments
cm3 ✔
= 11.48 g ✔
5.1.3 Acids, Bases and Buffers PhysicsAndMathsTutor.com
..........................................
COMMON ECF 4.592 OR 4.6 g AWARD 4
marks
use of 400 / 1000 twice
= 0.243
= 100.243 = 1.75 ✔
Examiner's Comments
Answer: 11.48 g
5.1.3 Acids, Bases and Buffers PhysicsAndMathsTutor.com
Total 17
1
a i Using a pH probe on a data logger OR pH meter 1
7
ii 2
Explanation:
Acid / H+ reacts with A− AND equilibrium (position) shifts ALLOW direction of equilibrium shift if
ii 4
towards HA (to give a red colour) equilibrium shown: HA ⇌ H+ + A−
i.e. ‘towards HA’ is equivalent to ‘to left’
i.e. ‘towards A−’ is equivalent to ‘to right’
Alkali / OH− reacts with HA/H+ AND equilibrium (position) shifts
towards A− (to give a yellow colour)
c i 4
Total 15
= 0.0302(mol dm−3)
CH3CH(OH)COO− + CH3CH2CH2COOH2+
State symbols NOT required
ALLOW 13.267
n(A−) = 9.25 × 10−3 (mol) ALLOW HA / acid and A−/salt throughout for
butanoate and butanoic acid
ii n(HA) = 0.0165 − 9.25 × 10−3 = 7.25 × 10−3 (mol) 4
10−3 OR 4.82
Total 16
1
2
9 1 mark for correct reactants AND products AND correct
positioning of + and − charges on products
Total 2
2
i 1 state symbols not required
0
Total 3
Total 3
5.1.3 Acids, Bases and Buffers PhysicsAndMathsTutor.com
Ka = 10−3.35 OR 4.47 × 10−4 (mol dm−3) Always ALLOW calculator value irrespective
ii of working as number may have been kept
3
i [H+] = √(Ka × [HNO2]) OR √(Ka × [HA]) in calculator.
OR √(Ka × 0.0450)
OR 4.48 × 10−3 (mol dm−3) Note: pH = 2.35 is obtained from all three
values above
pH = 2.35 (2 DP required) From no square root, pH = 4.70. Worth Ka
mark only.
Total 9
Equilibria, Energetics and Elements
Acids, Bases & Buffers
PhysicsandMathsTutor.com 1
5 × 10 −3 × 21.35
(b) moles of HCl = = 1.067 × 10–4 mol (1)
1000
1.067 × 10 4
moles of Ca(OH)2 = = 5.34 × 10–5 mol (1)
2
concentration of Ca(OH)2 = 40 × 5.34 × 10–5
= 2.136 × 10–3 mol dm–3 (1)
2 marks for 4.27 × 10–3/ 8.54 × 10–3 mol dm–3
(no factor of 4) 3
(d) 8 (1) 1
[9]
PhysicsandMathsTutor.com 2
(c) (i) HPO42– (1) 1
(ii) H3PO4 (1) 1
7. proton donor
partially dissociates
[2]
[HCOO – ] [H + ] [H + ] 2
8. Ka = / /[H + ] = ( Ka × [HA]) /
[HCOOH] [HCOOH]
[H + ] 2
1.58 × 10–4 = /
0.025
[H+] = √{(1.58 × 10–4) × (0.025) } = 1.99 × 10–3 mol dm−3
pH = −log[H+] = −log 1.99 × 10–3 = 2.70
5.4034 (no square root) with working would score 1 mark.
[3]
PhysicsandMathsTutor.com 3
(iii) Two points from:
Ka /pKa /acid strength/amount of dissociation
temperature (but not “temperature & pressure”)
ratio/amounts/concentrations of weak acid and
conjugate base/salt (or reverse ratio)
(not ………. concentration of base as it could imply
NaOH) 2 max
[5]
1400 × 65
10. Mass of HNO3 = / 910g
100
910
Moles of HNO3 = = 14.4
63
pH = –log[H+] = –log 14.4 = –1.16/1.2 calc –1.15836
pH from ignoring 65% pH = –1.35: with working, 2 marks.
[3]
0.1263 × 23.75
13. moles of NaOH = / 3.00 × 10–3 mol
1000
moles of acid = 3.00 × 10–3 mol
moles of acid in flask = 10 × 3.00 × 10–3 = 3.00 × 10–2 mol
mass 2.58
molar mass of compound = = = 86 4
n 3.00 × 10 – 2
PhysicsandMathsTutor.com 4
Molecular formula = C4H6O2
A 4 carbon carboxylic acid
(e.g. butanoic acid) shown (bod)
Any 2 possible isomers from:
CH2 = C(CH3)COOH
CH2 = CHCH2COOH
cis CH3CH = CHCOOH
trans CH3CH = CHCOOH 4
Accept structural formulae that are unambiguous.
[8]
[ H + (aq)][HCOO − (aq)]
(ii) Ka = (1) (state symbols not needed) 1
[ HCOOH(aq)]
[ H + (aq)] 2 (1.55 × 10 −3 ) 2
(iii) Ka = = (1)
[HCOOH(aq)] 0.015
= 1.60 × 10–4 (mol dm–3)(1)
pKa = –log Ka = –log (1.60 × 10–4) = 3.80 (1) 3
(1.55 × 10 −3 ) × 100
(iv) Percentage dissociating = = 10.3 % /
0.015
10% (1) 1
(working not required)
[8]
PhysicsandMathsTutor.com 5
(ii) n(HCOOH) = 0.0150 × 25.00/1000 = 3.75 × 10–4 (1)
volume of NaOH(aq) that reacts is 30 cm3 (1)
so [NaOH] = 3.75 × 10–4 × 1000/30 = 0.0125 mol dm–3 (1) 2
(iii) Kw = [H+(aq)][OH–(aq)] (1)
pH = –log(1 × 10–14/0.0125) = 12.10/12.1 (1)
(calc 12.09691001) 3
PhysicsandMathsTutor.com 6
17. HCl and CH3COOH have same number of moles/
release same number of moles H+ /
1 mole of each acid produce ½ mol of H2 (1)
PhysicsandMathsTutor.com 7
[C 6 H 5 O − ][H + ]
21. (i) Ka = (1) 1
[C 6 H 5 OH]
[ H + (aq)] 2
1.3 × 10–10 ≈ (1)
0.40
(‘=’ sign is acceptable)
22.
O-Na+
O-Na+
CH2(CH2)4CH3
/ NaOH /Na (1)
weak acid/base pair mixture formed (1) 2
On structure, 1 mark for O Na on either or both phenol groups.
[2]
PhysicsandMathsTutor.com 8
(ii) [H+] = 0.0075 mol dm–3
pH = –log(0.0075) = 2.12 / 2.1 (1) 1
[3]
PhysicsandMathsTutor.com 9
PMT
(ii) use of k = rate / [O3] [C2H4] = 1.0 × 10–12 / (0.5 × 10–7 × 1.0 × 10–8)
to obtain a calculated value (1)
k = 2 × 103 (1)
units: dm3 mol–1 s–1 (1) 3
(iii) rate = 1.0 × 10–12 /4 = 2.5 × 10–13 (mol dm–3 s–1) (1) 1
2. 1½O2(g) → O3(g)/
O2(g) + ½O2(g) → O3(g) (1)
NO is a catalyst (1) as it is (used up in step 1 and) regenerated in step 2/
not used up in the overall reaction(1)
allow 1 mark for ‘O/NO2 with explanation of regeneration.’ 3
[3]
rate 2.6
(iii) k= / (1)
[NO] [H 2 ] 0.10 2 × 0.20
2
[PCl 3 ][Cl 2 ]
10. (a) Kc = (1) 1
[PCl 5 ]
PMT
(b) (i) PCl5 > 0.3 mol dm–3 ; PCl3 and Cl2 < 0.3 mol dm–3 (1) 1
(ii) At start, system is out of equilibrium with too much PCl3
and Cl2 and not enough PCL5 /
0.3 × 0.3
= 0.3 is greater than Kc = 0.245 mol dm–3 (1) 1
0.3
5 × 10 −3 × 21.35
(b) moles of HCl = = 1.067 × 10–4 mol (1)
1000
1.067 × 10 4
moles of Ca(OH)2 = = 5.34 × 10–5 mol (1)
2
concentration of Ca(OH)2 = 40 × 5.34 × 10–5
= 2.136 × 10–3 mol dm–3 (1)
2 marks for 4.27 × 10–3/ 8.54 × 10–3 mol dm–3
(no factor of 4) 3
PMT
(d) 8 (1) 1
[9]
[H + (aq)][HCOO − (aq)]
(ii) Ka = (1) (state symbols not needed) 1
[HCOOH(aq)]
[H + (aq)] 2 (1.55 × 10 −3 ) 2
(iii) Ka = = (1)
[HCOOH(aq)] 0.015
= 1.60 × 10–4 (mol dm–3)(1)
pKa = –log Ka = –log (1.60 × 10–4) = 3.80 (1) 3
(1.55 × 10 −3 ) × 100
(iv) Percentage dissociating = = 10.3 % /
0.015
10% (1) 1
(working not required)
[8]
[HI] 2
16. (a) Kc = (1) 1
[H 2 ][I 2 ]
(b) (i) H2 I2 HI
0.30 0.20 0
PMT
0.32 2
(ii) Kc = = 18.28571429 (1)
0.14 × 0.04
= 18 (to 2 sig figs) (1)
no units (1)
(or ecf based on answers to (i) and/or (a)) 3
3.44 × 1000
(iii) [HI] = = 4.58/4.59 mol dm–3 (1)
750
pH = –log 4.59 = –0.66 (1) 2
[7]
rate 2.1 × 10 −9
k= / (1)
[H + ][CH 3COCH 3 ] 0.02 × 1.5 × 10 − 3
= 7.0 × 10–5 (1) dm3 mol–1 s–1 (1) 4
accept 7 × 10–5
PMT
N N O N N O
or
Look for atoms bonded together.
AND other lone pairs. 1
3.74
(d) mass C = 12 × = 1.02 g /
44.0
3.74
moles CO2 = = 0.085 mol (1)
44
2
mass H = × 0.918 = 0.102 g /
18
0.918
moles H2O = = 0.051 mol (1) 2
18
1.02 0.102
ratio C : H = : = 0.0850 : 0.102 = 5 : 6 / 10 : 12/
12 1
ratio CO2 : H2O = 5 : 3 / 10 : 6 (1)
mass O = 1.394 – (1.020 + 0.102) = 0.272 g
/ using 1.394 g eugenol and Mr = 164, shows that 1
molecule contains 2 atoms of O (1) 2
∴ molecular formula = C10H12O2 (1) 1
[13]
rate 7.10
(iii) k= = = 7.10 × 109 (1)
[O 2 ][ NO] 2
0.0010 × 0.0010 2
p(SO 3 ) 2
25. (a) Kp = (1)(1)
p(SO 2 ) 2 × p(O 2 )
1 mark for correct powers but wrong way up.
1 mark for square brackets 2
p(SO 3 ) 2
(c) 3.0 × 102 = (1)
10 2 × 50
p(SO3) = √(3.0 × 102 × 102 × 50) = 1225 kPa (1)
%(SO3) = 100 × 1225 /(1225 + 10 + 50) = 95% (1) 3
PMT
[C 6 H 5 O − ][H + ]
30. (i) Ka = (1) 1
[C 6 H 5 OH]
31.
O-Na+
O-Na+
CH2(CH2)4CH3
/ NaOH /Na (1)
weak acid/base pair mixture formed (1) 2
On structure, 1 mark for O Na on either or both phenol groups.
[2]
32. moles HCl in 23.2 cm3 = 0.200 × 23.2/1000 = 4.64 × 10–3 (1)
moles B in 25 cm3 = moles HCl = 4.64 × 10–3 (1)
moles B in 250 cm3 = 4.64 × 10–3 × 10 = 4.64 × 10–2 (1)
4.64 × 10–2 mol B has a mass of 4.32 g
molar mass of B = 4.32/4.64 × 10–2 = 93 g mol–1 (1)
93 – 16 = 77 (1)
Therefore B is phenylamine / C6H5NH2 (1) 6
There may be other valid structures that are amines. These can
be credited provided that everything adds up to 93.
Answer could be a primary, secondary or tertiary amines.
[6]
p(CH 3 OH)
(ii) Kp = (1)(1)
p(CO) × p(H 2 ) 2
1 mark for Kc / use of any [ ] /inverted/power missing. 2
(iv) calc value 2.7120546 × 10–3; answer and/or units ecf from (ii) 2
PhysicsAndMathsTutor.com
Question Answer Marks Guidance
(c) (ii) FIRST, CHECK THE ANSWER ON ANSWER LINE
IF answer = 2.9(0), award 3 marks
---------------------------------------------------------------------------------
–3
[C 2H5COOH] = 0.12(0) mol dm ALLOW HA for C2H5COOH and A– for C2H5COO–
pH = 2.59 2 marks
–log√ (1.35 x 10–5 × 0.480) Original conc
pH = 5.79 2 marks
–log(1.35 x 10–5 × 0.120) No √
pH = 5.19 1 mark
–log (1.35 x 10–5 × 0.480) Original conc, no √
pH = 4.87 0 marks
–log(1.35 x 10–5) = 4.87 –log Ka
PhysicsAndMathsTutor.com
Question Answer Marks Guidance
(d) (i) 2C2H5COOH + Na2CO3 → 2C2H5COONa + CO2 + H2O 1 IGNORE state symbols and use of equilibrium sign
FOR CO2 + H2O ALLOW H2CO3
ALLOW C2H5COO–Na+ OR C2H5COO– + Na+
BUT BOTH + and – charges must be shown
ALLOW NaC2H5COO
(d) (ii) H+ + OH– → H2O 1 ALLOW C2H5COOH + OH– → C2H5COO– + H2O
IGNORE state symbols
(e) (i) pH = –log 1.35 × 10—5 = 4.87 1 ONLY correct answer
DO NOT ALLOW 4.9 (Question asks for 2 DP)
PhysicsAndMathsTutor.com
Question Answer Marks Guidance
(e) (iii) CHECK WORKING CAREFULLY AS CORRECT NUMERICAL FULL ANNOTATIONS MUST BE USED
ANSWER IS POSSIBLE FROM WRONG VALUES
================================================== -----------------------------------------------------------------------------
ALLOW HA and A– throughout For n(Mg), 1 mark
Amount of Mg (1 mark) ALLOW ECF for ALL marks below from incorrect n(Mg)
6.075
n(Mg) = 24.3 = 0.25(0) mol ECF ONLY available from concentrations that have
------------------------------------------------------------------------------------ • subtracted 0.50 OR 0.25 from 1 for [C2H5COOH]
Moles/concentrations (2 marks) • added 0.50 OR 0.25 to 1 for [C2H5COO–]
i.
n(C2H5COOH) = 1.00 – (2 × 0.25) = 0.50 (mol) For moles/concentration 1 mark (1 mark lost)
1. n (C2H5COOH) = 0.75 AND n(C2H5COO–) = 1.25
(C2H5COO–) = 1.00 + (2 × 0.25) = 1.50 (mol) 2. n(C2H5COOH) = 0.50 AND n(C2H5COO–) = 1.25
3. n(C2H5COOH) = 0.75 AND n(C2H5COO–) = 1.50
-n--------------------------------------------------------------------------------------- ----------------------------------------------------------------------------
[H+] and pH (1 mark) ALLOW ECF ONLY for the following giving 1 additional
0.50 mark and a total of 3 marks
+
] = 1.35 × 10 × 1.50 OR 4.5 × 10–6 (mol dm–3)
–5
0.75
[H 1. [H +] = 1.35 × 10–5 × 1.25 pH = –log 8.1 × 10–6 = 5.09
pH = –log 4.5 × 10–6 = 5.35 2 dp required 4
0.50
2. [H +] = 1.35 × 10–5 × 1.25 pH = –log 5.4 × 10–6 = 5.27
NOTE: IF there is no prior working,
0.50
ALLOW 4 MARKS for [H+] =1.35 × 10–5 × 1.50 AND pH = 5.35 0.75
3. [H +] = 1.35 × 10–5 × 1.50 pH = –log 6.75 × 10–6 = 5.17
IF the ONLY response is pH = 5.35, award 1 mark ONLY
Award a maximum of 1 mark (for n(Mg) = 0.25 mol) for:
pH value from Ka square root approach (weak acid pH)
pH value from Kw /10–14 approach (strong base pH)
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
ALLOW alternative approach based on Henderson–Hasselbalch equation for final 1 mark
1.5 0.5
pH = pKa + log OR pKa – log pH = 4.87 + 0.48 = 5.35 ALLOW –log Ka for pKa
0.5 1.5
Total 16
PhysicsAndMathsTutor.com
Question Answer Marks Guidance
2 (a) CH3COOH + H2O ⇌ H3O+ + CH3COO– IGNORE state symbols (even if incorrect)
Acid 1 Base 2 Acid 2 Base 1
2 ALLOW 1 AND 2 labels the other way around.
ALLOW ‘just acid’ and ‘base’ labels if linked by lines so that it is
clear what the acid–base pairs are
ALLOW A and B for ‘acid’ and ‘base’
PhysicsAndMathsTutor.com
(b) (ii) FIRST, CHECK THE ANSWER ON ANSWER LINE IF there is an alternative answer, check to see if there is any
ECF credit possible using working below.
IF answer = 1.15 10–11, award 2 marks
-------------------------------------------------------------
--------------------------------------------------------------------–
ALLOW 2 SF: 8.7 10–4 up to calculator value of
+ –3.06 –4 –3
[H ] = 10 = 8.71 10 (mol dm ) 8.7096359 10–4 correctly rounded
ALLOW Ca(CH3COO)2
ALLOW (CH3COO–)2Ca2+
BUT DO NOT ALLOW if either charge is missing or incorrect
PhysicsAndMathsTutor.com
(c) (ii) ALLOW names: ethanoic acid for CH3COOH
solution contains CH3COOH AND CH3COO– 1 ethanoate for CH3COO–
PhysicsAndMathsTutor.com
(c) (iii) Quality of written communication, QWC FULL ANNOTATIONS MUST BE USED
2 marks are available for explaining how the equilibrium --------------------------------------------------------------------------------
system allows the buffer solution to control the pH on addition Note: If there is no equilibrium equation then the two
of H+ and OH- (see below) subsequent equilibrium marks are not available: max 2
------------------------------------------------------
CH3COOH ⇌ H+ + CH3COO– DO NOT ALLOW HA ⇌ H+ + A–
DO NOT ALLOW more than one equilibrium equation.
----------------------------------------------------- -----------------------------------------------------
ALLOW response in terms of H+, A– and HA
CH3COOH reacts with added alkali
OR CH3COOH + OH– IF more than one equilibrium shown, it must be clear which one
OR added alkali reacts with H+ is being referred to by labeling the equilibria.
OR H+ + OH–
Equilibrium right OR Equilibrium CH3COO– (QWC) ALLOW weak acid reacts with added alkali
DO NOT ALLOW acid reacts with added alkali
PhysicsAndMathsTutor.com
(d) FULL ANNOTATIONS MUST BE USED
--------------------------------------------------------------------------------
FIRST, CHECK THE ANSWER ON ANSWER LINE IF there is an alternative answer, check to see if there is any
ECF credit possible.
IF answer = 11.48 OR 11.5 (g), award 5 marks
Incorrect use of [H+] = √( [CH3COOH] Ka ) scores zero
--------------------------------------------------------------------– BUT IGNORE if an alternative successful method is present
[H+] = 10–5 (mol dm–3)
Incorrect use of Kw, 1 max for [H+] = 10–5 (mol dm–3)
BUT IGNORE if an alternative successful method is present
-------------------------------------------------------------------------- ------------------------------------------------------------------------
1.75 105 ALLOW n(CH3COONa/CH3COO–)
[CH3COO–] = 0.200 = 0.350 mol dm–3
10 –5
1.75 105
= 0.08 = 0.14(0) (mol)
10–5
n(CH3COONa/CH3COO–) in 400 cm3
400 Note: There is no mark just for
= 0.350 = 0.14(0) (mol)
1000 400
n(CH3COOH) in 400 cm3 = 0.200 = 0.08 (mol)
1000
------------------------------------------------------------------------ -----------------------------------------------------------
mass CH3COONa = 0.140 82.0 = 11.48 OR 11.5 (g) As alternative for the 4th and 5th marks, ALLOW:
5
mass of CH3COONa in 1 dm3 = 0.350 82.0 = 28.7 g
For ECF, n(CH3COONa/CH3COO–) must have been 400
calculated in step before mass of CH3COONa in 400 cm3 = 28.7 = 11.48 g
1000
-------------------------------------------------------------
COMMON ECF
4.592 OR 4.6 g AWARD 4 marks
use of 400/1000 twice
PhysicsAndMathsTutor.com
ALLOW variants of Henderson–Hasselbalch equation.
[CH3COO ]
log = pH – pKa = 5 – 4.757 = 0.243
[CH3COOH]
[CH3COO ]
= 100.243 = 1.75
[CH3COOH]
Total 17
PhysicsAndMathsTutor.com
Question
uest er Marks Guidance
3 (a) 5 ANNOTATE WITH TICKS AND CROSSES, etc
HCl is a strong acid AND HClO is a weak acid ALLOW HCl completely dissociates
AND HClO partially dissociates
HCl:
pH = –log 0.14 = 0.85 (2 DP required) ALLOW HCl H+ + Cl AND HClO ⇌ H+ + ClO–
PhysicsAndMathsTutor.com
Question
uest er Marks Guidance
(b) 2 IGNORE state symbols
2Al + 6CH3COOH 2(CH3COO)3Al + 3H2 ALLOW correct multiples, e.g.:
Al + 3CH3COOH (CH3COO)3Al + 1.5H2
ALLOW any unambiguous formula for (CH3COO)3Al,
i.e. (CH3CO2)3Al, Al(CH3CO2)3, (CH3COO–)3Al3+, etc.
Note: IF charges are shown, they must be correct with
both – and 3+ shown
--------------------------------------------------------------------
PhysicsAndMathsTutor.com
Question
uest er Marks Guidance
(d) (i) 7 ANNOTATE WITH TICKS AND CROSSES, etc
QWC: Equilibrium shifts forming HCOO– OR H+ DO NOT ALLOW this mark if there is no equilibrium
OR (HCOOH) Equilibrium right system shown, e.g. HCOOH ⇌ H+ + HCOO– is absent
QWC: Equilibrium shifts forming HCOOH DO NOT ALLOW this mark if there is no equilibrium
OR (HCOOH) Equilibrium left system shown, e.g. HCOOH ⇌ H+ + HCOO– is absent
PhysicsAndMathsTutor.com
Question
uest er Marks Guidance
(d) (ii) HCOOH reacts with NaOH forming HCOO–/HCOONa 6 ANNOTATE WITH TICKS AND CROSSES, etc
OR DO NOT ALLOW just ‘methanoate/HCOO– forms’
HCOOH + NaOH HCOONa + H2O formulae or names of reactants also required
Equilibrium sign allowed
ALLOW HCOOH + OH– HCOO– + H2O
IGNORE conjugate base/salt forms
(Some) HCOOH/(weak) acid remains
OR HCOOH/(weak) acid is in excess IGNORE HCOOH has been partially neutralised
Calculation
CHECK THE ANSWER IF answer = 3.99, award all four calculation marks
n(HCOOH) OR [HCOOH] Note: There must be a clear statement that 0.24 and 0.4
= 0.24(0) (mol / mol dm–3) apply to moles or concentrations of HCOOH and HCOO–.
DO NOT ALLOW these values if unlabelled
n(HCOO–) OR [HCOO–] OR [HCOONa]
= 0.4(00) (mol / mol dm–3)
PhysicsAndMathsTutor.com
Question Answer Marks Guidance
4 (a) (i) +
[H ][CH3 (CH2 )2COO ] – ALLOW CH 3 CH 2 CH 2 COOH OR C 3 H 7 COOH in expression
(K a =) 1
[CH3 (CH2 )2COOH] DO NOT ALLOW use of HA and A– in this part.
DO NOT ALLOW:
[H+ ][CH3 (CH2 )2COO – ] [H+ ]2
= : CON
[CH3 (CH2 )2COOH] [CH3 (CH2 ) 2COOH]
(iii) FIRST, CHECK THE ANSWER ON ANSWER LINE IF alternative answer to more or fewer decimal places, check
IF answer = 2.71 award 3 marks calculator value and working for 1st and 2nd marks
-------------------------------------------------------------------- ------------------------------------------------------------
ALLOW use of HA and A– in this part
[H+] = [K a ][CH3 (CH2 )2COOH] OR 1.51 10 5 0.250
Calculator: 1.942935923 x 10–3
ALLOW use of calculated K a value, either calculator value or
rounded on script.
[H+] = 1.94 x 10–3 (mol dm–3) pH must be to 2 decimal places
PhysicsAndMathsTutor.com
Question Answer Marks Guidance
(b) (i) Mg + 2H+ Mg2+ + H 2 1 IGNORE state symbols
ALLOW Mg + 2 CH 3 (CH 2 ) 2 COOH
2CH 3 (CH 2 ) 2 COO– + Mg2+ + H 2
DO NOT ALLOW on RHS: (CH 3 (CH 2 ) 2 COO–) 2 Mg2+
Ions must be shown separately
(c) (i) CH 3 (CH 2 ) 2 COONa OR CH 3 (CH 2 ) 2 COO– forms ALLOW names throughout
OR ALLOW ‘sodium salt of butanoic acid’
CH 3 (CH 2 ) 2 COOH + OH– CH 3 (CH 2 ) 2 COO– + H 2 O ALLOW CH 3 (CH 2 ) 2 COOH + NaOH CH 3 (CH 2 ) 2 COONa +
H2O
CH 3 (CH 2 ) 2 COOH is in excess OR acid is in excess DO NOT ALLOW just ‘forms a salt/conjugate base’
OR some acid remains 2 i.e. identity of product is required
PhysicsAndMathsTutor.com
Question Answer Marks Guidance
(c) (ii) Moles (2 marks) ANNOTATIONS MUST BE USED
amount CH 3 (CH 2 ) 2 COOH = 0.0100 (mol) --------------------------------------------------------------------------------
ALLOW HA and A– throughout
amount CH 3 (CH 2 ) 2 COO– = 0.0025 (mol) 2
Mark by ECF throughout
Concentration (1 mark)
[CH 3 (CH 2 ) 2 COOH] = 0.100 mol dm–3
AND
[CH 3 (CH 2 ) 2 COO–] = 0.025 mol dm–3 1
[H+] and pH (2 marks) ONLY award final 2 marks via a correct pH calculation via
0.100 [CH3 (CH2 ) 2COOH]
[H+] = 1.51 105 = 6.04 x 10-–5 (mol dm–3) Ka using data derived from that in the
0.025 [CH3 (CH2 )2COO ]
2 question (i.e. not just made up values)
-–5
pH = –log 6.04 x 10 = 4.22 pH to 2 DP
ALLOW alternative approach based on Henderson–Hasselbalch equation for final 2 marks
0.025 0.100
pH = pK a + log OR pK a – log pH = 4.82 – 0.60 = 4.22 ALLOW –logK a for pK a
0.100 0.025
TAKE CARE with awarding marks for pH = 4.22 Common errors
There is a mark for the concentration stage. pH = 4.12
If this has been omitted, the ratio for the last 2 marks use of initial concentrations: 0.250 and 0.050 given in question.
will be 0.0100 and 0.0025. 4 marks max. Award last 3 marks for:
0.250/2 AND 0.050/2 = 0.125 AND 0.025
Common errors 0.125
pH = 5.42 1.51 105 = 7.55 x 10-–5 (mol dm–3)
0.025
As above for 4.22 but with acid/base ratio pH = –log[H+] = 4.12
inverted.
Award 4 OR 3 marks Award last 2 marks for:
0.250
Award zero marks for: 1.51 105 = 7.55 x 10-–5 (mol dm–3)
4.12 from no working or random values 0.050
pH value from K a square root approach (weak acid pH) pH = –log[H+] = 4.12
pH value from K w /10–14 approach (strong base pH) pH = 5.52
As above for 4.12 but with acid/base ratio inverted.
Award 2 OR 1 marks as outlined for 4.12 above
PhysicsAndMathsTutor.com
Question Answer Marks Guidance
(d) HCOOH + CH 3 (CH 2 ) 2 COOH ⇌ State symbols NOT required
HCOO– + CH 3 (CH 2 ) 2 COOH 2 + ALLOW 1 and 2 labels the other way around.
ALLOW ‘just acid’ and ‘base’ labels throughout if linked by
lines so that it is clear what the acid-base pairs are
acid 1 base 2
base 1 acid 2 2 For 1st mark, DO NOT ALLOW COOH–
(i.e. H at end rather than start)
CARE: but within 2nd mark ALLOW COOH– by ECF
Both + and – charges are required for the products in
the equilibrium IF proton transfer is wrong way around then
DO NOT AWARD the 2nd mark from an equilibrium ALLOW 2nd mark for idea of acid–base pairs, i.e.
expression that omits either charge HCOOH + CH 3 (CH 2 ) 2 COOH ⇌
HCOOH 2 + + CH 3 (CH 2 ) 2 COO–
base 2 acid 1
acid 2 base 1
Total 16
PhysicsAndMathsTutor.com
Quest
Question er Mark Guidance
1 (a) (i) proton donor 1 ALLOW H+ donor
(ii) (the proportion of) dissociation ALLOW a weak acid partly dissociates
ALLOW a strong acid totally dissociates
ALLOW ionisation for dissociation
ALLOW the ability to donate a proton
(iii) weakest: CH3COOH acetic acid ALLOW correct order using any identifier from the table,
C6H5COOH benzoic acid ie, common name, systematic name, structural formula OR pKa
CH3CHOHCOOH lactic acid value
strongest: CH3COCOOH pyruvic acid 1
(iv) C6H5COOH2+ + CH3CHOHCOO– BOTH products AND correct charges required for mark
1 Mark ECF from incorrect order in (iii)
See response from (iii) below response to (iv)
PhysicsAndMathsTutor.com
Quest
Question er Mark Guidance
(b) (i) All species AND balancing required for the mark
2CH3COCOOH + Ca(OH)2 → (CH3COCOO)2Ca + 1 ALLOW (CH3COCOO–)2Ca2+
2H2O ALLOW equation showing 2CH3COCOO– + Ca2+
IF charges shown, charges must balance,
Note: pyruvic acid must have been used here and e.g. DO NOT ALLOW (CH3COCOO–)2Ca
formula of pyruvic acid and pyruvate must be correct IGNORE state symbols if shown
ALLOW multiples ALLOW equilibrium sign
(ii) H+ + OH– ⎯→ H2O 1 ALLOW multiples but not same species on both sides
ALLOW equilibrium sign
IGNORE state symbols if shown
ALLOW H3O+ + OH– ⎯→ 2H2O
ALLOW CH3COCOOH + OH– ⎯→ CH3COCOO– + H2O
(c) FIRST, CHECK THE ANSWER ON ANSWER LINE IF there is an alternative answer, check to see if there is any ECF
IF answer = 2.11, award 4 marks credit possible using working below
-------------------------------------------------------------------- -------------------------------------------------------------
IF ECF, ANNOTATE WITH TICKS AND CROSSES, etc
Ka = 10–pKa
= 10–2.39 OR 0.00407 ALLOW 0.0041 to calculator value: 0.004073802
+ –
Ka = [H ] [CH3COCOO ] (ALLOW use of HA,H+ and A–)
[CH3 COCOOH] IF the pKa of a different weak acid has been used
use ECF from 2nd marking point
PhysicsAndMathsTutor.com
Quest
Question er Mark Guidance
(d) (i) O O 1 ALLOW correct structural OR displayed OR skeletal formula OR
recognisable mixture of formulae
C C
DO NOT ALLOW molecular formula but
H O O H ALLOW (COOH)2 OR (CO2H)2
O O
C C
OH
ALLOW OH BUT not O–H–C
PhysicsAndMathsTutor.com
Quest
Question er Mark Guidance
(e) Chemicals (1 mark) ANNOTATE WITH TICKS AND CROSSES, etc
lactic acid / CH3CHOHCOOH AND
(sodium) lactate / CH3CHOHCOO– (Na+) ALLOW any lactate salt
ALLOW lactic acid AND NaOH
OR lactic acid AND OH–
---------------------------------------------------------------
FOR ALTERNATIVE using Henderson–Hasselbalch equation,
Concentrations (4 marks) SEE PAGE 11
--------------------------------------------------------------
If another weak acid has been selected and salt has been
selected, allow ECF for remainder of question SEE PAGE 12
EITHER --------------------------------------------------------------
[H+(aq)] = 10–3.55 OR 2.8 x 10–4 ALLOW 2.8 x 10–4 up to calculator value of 2.81838 x 10–4
OR 2.82 x 10–4 (mol dm–3) separate marking point ALLOW 0.00028, etc
Ka = 10–3.86 OR 1.4 x 10–4 OR 1.38 x 10–4 (mol dm–3) ALLOW 1.4 x 10–4 up to calculator value of 1.38038 x 10–4
ALLOW 0.00014, etc
separate marking point
ALLOW use of CH3CHOHCOOH AND CH3CHOHCOO–(Na+)
ALLOW use of acid AND salt
[HA] [H+ ] [A − ] K
= OR = +a calculated value of [H+ ]
[A − ] Ka [HA] [H ] ALLOW value from
calculated value of K a
[HA] 2.8 × 10–4 2 [A − ] 0.5 ALLOW 2SF up to calculator value of 2.041742129 correctly
−
= OR OR 2 OR = OR rounded but ALLOW 2 if 2.8 x 10–4 and 1.4 x 10–4 used
[A ] 1.4 × 10 –4
1 [HA] 1
ALLOW 2 mol dm–3 HA AND 1 mol dm–3 A–
0.5 OR any concentration ratio of 2(acid) : 1(salt)
This marking point subsumes previous marking point
ONLY ALLOW 2SF up to calculator value of 0.489778819 correctly
Comment (1 mark) rounded but ALLOW 0.5 if 2.8 x 10–4 and 1.4 x 10–4 used
Magic tang/taste could come from other 6
chemicals/substances in the sweet
OR
The buffer would have the same taste/tang as the
magic tang
PhysicsAndMathsTutor.com
Quest
Question er Mark Guidance
ALTERNATIVE approach for concentrations using
Henderson–Hasselbalch equation (4 marks) ALLOW use of CH3CHOHCOOH AND CH3CHOHCOO–(Na+)
ALLOW use of acid AND salt
[A − ] [A − ] [HA] [HA]
pH = pKa + log OR –logKa + log ALLOW pH = pKa – log − OR –logKa – log −
[HA] [HA] [A ] [A ]
[A − ] [HA]
log = 3.55 – 3.86 (subsumes previous ALLOW log = 3.86 –3.55 (subsumes previous mark)
[HA] [A − ]
mark)
[HA]
− ALLOW log = 0.31 (subsumes previous mark)
log
[A ]
= –0.31 (subsumes previous mark) [A − ]
[HA]
[HA] 2.04 2
ALLOW = 100.31 = OR OR 2
[A − ] 0.490 −
[A ] 1 1
= 10–0.31 = OR 0.490
[HA] 1 [A − ]
For , ALLOW 2 SF up to calculator value of 0.48978819
[HA]
[HA]
For , ALLOW 2 SF up to calculator value of 2.041737945
[A − ]
but ALLOW 2 if 10–0.31 used
PhysicsAndMathsTutor.com
Quest
Question er Mark Guidance
(e) SUMMARY OF 4(e) MARKING POINTS FOR EACH POSSIBLE ACID CHOSEN
FIRST, CHECK THE ANSWER ON ANSWER LINE: IF answer is correct for weak acid chosen, award MP2–MP5
IF there is an alternative answer, check to see if there is any ECF credit possible using working below
lactic
a yruvic acetic benzoic
pKa 3.86
.86 4.19
lactic AND lactate
MP1 No mark No mark No mark
OR lactic acid AND OH–
MP2: [H+] 10–3.55 OR 2.82 x 10–4 ( calc: 2.81838 x 10–4 )
MP3: Ka 10–3.86 OR 1.38 x 10–4 10–2.39 OR 4.07 x 10–3 10–4.76 OR 1.74 x 10–5 10–4.19 OR 6.46 x 10–5
calc: 1.380384265 x 10–4 4.073802778 x 10–3 1.737800829 x 10–5 6.45654229 x 10–5
MP5:
[HA] 2.82 × 10 –4 2.82 × 10 –4 2.82 × 10 –4 2.82 × 10 –4
OR 2.04 OR 0.0693 OR 16.2 OR 4.37
[A − ] 1.38 × 10 –4 4.07 × 10 –3 1.74 × 10 –5 6.46 × 10 –5
calc: 2.041737945 calc: 0.069183097 calc: 16.21810097 calc: 4.365158322
TAKE CARE: Calc values are completely unrounded and may differ between brands of calculator
Use actual candidate values at each stage using rounding to 2 or more SF.
MP5: calculated using 3 SF from MP2 and MP3
calc values for MP5 are completely unrounded (using calculator values from MP2 and MP3)
Be slightly flexible as candidates may have written down rounded values but carried on with calculator values
– This appr ach is ACCEPTABLE
Total 20
PhysicsAndMathsTutor.com
Question Expected Answers Marks Additional Guidance
2 a 4 ALLOW C2H5 throughout question
measured pH > 1 OR [H+] < 0.1 (mol dm–3) ALLOW [H+] < [CH3CH2COOH] OR [H+] < [HA]
ALLOW measured pH is higher than expected
ALLOW measured pH is not as acidic as expected
ALLOW a quoted pH value or range > 1 and < 7
OR between 1 and 7
[H+] = 10–pH ALLOW [H+] = antilog –pH OR [H+] = inverse log –pH
[H+ ]2 [H+ ]2
Calculate Ka from IF Ka is NOT given and Ka = is shown, award mark for Ka also
0.100 0.100
[H+ ]2
(i.e. Ka = is automatically awarded the last 2 marks)
0.100
b Marks are for correctly calculated values. 2 ALLOW 3.467368505 × 10–14 and correct rounding to 3.5 × 10–14
Working shows how values have been derived.
[H+] = 10–13.46 = 3.47 x 10–14 (mol dm–3) ALLOW 0.28840315 and correct rounding to 0.29,
i.e. ALLOW 0.288
[OH–] = 1.0 × 10–14 = 0.29 (mol dm–3)
3.47 × 10–14 ALLOW alternative approach using pOH:
PhysicsAndMathsTutor.com
Question Expected Answers Marks Additional Guidance
c 7 ANNOTATIONS MUST BE USED
Propanoic acid reacts with sodium hydroxide ALLOW C2H5 throughout question
forming propanoate ions/sodium propanoate ALLOW Adding NaOH forms propanoate ions/sodium propanoate
OR (imples that the NaOH is added to the propanoic acid)
CH3CH2COOH + NaOH CH3CH2COONa + H2O
Added acid
CH3CH2COO– reacts with added acid ALLOW conjugate base reacts with added acid
OR [H+] increases DO NOT ALLOW salt reacts with added acid
5
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d HNO3 + CH3CH2COOH ⇌ CH3CH2COOH2+ + NO3
– 2 State symbols NOT required
acid 1 base 2 acid 2 base 1 ALLOW 1 AND 2 labels the other way around.
ALLOW ‘just acid’ and ‘base’ labels throughout if linked by lines so that
it is clear what the acid–base pairs are.
IF proton transfer is wrong way around then ALLOW 2nd mark for idea
of acid–base pairs, i.e.
HNO3 + CH3CH2COOH ⇌ CH3CH2COO– + H2NO3+
base 2 acid 1 base 1 acid 2
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3 a A strong acid completely dissociates ALLOW ionises for dissociates
AND
a weak acid partially dissociates 1
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i.e., NO ECF from impossible chemistry
pH = –log 1.25 × 10–13 = 12.90 DO NOT ALLOW 12.9 not two decimal places
--------------------------------------------- ---------------------------------------
pOH variation (also worth 3 marks) 3 COMMON ERRORS
[OH–] = 2 × 0.04(00) = 0.08(00) (mol dm–3) 12.60 no × 2 for [OH–]
12.6 no × 2 for [OH–] AND 1 DP only
pOH –log 0.08(00) = 1.10 12.30 ÷ 2 [OH–]
12.3 ÷ 2 [OH–] AND 1 DP only
pH = 14.00 – 1.10 = 12.90 1.40 NO marks
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d ANNOTATIONS MUST BE USED
Equilibrium Equilibrium sign is required
H2CO3 ⇌ H+ + HCO3– IGNORE HA ⇌ H+ + A–
DO NOT ALLOW H2CO3 ⇌ 2H+ + CO32–
DO NOT ALLOW NaHCO3 ⇌ Na+ + HCO3–
IGNORE H2O + CO2 ⇌ H2CO3
Added alkali IF more than one equilibrium shown, it must be clear which
H2CO3 reacts with added alkali equilibrium is being referred to
OR H2CO3 + OH–
OR added alkali reacts with H+ ALLOW added alkali reacts with weak acid
OR H+ + OH–
Quality of Written Communication
Mark is for linking the action of the buffer in controlling
added alkali and hence pH
Equilibrium right
OR equilibrium shifts forming H+ OR HCO3–
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Added acid 5 HCO3– is required for this mark BUT …
HCO3– reacts with added acid ALLOW added acid reacts with conjugate base ONLY if
HCO3– is present in equilibrium with H2CO3
Equilibrium left DO NOT ALLOW salt reacts with added acid
OR equilibrium shifts forming H2CO3
d ii FIRST, CHECK THE ANSWER ON ANSWER LINE IF there is an alternative answer, check to see if there is any
IF answer = 6.6 : 1 OR 1 : 0.15 ECF credit possible using working below
CHECK ratio is HCO3– : H2CO3 and award 5 marks. -----------------------------------------------------------
IF answer = 0.15 : 1 , ANNOTATIONS MUST BE USED
CHECK ratio is H2CO3 : HCO3– and award 4 marks FOR ALTERNATIVE using Henderson–Hasselbalch
-------------------------------------------------------------------- equation below
In blood at pH 7.40, --------------------------------------------------------------
[H+] = 10–pH = 10–7.40 = 3.98 × 10–8 (mol dm–3) ALLOW 3.98 × 10–8 up to calculator value of
[H+ ] [HCO3– ] 3.98 10 –8 10.5 3.981071706 × 10–8 correctly rounded
Ka = =
[H2CO3 ] 1
OR Ka = 4.18 × 10 (mol dm–3)
–7
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–
[HCO ]
At pH = 7.20, log 3
= pH – pKa = 7.20 – 6.38 = 0.82 (subsumes previous mark)
[H2CO3 ]
[HCO3– ] 6.6
= 100.82 = OR 6.6 : 1
[H2CO3 ] 1
Total 22
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Question Answer Mark Guidance
1 (a) (i) HOCH 2 COOH + NaOH HOCH 2 COONa + H 2 O 1 ALLOW: HOCH 2 COOH + OH– HOCH 2 COO– + H 2 O
ALLOW: H+ + OH– H 2 O
DO NOT ALLOW molecular formulae
(cannot see which OH has reacted)
(ii) FIRST, CHECK THE ANSWER ON ANSWER LINE IF there is an alternative answer, check to see if there is any
IF answer = 0.142 (mol dm–3), award 2 marks ECF credit possible using working below
-------------------------------------------------------------------- -------------------------------------------------------------
25.0 ANNOTATE WITH TICKS AND CROSSES, etc
amount of HOCH 2 COOH = 0.125
1000
= 0.003125 (mol) ALLOW 3.125 × 10–3 mol
1000 1000
concentration NaOH = 0.003125 × ALLOW ECF: answer above ×
22.00 22.00
= 0.142 (mol dm–3) 2 ALLOW 2 SF: 0.14 to calculator value: 0.142045454
--------------------------------------------------------------------- -----------------------------------------------------------------------------
If candidate has written in (a)(i): HOCH 2 COOH + 2NaOH,
mark by ECF:
1000
concentration NaOH = 2 × 0.003125 ×
22.00
= 0.284 (mol dm–3)
(iii) Vertical section matches the (pH) range (of the ALLOW stated pH range for vertical section at about
indicator) 7–10, 6–10, etc
OR colour change (of the indicator) 1 ie ALLOW ‘pH range must be about 7–10’
OR end point (of the indicator) ALLOW ‘pH changes rapidly’ for vertical section
ALLOW ‘equivalence point’ for vertical section,
ie ALLOW equivalence point matches the (pH) range, etc
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Quest
Question er Mark Guidance
(b) (i) H HOCH2COO
+ – IGNORE state symbols
(K a =)
1 2
HOCH2COOH H+
IGNORE in (i) but ALLOW in (ii)
HOCH2COOH
(ii) FIRST, CHECK THE ANSWER ON ANSWER LINE IF there is an alternative answer, check to see if there is any
IF answer = 1.46 x 10–4, award 2 marks ECF credit possible using working below
THEN IF units are mol dm–3, award 1 further mark UNITS can be credited with no numerical answer
--------------------------------------------------------------------- ----------------------------------------------------------------------
ANNOTATE WITH TICKS AND CROSSES, etc
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Question Answer Mark Guidance
(c) ONE mark for equilibrium expression ANNOTATE WITH TICKS AND CROSSES, etc
equilibrium: HOCH 2 COOH ⇌ H+ + HOCH 2 COO– 1 DO NOT ALLOW H+, A– and HA
ALLOW < – > as alternative for equilibrium sign
--------------------------------------------------------------------- ---------------------------------------------------------------------
Four marks for action of buffer ALLOW response in terms of H+, A– and HA
Equilibrium responses must refer back to a written equilibrium:
IF more than one equilibrium shown, assume correct one
HOCH 2 COOH reacts with added alkali ALLOW weak acid reacts with added alkali
DO NOT ALLOW acid reacts with added alkali
OR HOCH 2 COOH + OH–
OR added alkali reacts with H+
OR H+ + OH–
HOCH 2 COO–
OR Equilibrium right
HOCH 2 COO– reacts with added acid ALLOW conjugate base reacts with added acid
HOCH 2 COOH DO NOT ALLOW salt/base reacts with added acid
OR Equilibrium left 4
--------------------------------------------------------------------- ---------------------------------------------------------------------
Two marks for preparation of buffer
Ammonia reacted with an excess of glycolic acid
OR some glycolic acid remains ALLOW as products HOCH 2 COO– + NH 4 +
HOCH 2 COOH + NH 3 HOCH 2 COONH 4 2 ALLOW ⇌ sign instead of
(d) Base 1 + Acid 2 ⇌ Acid 1 + Base 2 ALLOW: Base 2 + Acid 1 ⇌ Acid 2 + Base 1
1st mark for identifying acids and bases.
2nd mark for correct pairing (ie numbers) 2
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(e) 2HSCH 2 COO– + R–S–S–R ALLOW (SCH 2 COO–) 2
–OOCCH 2 S–SCH 2 COO– + 2 SH ALLOW equation with ammonium salt, ie:
Total 20
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Question Answer Mark Guidance
2 (a) (i) (K w = ) [H+(aq)] [OH–(aq)] 1 IGNORE state symbols
ALLOW [H 3 O+(aq)] [OH–(aq)]
(ii) FIRST, CHECK THE ANSWER ON ANSWER LINE IF there is an alternative answer, check to see if there is any
IF answer = 2.3 × 10–10 (mol dm–3), award 2 marks ECF credit possible using working below
IF answer = 2.34 × 10–10 (mol dm–3) , award 1 mark ANNOTATE WITH TICKS AND CROSSES, etc
--------------------------------------------------------------------- ----------------------------------------------------------------------
------ ALLOW 4.3 × 10–5 up to calculator: 4.265795188 × 10–5
[H+] = 10–pH = 4.27 × 10–5 (mol dm–3) ALLOW 0.0000427
(b) (i) Endothermic because K w increases with 1 Endothermic AND reason required for the mark
temperature ALLOW Endothermic because increasing temperature shifts
equilibrium/reaction to the right
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(b) (iii) (Work is) inaccurate OR invalid Response requires reason for inaccuracy/invalidity in terms of
because K w varies with temperature 1 Kw
ALLOW incorrect with reason
IGNORE unreliable
ALLOW inaccurate because wrong K w was used
For K w varies with temperature, ALLOW equilibrium shifts with
temperature
Amounts of acid AND alkali stated Amounts could be expressed as amounts, moles, volumes OR
concentrations
Q
∆H neut = –energy change 6 ALLOW ‘–‘ sign shown in earlier part, ie ∆H neut =
n
ALLOW a statement linking ∆H with temperature change, ie:
IF temperature increases, ∆H neut is –ve
OR IF temperature decreases, ∆H neut is +ve
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(d) ANNOTATE WITH TICKS AND CROSSES, etc
Lattice enthalpy
Lattice enthalpy of KF is more negative than RbF ALLOW lattice enthalpy of KF > lattice enthalpy of RbF
OR
K+ has greater attraction for F– ALLOW more energy needed to separate K+ AND F–
IGNORE KF has stronger bonds
Hydration enthalpy
∆H(hydration) of K+ is more negative than Rb+ ALLOW ∆H(hydration) of K+ > ∆H(hydration) of Rb+
OR
K+ has greater attraction for H 2 O ALLOW more energy needed to separate K+ AND H 2 O
IGNORE K+ has a stronger bond to H 2 O
Enthalpy change of solution
Idea that ∆H(solution) is affected more by lattice ALLOW a correct attempt to link the contribution of lattice
enthalpy than by hydration enthalpy 4 enthalpy and hydration enthalpy to ∆H(solution), ie lattice
enthalpy is a more important factor than hydration enthalpy
IGNORE ∆G is negative
Total 20
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