Download 1. (a) Propan-1ol, C2H5CH2OH can be oxidised to propanoic acid

1.
(a)
Propan-1ol, C2H5CH2OH can be oxidised to propanoic acid, C2H5COOH.
(i)
State the names of the reagents necessary for this conversion.
............................................................................................................................
............................................................................................................................
(2)
(ii)
5.67 of propan-1-o1 was oxidised in a reaction with a 64% yield. Calculate the
mass of propanoic acid produced.
(3)
(b)
Propanoic acid is a weak acid. It ionises according to the equation:
C2H5COOH(aq)+H2O(l)
(i)
C2H5COO–(aq) + H3O+(aq)
H
= +6kJ mol–1
State and explain the effect on the position of this equilibrium of:
an increase in temperature;
............................................................................................................................
............................................................................................................................
(2)
an addition of solid sodium propanoate.
............................................................................................................................
............................................................................................................................
(2)
(ii)
How does the addition of sodium propanoate affect the pH of the solution of
propanoic acid?
............................................................................................................................
(1)
Maltby Academy
1
(c)
(i)
Sketch, with reasonable accuracy, how the pH changes during the titration of
25cm3 of a weak acid, such as propanoic acid, with sodium hydroxide solution of
the same concentration.
14
12
10
pH
8
6
4
2
0
0
10
20
30
40
50
volume NaOH/cm³
(ii)
The table contains some data about three indicators.
Indicator
pKind
Acid colour
Alkaline colour
Bromophenol blue
4.0
yellow
blue
Bromothymol blue
7.0
yellow
blue
Thymol blue
8.9
yellow
blue
State which of these indicators would be best for this titration. Give a reason for
your choice.
Indicator ...........................................................................................................
Reason ..............................................................................................................
...........................................................................................................................
(2)
Maltby Academy
2
(d)
The standard enthalpy change of neutralisation of some acids with sodium hydroxide is
tabulated below:
(i)
Acid
H /kJ mol–1
Propanoic acid, C2H5COOH
–51
Hydrocyanic acid, HCN
–12
Hydrochloric acid, HCl
–57
Nitric acid, HNO3
–57
Why are the values for the enthalpy change of neutralisation of the two strong acids
the same?
..........................................................................................................................
..........................................................................................................................
(1)
(ii)
Why is the enthalpy change of neutralisation of hydrocyanic acid so much less than
that of hydrochloric acid?
..........................................................................................................................
..........................................................................................................................
(2)
(Total 19 marks)
2.
Sorbic acid has the formula
CH3–CH=CH–CH=CH–COOH
Together with its potassium salt sorbic acid is used in foodstuffs as an inhibitor of fungal and
mould growth.
Maltby Academy
3
(a)
Sorbic acid can show geometric isomerism.
(i)
Explain carefully why sorbic acid shows geometric isomerism.
(2)
(ii)
State, with a reason, the number of geometric isomers that sorbic acid has.
(2)
(b)
Sorbic acid and potassium sorbate together form a buffer system.
(i)
What is a buffer system?
(2)
(ii)
Write the chemical equations that are relevant to the functioning of the mixture as a
buffer, and use them to explain how buffering action arises. You may represent
sorbic acid as RCOOH and potassium sorbate as RCOOK.
(5)
(c)
Sorbic acid will react with a solution of bromine in tetrachloromethane.
(i)
Write an equation for the complete reaction of sorbic acid with bromine.
(1)
(ii)
A 7.00g sample of sorbic acid was reacted with a solution of bromine in
tetrachloromethane; 125cm3 of the solution was required. Calculate the
concentration of the bromine solution in mod dm–3.
(4)
(d)
The bromine-containing product from (c)(i) can be reacted with sodium hydroxide in
aqueous ethanol to give a substitution reaction.
(i)
Explain why aqueous ethanol is used as the solvent rather than, say, ethanol alone.
(2)
(ii)
Write the structural formula of the product.
(1)
Maltby Academy
4
(e)
Potassium sorbate can function as a soap.
(i)
Explain why this is so.
(1)
(ii)
Soaps are in general made by the alkaline hydrolysis of fats. Write an equation to
illustrate this reaction.
(3)
(Total 25 marks)
3.
(a)
(i)
Calculate the concentration, in mol dm–3, of a solution of hydrochloric acid, HCl,
which has a pH of 1.13.
(1)
(ii)
Calculate the concentration, in mol dm–3 , of a solution of chloric(l) acid, HOCl,
which has a pH of 4.23.
Chloric(l) acid is a weak acid with Ka = 3.72 × 10–8 mol dm–3.
(4)
Maltby Academy
5
(b)
The pH of 0.100 mol dm–3 sulphuric acid is 0.98.
(i)
Calculate the concentration of hydrogen ions, H+, in this solution.
(1)
(ii)
Write equations to show the two successive ionisations of sulphuric acid, H2SO4, in
water.
......................................................................................................................
......................................................................................................................
(2)
(iii)
Suggest why the concentration of hydrogen ions is not 0.20 mol dm–3 in 0.100 mol
dm–3 sulphuric acid.
......................................................................................................................
......................................................................................................................
(1)
(c)
Many industrial organic reactions produce hydrogen chloride as an additional product.
This can be oxidised to chlorine by the Deacon process:
4HCl(g) + O2(g)
2Cl2(g) + 2H2O(g)
H = –115 kJ mol–1.
0.800 mol of hydrogen chloride was mixed with 0.200 mol of oxygen in a vessel of
volume 10.0 dm3 in the presence of a copper(I) chloride catalyst at 400 ºC. At
equilibrium it was found that the mixture contained 0.200 mol of hydrogen chloride.
(i)
Write an expression for the equilibrium constant Kc.
(1)
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6
(ii)
Calculate the value of Kc at 400 ºC.
(4)
(d)
State and explain the effect, if any, on the position of equilibrium in (c) of:
(i)
decreasing the temperature;
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(ii)
decreasing the volume;
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(iii)
removing the catalyst.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 20 marks)
Maltby Academy
7
4.
(a)
Benzocaine, C9H11O2N, is an aromatic compound which is used commercially in creams
to alleviate sunburn.
 Benzocaine reacts with dilute acids to form the ion C9H12O2N+ and with ethanoyl
chloride to form C11H13O3N.
 When benzocaine is heated under reflux with aqueous sodium hydroxide and the
solution obtained is neutralised, two compounds X and Y are formed.
 X has a formula of C7H7O2N and is a solid with a melting temperature of 190 ºC. It
is soluble in water.
 Y is a volatile liquid with a formula C2H6O which gives steamy fumes with
phosphorus pentachloride.
 X reacts with sodium hydrogencarbonate solution to give a gas which turns lime water
milky. It also reacts with a solution of sodium nitrite and hydrochloric acid between
0 ºC and 5 ºC to produce a substance which reacts with phenol to give an orange
precipitate, Z.
These reactions are summarised as follows.
H + (aq)
C 9 H 12O2 N +
C9 H 11 O2 N
Benzocaine
CH3 COCl
C11 H 13O 3 N
1. NaOH(aq) heat/reflux
2. HCl(aq) until neutral
C7H7O 2N
X
NaHCO 3 (aq)
C 7 H 6 O 2 NNa
(i)
+
C 2 H 6O
Y
1. HNO 2 between 0 ºC and 5 ºC
2. phenol
orange ppt. Z
Deduce a structural formula for benzocaine and explain its three reactions shown
above. You may either describe the types of reaction or write the equations for the
reactions.
(6)
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8
(ii)
Write equations for the two reactions of X. Include in your answer the structural
formula of Z.
(3)
(iii)
Explain why substance X has a fairly high melting temperature and why it is
soluble in water.
(3)
(b)
Substance X is a weak monobasic acid and for the purpose of the remainder of this
question you may write its formula as HA.
X has a relative molecular mass of 137, with a pKa value of 4.92 at 25 ºC.
(i)
Calculate the pH of a solution containing 21.37g of X per dm3 at a temperature of
25 ºC.
(4)
(ii)
50.0 cm3 of this solution was mixed with 50.0 cm3 of a 0.100 mol dm–3 solution of
sodium hydroxide. Calculate the concentration of the salt of X produced, and the
concentration of the acid X left unreacted.
Hence calculate the pH of the mixed solution.
(4)
(c)
The standard enthalpy change at 25 ºC for the neutralisation of a strong acid by a strong
base is –57.2 kJ mol–1.
The standard enthalpy change for the ionisation of the weak acid HA in water is
+8.3 kJ mol–1.
(i)
Write the ionic equation for the neutralisation of a strong acid by a strong base and
hence calculate the standard enthalpy of neutralisation of the acid HA.
(3)
(ii)
State and explain how the value of Ka of the acid X and hence the pH of the
solution in (b)(i) would change if the temperature of the solution were increased.
(2)
(Total 25 marks)
5.
(a)
The acid dissociation constant, Ka for ethanoic acid is 1.80 × 10–5 mol dm–3.
(i)
Write an equation for the dissociation of aqueous ethanoic acid.
......................................................................................................................
(1)
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9
(ii)
Give the expression for the acid dissociation constant, Ka, for aqueous ethanoic
acid.
(1)
(iii)
Calculate the pH of 2.00 mol dm–3 ethanoic acid.
(3)
(b)
When ethanoic acid and propan-1-ol are heated together in the presence of a little
concentrated sulphuric acid an ester is produced.
(i)
Draw the structural formula of the ester produced in this reaction.
(2)
(ii)
Give the name of the ester.
......................................................................................................................
(1)
(iii)
State the function of the sulphuric acid.
......................................................................................................................
(1)
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10
(c)
6.00g of ethanoic acid was mixed with 6.00g of propan-1-ol and 0.00100mol of
concentrated sulphuric acid was added. The mixture was heated under reflux until no
further reaction was detectable. The mixture was then cooled rapidly to room temperature
and titrated with 1.00 mol dm–3 sodium hydroxide solution. 35.0 cm3 of the 1.00 mol
dm–3 sodium hydroxide solution was required for complete neutralisation.
(i)
Calculate the volume of sodium hydroxide solution that would have been needed if
only the 0.00100mol of concentrated sulphuric acid had been in the flask.
(2)
(ii)
Calculate the volume of sodium hydroxide solution that must have been needed to
react with the ethanoic acid remaining in the cooled mixture.
(1)
(iii)
Calculate the amount (in moles) of ethanoic acid remaining.
(1)
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11
(iv)
Hence, calculate the amount (in moles) of propan-1-ol, ester and water in the final
cooled mixture.
(3)
(v)
Calculate the value of the equilibrium constant under these conditions. The
volume of the reaction mixture should be taken as 100 cm3.
(2)
(d)
The reverse of esterification is hydrolysis. In experiments to study the hydrolysis of this
ester with aqueous sodium hydroxide, the following data were obtained.
Maltby Academy
Experiment
Concentration of
ester
/mol dm–3
Concentration of
sodium hydroxide
/mol dm–3
Rate of reaction
/mol dm–3 s–1
1
0.1
1.0
6 × 10–4
2
0.1
2.0
1.2 × 10–3
3
0.2
2.0
2.4 × 10–3
12
(i)
Determine the order of the reaction with respect to the:
ester;..............................................................................................................
(1)
sodium hydroxide...........................................................................................
(1)
(ii)
Calculate the value of the rate constant, k, for this reaction including its units.
(3)
(Total 23 marks)
6.
(a)
When solid sodium nitrate is heated with concentrated sulphuric acid the reaction shown
below occurs:
(i)
NaNO3 + H2SO4
HNO3 + NaHSO4
..........
..........
..........
..........
Identify the two conjugate acid-base pairs in this reaction. Write your answer in
the spaces below the equation.
(2)
(ii)
The nitric acid produced is evolved as a gas under these conditions. Explain the
effect of this on the position of equilibrium on heating the mixture in an open
container.
......................................................................................................................
......................................................................................................................
(2)
Maltby Academy
13
(iii)
Nitric acid molecules in the vapour phase can be represented as
O
N
O
OH
Draw a diagram of the likely shape of this molecule and justify your prediction.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(b)
Dilute sulphuric acid is a strong acid but very concentrated sulphuric acid is weak acid.
Explain the meanings of the terms weak and dilute as applied to acids.
Weak.......................................................................................................................
................................................................................................................................
Dilute.......................................................................................................................
................................................................................................................................
(2)
(c)
Heating solid sodium nitrate, NaNO3, to decomposition requires a temperature of about
550°C and produces oxygen as the only gaseous product. Solid magnesium nitrate,
Mg(NO3)2, decomposes on heating at a much lower temperature.
(i)
Write an equation for the decomposition of sodium nitrate.
......................................................................................................................
(1)
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14
(ii)
Relate the ease of decomposition of the two nitrates to the relative sizes and
charges of the cations.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(Total 13 marks)
7.
(a)
What is meant by the term weak acid?
....................................................................................................................................
....................................................................................................................................
....................................................................................................................................
(2)
(b)
A weak acid, represented by HA, dissociates in water according to the equation:
HA(aq) + H2O(l)
H3O+(aq) + A–(aq)
Write an expression for the dissociation constant, Ka, for HA.
(1)
Maltby Academy
15
(c)
25 cm3 of 1.00 mol dm–3 aqueous HA, was titrated with 1.00 mol dm–3 aqueous sodium
hydroxide and the pH measured throughout. The titration curve is shown below.
14
12
pH 10
8
6
4
2
0
10
20
30
40
50
3
Volume of sodium hydroxide added / cm
Use the titration curve to find:
(i)
the value of the pH at the end point of the titration.
............................................................................................................................
(1)
(ii)
the pH of an aqueous solution of the salt NaA.
............................................................................................................................
(1)
(iii)
the value of pKa for the acid HA and, hence the value Ka.
pKa ...................................................................................................................
Ka .....................................................................................................................
(2)
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16
(d)
Some of the solutions made during this titration would act as buffer solutions.
(i)
What is meant by the term buffer solution?
............................................................................................................................
............................................................................................................................
............................................................................................................................
(2)
(ii)
Use the titration curve to find:
the range of pH values over which this mixture acts as a buffer;
from ....................................................
to ....................................................
(1)
the pH of the most efficient buffer solution.
............................................................................................................................
(1)
(e)
Suggest, with reasoning, whether methyl orange or phenolphthalein would be the better
indicator for this titration.
Choice ........................................................................................................................
Reasoning ..................................................................................................................
....................................................................................................................................
....................................................................................................................................
(3)
(f)
Explain why, as the titration proceeds, the flask becomes warm but not as warm as it
would in a similar titration using 1.00 mol dm–3 solutions of hydrochloric acid and
sodium hydroxide.
....................................................................................................................................
....................................................................................................................................
....................................................................................................................................
(3)
Maltby Academy
17
(g)
A different monobasic weak acid has a dissociation constant of 1.8 × 10– 5 mol dm–3.
(i)
Define pH.
............................................................................................................................
(1)
(ii)
Calculate the pH of a 1.00 mol dm–3 aqueous solution of this acid.
(3)
(Total 21 marks)
8.
Ethanoic acid, CH3COOH, is a weak acid which can be used, with its salts, to make buffer
solutions.
(a)
Explain what is meant by the term weak acid.
....................................................................................................................................
....................................................................................................................................
(1)
(b)
Explain what is meant by the term buffer solution.
....................................................................................................................................
....................................................................................................................................
(2)
Maltby Academy
18
(c)
An aqueous solution of ethanoic acid of concentration 1.00 mol dm–3 has a pH of 2.8.
Sketch, with care, how the pH changes during the titration of 25.0 cm3 1.00 mol dm–3
aqueous ethanoic acid with aqueous sodium hydroxide of the same concentration.
14
pH
12
10
8
6
4
2
0
0
10
20
30
40
50
Volume of alkali added/cm 3
(4)
(d)
Indicate on your sketch the portion of the curve where the mixture is behaving as a buffer.
(1)
(e)
(i)
Write an expression for the dissociation constant, Ka, for ethanoic acid.
(1)
(ii)
Explain how the pKa of ethanoic could be found from the graph.
............................................................................................................................
............................................................................................................................
(2)
(Total 11 marks)
9.
(a)
(i)
Maltby Academy
Reagent: potassium dichromate (VI)/potassium manganate (VII) (1) or
19
formulae sulphuric acid or hydrochloric acid
(1) or formulae
If potassium manganate(VII) chosen not HCl or conc H2SO4 for second mark
‘Acidfied dichromate’ or H+ / Cr2O72–(1)
2
(ii)
OR
(b)
(i)
(ii)
amount of propanol = 5.67/60 = 0.0945 mol (1)
amount of propanoic acid produced = 0.64 × 0.0945
= 0.06048 mol (1)
yield of propanoic acid = 74 × 0.06048 = 4.5 / 4.48 / 4.476 g
(1)
by mass ratio: ratio acid/alcohol = 74/60 = 1.23 (1)
100% yield = 1.23 × 5.67g = 6.99 g (1)
64% yield = 6.99 g × 0.64 = 4.5 / 4.48 / 4.476 g (1)
3
increase in temperature:
(position of ) equilibrium goes to the right (1)
as endothermic left to right (1)
on the addition of sodium propanaoate
the position of equilibrium goes to left (1)
higher concentration of / more propanoate ions
or
sodium propanoate produces propanoate ions (1)
4
pH rises (consequential on above) (1)
1
14
start pH 2 to 4 (1)
12
correct general shape (1)
9
vertical 6/7.6 to 10/12 (1)
6
(c)
(i)
at 25 cm3 (1)
3
4
10
(ii)
20
30
40
50
indicator : thymol blue (1) consequential on vertical part of graph
reason: pH change sharp around pKin value / its colour changes around end point
pH / band pH8 to 10 shown on graph (1)
2
Maltby Academy
20
(d)
(i)
(ii)
fully dissociated and reactions identical
OR
H+ + OH–  H2O (1)
HCN weak acid / partially dissociated
Hionisation of HCN endothermic (1)
1
(1)
2
[19]
10.
(a)
(i)
Restricted / no rotation about the C=C bond (1)
a
d
C
C
b
(ii)
(b)
(i)
(ii)
in the case of sorbic acid a = b and d = e (1)
e
The four groups do not need to be different
2
Four (1)
two for each double bond (1)
2
A system that resists changes of pH (1)
on the addition of small amounts of H+ or OH–
RCOOH
RCOO–
(1)
2
+ H+ (1)
This equation must show
RCOOK  RCOO–
+ K+ (1) or high concentration of sorbate ion from the
complete ionisation of the salt
This equation, if given, must not show
first equilibrium suppressed (1)
When H+ added : RCOO–
+ H+  RCOOH
(1)
When OH– added : RCOOH + OH–  RCOO– + H2O reducing the effect of
the addition of the OH– (1)
Judge quality of language at this point but show it at the end of the question
as a whole (1)
4max
+
Q of L
(c)
(i)
CH 3 CH
CH
CH
CHCOOH + 2 Br 2
product can be shown as CH3(CHBr)4COOH
Maltby Academy
CH3 CH
CH
CH
CHCOOH
Br
Br
Br
Br
(1)
1
21
(ii)
Molar mass of sorbic acid = (6 × 12) + 8 + (2 × 16) = 112 g mol–1
(1)
amount of sorbic acid = 7.00/112 = 0.0625 mol (1)
amount of bromine
(d)
(i)
(ii)
=
0.0625 × 2 = 0.125 mol (1)
concentration of bromine solution = 0.125/0.125
= 1.00 mol dm–3 (1)
4
Aqueous ethanol favours substitution whereas pure ethanol
favours elimination (1)
1
CH3 CH
CH
CH
CHCOOH
OH
OH
OH
OH
(1)
can also show ----COO –
This is consequential on (c)(i)
(iii)
1
Large number of OH groups on product will hydrogen bond with water so
soluble (1)
Non–polar chain in bromo– compound inhibits solubility/ cannot hydrogen bond
extensively with solvent /only one –OH group / intermolecular bonds stronger than
bonds formed with water (1)
2
(e)
(i)
Long hydrophobic chain with polar end (1)
1
C17 H 35 COOCH 2
C17 H 35 COOCH 2
(ii)
+ 3NaOH
C17 H 35 COOCH 2
(1)
CH 2 OH
3 C17 H35 COONa + CHOH
(1)
CH 2 OH
Any plausible fat will do but it must be an ester of glycerol for full marks. If ester
is not of glycerol but has at least two acid groups max 2
Use of R in place of Cl7H35 is acceptable
3
[25]
Maltby Academy
22
11.
(a)
(i)
(ii)
HCl: pH = 1.13
 [H+] = 0.074 mol dm–3
 [HCl] = 0.074 mol dm–3 [0.074 to 0.07413]
1
HOCl: pH = 4.23
 [H+] = 5.89 × 10–5 mol dm–3 (1)
Ka = [H+] [OCl–] (1)
[HOCl]
[H+] = [OCl–] (1) or implied later in calculation
[HOCl] = [H+]2/ Ka = 0.0932 mol dm–3 (1)
(b)
(i)
[H+] = 0.10 / 0.1047 / 0.105 (1)
(ii)
H2SO4  H+ + HSO4–(1)
 or
+
–
H2SO4 + H2O  H3 O + HSO4 ignore state symbols
HSO4–
H+ + SO42– (1)
H2SO4 + H2O
(c)
H3O+ + SO42– ignore state symbols
second ionisation suppressed by the first ionisation (1)
(i)
Kc 
[Cl 2 ] 2  [H 2 O] 2
(i)
(ii)
4
1
[HCl]4  [O 2 ]
4HCl
+
O2
2Cl2
equilibrium mols
0.20
[ ] eq ÷ 10 (1)
0.020
Kc = [0.030]2 x [0.030]2
[0.020]4 x [0.005]
(d)
Must be
(iii)
(ii)
4
+
2H2O
0.050 (1) 0.30 and 0.30 (1)
0.0050
0.030
0.030
= 1010 or 1012 or 1013 or 1012.5
(mol –1 dm3) (1)
As reaction (left to right) is exothermic (1)
Decrease in temperature drives equilibrium to from left to right (1)
4
2
As more (gas) molecules on the left (1), equilibrium is driven from left to right
(1)
2
(iii)
A catalyst has no effect (1)
As it only alters the rate of the reaction not the position of equilibrium / it alters the
rate of the forward and reverse reactions equally (1)
2
[20]
Maltby Academy
23
12.
(a)
(i)
or H2N – C6H4 – CO2C2H5
Benzocaine is
H2 N
CO 2 C 2 H 5
(1)
(1)
(NH2 group in any position on ring) (1 for each group);
Either:
RNH2 + H+  RNH3+ (1)
RNH2 + CH3COCl  RNHCOCH3 + HCl (1)
RCO2C2H5 + NaOH  RCO2Na + C2H5OH (1)
RCOONa + HCl  RCO2H + NaCl (1)
Or:
amines are bases and react with acids (1)
Amines react with acid chlorides to form amides (1)
Esters are hydrolysed by alkali to salt and alcohol (1) The carboxylic acid is
formed on acidification. (1)
6
(ii)
RCOOH + NaHCO3  RCOONa + H2O + CO2 (1)
CO 2 H + HNO2 + H +
NH2
+
N2
CO 2 H + 2H 2 O
(1)
HO2 C
+
N2 +
OH
HO 2 C
N
N
OH
(1)
(iii)
(b)
(c)
(i)
Coupling in any position on phenol ring
3
Substance X forms a zwitterion / dipolar molecule/ compound of formula
+
H3N–
–COO– (1)
Ionic substances tend to be soluble in water and have high melting
temperatures (1)
Quality of language (1)
3
[HA] = 21.37g / 137g mol–1 (1) = 0.156 mol dm–3
Ka = 1.20 × 10–5 mol dm–3 (1)
[H+] = [A–] =  (Ka × [HA]) (1) = 1.37 × 10–3 mol dm–3
pH = 2.86 (1)
4
(ii)
[A–] = 0.0500mol dm–3 (1)
[acid] = (moles of original acid – moles NaOH)/0.100 (1) = 0.0280 mol dm–3
[H+] = Ka [HA]
[A–]
= 1.20 × 10–5 × (0.0280/0.0500) (1) = 6.72 × 10–6
pH = –lg[6.72 × 10–6 ] (1) = 5.17
Alternative answer route via Henderson Equation acceptable
4
(i)
Strong acid/strong base:
Maltby Academy
H+ + OH–  H2O or H3O+ + OH–  2H2O (1)
24
HA + H2O  H3O+ + A– H = + 8.3 kJ mol–1
H3O+ + OH–  2H2O
H = – 57.2 kJ mol–1
HA + OH–  A– + H2O H = + 8.3 – 57.2 = – 48.9 kJ mol–1 (2)
(ii)
Answer alone scores 1 mark only
3
Ionisation is endothermic (1) Ka increases and pH falls (1)
2
[25]
13.
(a)
(i)
CH3COOH + H2O
CH3COO– + H3O+ (1)
If no water present (0)
(ii)
(iii)
1
[H 3 O  ][CH 3 COO – ]
Ka =
(1)
[CH 3 COOH]
allow [H+] instead of [H3O+]
1
[H+]2 = Ka [acid] / some recognition that [H+] = [anion] (1)
[H+] = 1.8 × 10–5 × 2.00 (1)
pH = –log10 (6 × 10–3) = 2.2(2) (1)
or
pH = ½ pKa – ½ log10 [acid] (1)
= ½ × 4.74 - ½ × 0.3 (1) = 2.2(2) (1)
3
O
C
O
C
H
H
O
C
C
O
H
(1)
H
H
H
C
C
C
H
H
H
H
(2)
(b)
(i)
or
2
O
CH 3
(c)
C O CH 2
CH 2
CH 3
(ii)
propyl ethanoate (1)
1
(iii)
catalyst (1)
1
(i)
0.001 × 1000 (1) × 2 (1) = 2 cm3
1cm3 scores 1 mark
2
(ii)
33 or (35 – ans on (i)) (1)
1
(iii)
33 cm3 = 3.3 × 10–2 mol NaOH = 3.3 × 10–2 mol remaining ethanoic acid (1)
1
(iv)
3.3 × 10–2 mol propanol (1)
6.7 × 10–2 mol ester (1)
Maltby Academy
25
6.7 × 10–2 mol water (1)
(v)
(d)
Keq =
3
0.67  0.67
= 4.1(1) scores (2 marks)
0.33 0.33
or
6.7  10 –2  6.7  10 –2
= 4.12 (1)
3.3  10 – 2  3.3  10 – 2
+ Some reference to volume or 100 cm3 used
e.g volumes cancel (1)
2
(i)
1st (1) and 1st (1)
2
(ii)
6 × 10–4 = k × 1 × 0.1 (1)
k = 6 × 10–3 (1) mol–1 dm3 sec–1(1)
Consequential on (i)
3
[23]
14.
(a)
(i)
NaNO3
Base2
+
H2SO4
Acid1
HNO3 +
Acid2
NaHSO4
Base1 (2)
There must be some attempt to link the acid/base pairs for second mark
If candidate writes the actual ions they must link the pairs
e.g
HNO3 and NO3–(1) H2SO4 and HSO4–(1)
2
(ii)
Nitric acid escapes / leaves the system (1)
Moves to the right / completion / the system never reaches equilibrium (1) 2
(iii)
HO
O
N
(1)
O
an unclear diagram can be rescued by a correct label
 Three centres of negative charge / three separate groups of bonding electrons / 3
sigma bond pairs / one double bond electrons repel more than the single bond
electrons leading to angles not all at 120° / localised double bond pairs same as
single bond pairs (1)
 Some argument or reasoning based on electron pair repulsion / electron pairs
or bond pairs repel as far apart as possible / equally (1)
NOT ‘bonds’ repel / atoms repel
(b)
3
The candidate must explain that weak refers to dissociation
and that dilute refers to concentration.
e.g weak - does not fully dissociate (into ions) (1)
dilute - low concentration (of moles per unit volume) (1)
Maltby Academy
2
26
(c)
(i)
2NaNO3  2NaNO2
(ii)
The (cat)ion size decreases from Na+ to Mg2+ and the charge increases or the
charge density increases (1)
The polarising power of the ion increases from Na+ to Mg2+ or a description of
what is happening during polarisation (1)
The more polarised/distorted the anion the easier it is to decompose the nitrate (1)
+
O2 (1)
1
The final mark may be gained by mention of ‘the magnesium ion distorts the
electron cloud more / weakens the N-O bond more
Or
Trend in charge and size (1)
Comparison of lattice energies of the nitrate and the product (1)
Balance in favour of one with sodium and other with magnesium (1)
3
[13]
15.
(a)
only partially dissociated / ionised / not fully dissociated (1)
into H+ ions / H3+O / proton donor (1)
2
(b)
Ka =
[H 3 O  ][A – ]
(1)
[HA]
1
(c)
(i)
9.0 to 9.4 (1)
1
(ii)
9.0 to 9.4 (1)
or same answer as (c)(ii)
1
pKa = 5.6 (1)
Ka = 2.5 × 10–6 (1) consequential
2
(a solution that) resists change in pH / retains an almost
constant pH (1)
on addition of small quantities of acid or alkali (1)
2
5.2 to 5.8 (1)
5.5 or 5.6 (1) or answer from (c) (iii) based on misreading scale
of graph, eg. 4.8
2
Phenolphthalein (1)
indictor changes colour between pH 7 and 10 this is vertical part of graph (1)
methyl orange would change in acid / give pH between pH4 and pH6 (1)
n. b. must be +ve statement about methyl orange
3
exothermic reaction / heat (energy) released during reaction (1)
HCl is strong acid fully ionised (1)
this is weak acid so some energy used for dissociation (1)
3
(iii)
(d)
(i)
(ii)
(e)
(f)
Maltby Academy
27
(g)
(i)
pH = log(10) [H+] or in words
1
(ii)
1.8 × 105 =
[H  ] 2
(1)
1
[H+] = 1.8 × 105 = 4.24 × 103 (1)
pH = log (4.24 × 103)
= 2.37/2.4 (1) 2 to 4 sig. figs.
3
[21]
16.
(a)
Few molecules dissociate (into protons) / partially dissociated / ionised (1)
Not fully dissociated scores zero
(b)
Maintains an almost constant pH / resists change in pH (1)
with the addition of small amounts of acid or alkali (1)
Maltby Academy
1
2
28
14
12
pH 10
8
6
4
(c)
2
0
(d)
(e)
10
20
30
40
50
Volume of alkali added / cm 3
starting pH (1) at 2.8
endpoint (1) vertical between 6 and 11 including 7-10
vertical (1) at 25 cm3
general shape (1) finish above 12
4
Almost horizontal area marked on graph (1)
1
(i)
Ka 
[CH 3 COO  ][H ]
(1)
[CH 3 COOH]
or
Ka 
(ii)
[CH 3 COO  ][H3 O  ]
(1)
[CH 3 COOH]
pH = pKa at half way to neutralisation point = 12.5 cm3 (1)
This could be shown on the graph
because pH = pKa when [CH3COO] = [CH3COOH] (1)
1
2
[11]
Maltby Academy
29