Doc Brown's
GCE AS A2 A Level Chemistry
 Advanced
Level Chemistry Revision on Volumetric Titrations
GCE A
Level ASA2 IB Chemistry Volumetric Analysis
Acidbase and other nonredox volumetric titration
calculation answers to PART 1 Questions 1 to 20
All my advanced A level organic chemistry notes
The basics
of how to do volumetric titrations and calculations
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PART 1
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PART 2 Questions
PART 2 Question Answers * Redox
Titration Q's *
Qualitative
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ILLUSTRATIONS OF ACIDALKALI
TITRATIONS and SIMPLE STARTER CALCULATIONS
I
DO MY BEST TO CHECK MY CALCULATIONS, as you yourself should do, BUT I
AM HUMAN! AND IF YOU THINK THERE IS A 'TYPO' or CALCULATION ERROR PLEASE
EMAIL ME ASAP TO SORT IT OUT!
The
nonredox titration question ANSWERS to Q1 to Q20
Most of the answers have been rounded up or rounded
down to three significant figures (3sf)
Q1
ANSWERS (a) NaOH_{(aq)} + HCl_{(aq)} ==> NaCl_{(aq)} + H_{2}O_{(l)}
(b) (i) pipette (ii) burette
(c) everything with distilled water, then pipette with a little of the NaOH(aq) and the burette with a little of the HCl(aq)
(d) pink to colourless, the first drop of excess acid removes the pink alkaline colour of phenolphthalein
(e) moles sodium hydroxide neutralised:
0.250 x 25.0/1000
=
0.00625 mol NaOH
(remember: moles =
molarity x volume in dm^{3} and its two rearrangements and 1 dm^{3}
= 1000 cm^{3})
(f) moles HCl = moles NaOH (equation)
=
0.00625 mol HCl (in 22.5 cm^{3})
(g) concentration hydrochloric acid = 0.00625 x
1000 ÷ 22.5
= 0.278 mol dm^{3} (3sf)
(scaling up to 1 dm^{3} =
1000 cm^{3} to get the molarity)
Another way to work it out
is 22.5 cm^{3} = 22.5
÷ 1000 = 0.0225 dm^{3}
Therefore molarity =
0.00625 ÷ 0.0225 = 0.278 mol dm^{3}
Q2
ANSWERS (a) Ba(OH)_{2(aq)} + 2HCl_{(aq)} ==> BaCl_{2(aq)} + 2H_{2}O_{(l)}
(b) formula mass of Ba(OH)_{2} = 171, moles = 2.74
÷ 171 = 0.016 mol in 100 cm^{3},
(scaling up x 10)
therefore 0.16 mol in 1000 cm^{3}, so
molarity of Ba(OH)_{2} = 0.16 mol dm^{3}
(c) moles Ba(OH)_{2} used in titration = 0.16
x 20/1000
= 0.0032 mol
(d) moles HCl titrated = 2 x moles of Ba(OH)_{2} used (2 : 1 in equation)
2 x .0032
= 2 x 0.0032 =
0.0064 mol HCl in 18.7 cm^{3} of the acid solution,
18.7 cm^{3} = 0.0187 dm^{3}
(e) therefore
molarity of HCl_{(aq)} = 0.0064/0.0187
=
0.342 mol dm^{3}
Q3
ANSWERS (a) 2NaOH_{(aq)} + H_{2}SO_{4(aq)}
==> Na_{2}SO_{4(aq)} + 2H_{2}O_{(l)}
(b) moles H_{2}SO_{4} = 4.90 ÷ 98 = 0.050 mol in 200cm^{3}
scaling up to get molarity of the sulphuric acid solution, 0.050 x 1000
÷ 200
= 0.25 mol dm^{3}
(c) moles of sulphuric acid neutralised =
0.250 x 20.7/1000
= 0.005175 mol
(d) moles of sodium hydroxide neutralised = 2 x 0.005175
=
0.01035 mol (2 : 1 in equation)
(e) concentration of the sodium hydroxide
= 0.01035 x 1000
÷ 10
= 1.035 mol dm^{3} (molarity 1.04, 3sf)
Q4
ANSWERS (a) Mg(OH)_{2(aq)} + H_{2}SO_{4(aq)}
==> MgSO_{4(aq)} + 2H_{2}O_{(l)}
(b) moles of sulphuric acid neutralised =
0.100 x 4.5/1000
= 0.00045 mol
(c) moles of magnesium hydroxide neutralised also
= 0.00045 (1:1 in equation) in 100 cm^{3}
(d)
concentration of the magnesium hydroxide = 0.00045 x 1000
÷ 100
= 0.0045 mol dm^{3}
(scaling up to 1000cm^{3} = 1dm^{3},
to get molarity)
(e)
molar mass of Mg(OH)_{2}
=
58.3
so concentration of the magnesium
hydroxide = 0.0045 x 58.3
= 0.26 g dm^{3}
(= g per 1000 cm^{3}),
so concentration = 0.26
÷ 1000
= 0.00026 g cm^{3}
Q5
ANSWERS (a)(i) The magnesium oxide is insoluble
in water and so cannot be titrated directly by a volumetric analysis method. By
dissolving in excess acid you are sure to get all the MgO into solution,
and then determine the excess acid, from which you can derive the amount of MgO
that dissolved.
(a)(ii) MgO_{(s)} + 2HCl_{(aq)} ==> MgCl_{2(aq)} +
H_{2}O_{(l)}
NaOH_{(aq)} + HCl_{(aq)}
==> NaCl_{(aq)} + H_{2}O_{(l)}
(b) moles of hydrochloric acid added to the magnesium oxide =
2 x 100/1000
= 0.20 mol HCl
(c) moles of excess hydrochloric acid titrated = 19.7
÷ 1000 x 0.200
= 0.00394 mol HCl
{mole ratio NaOH : HCl is 1 : 1 from equation
(ii)}
(d) moles of hydrochloric acid reacting with the magnesium oxide = 0.20  0.00394
=
0.196 mol HCl
(e) mole MgO reacted
= 0.196 ÷ 2 = 0.098 {1: 2 in equation (i)}
the formula mass of MgO = 40.3
therefore mass of MgO reacting with acid = 0.098 x 40.3
=
3.95 g
(f) % purity
= 3.95
÷ 4.06 x 100
= 97.3% MgO
(g) Mg(OH)_{2} from MgO + H_{2}O, MgCO_{3} from the original mineral source, both of these compounds react with acid and would lead to a false titration value.
Q6
ANSWERS (a) CaCO_{3(s)} + 2HCl_{(aq)} ==> CaCl_{2(aq)} + H_{2}O_{(l)} + CO_{2(g)}
(b) moles of hydrochloric acid was spilt = 2.00 x 10.0
=
20 mol HCl
(c) moles of calcium carbonate to neutralise the acid = 20
÷ 2
= 10.0 mol CaCO_{3} (1:2 in equation)
(d) formula mass of CaCO_{3} = 100,
so mass of limestone powder needed to neutralise the acid = 100 x 10
=
1000g CaCO_{3}
(e) the neutralisation reaction is
MgO + H_{2}SO_{4}
==> MgSO_{4} + H_{2}O,
moles H_{2}SO_{4} = 1000 x 2 = 2000 mol acid,
2000 mol MgO needed (1:1 in equation),
mass MgO needed = 2000 x 40.3
=
80600 g or 80.6 kg
Q7
ANSWERS (a) 2NaOH_{(aq)} + H_{2}SO_{4(aq)} ==> Na_{2}SO_{4(aq)} + 2H_{2}O_{(l)}
(b) moles of sodium hydroxide used in the titration = 25.0
x 1/1000 = 0.025 mol NaOH
(c) mol H_{2}SO_{4} = mol NaOH ÷ 2 = 0.0125 mol in 20.0 cm^{3},
so scaling up to 1000 cm^{3} to get molarity of diluted acid = 0.0125 x 1000
÷ 20 = 0.625 mol dm^{3}
(or molarity = 0.0125 mol/0.02
dm^{3} = 0.625 mol dm^{3})
(d) scaling up from 50 to 1000 cm^{3}, gives the concentration of the original concentrated sulphuric acid solution,
= 0.625 x 1000
÷ 50
= 12.5 mol dm^{3}
Q8
ANSWERS (a) NaHCO_{3(s)} + HCl_{(aq)} ==> NaCl_{(aq)} + H_{2}O_{(l)} + CO_{2(g)}
(b) mol = molarity x volume in dm^{3},
mol acid = 0.200 x 23.75/1000 = 4.75 x 10^{3 }mol HCl
from equation HCl:NaHCO_{3}
is 1:1 by ratio
so mol HCl
= mol NaHCO_{3}
= 4.75 x 10^{3}
(c) mass = mol x formula mass, f.
mass NaHCO_{3} = 23 + 1 + 12 + (3 x 16) = 84
mass NaHCO_{3} =
4.75 x 10^{3} x 84
= 0.399 g
% purity of NaHCO_{3}
= 0.399 x 100/0.40
= 99.75% (99.8%, 3sf)
(d)(i) mol = molarity x volume in dm^{3},
mol acid = 0.200 x 20.00/1000 = 4.00 x 10^{3 }mol HCl
from equation above HCl : MHCO_{3}
is 1 : 1 by ratio
so mol HCl
= mol MHCO_{3}
= 4.00 x 10^{3}
(ii) to get to the relative
formula mass
moles = mass / M_{r},
M_{r} = mass / mol = 0.400 / 4.00 x 10^{3}
= 100
(iii) for MHCO_{3}, M_{r}(HCO_{3})
= 1 + 12 + 48 = 61, A_{r} (M) = 100  61 = 39,
which, from the periodic table
relative atomic mass values, corresponds to potassium
So the formula of the group 1
hydrogencarbonate titrated was KHCO_{3}
Q9
ANSWERS (a)(i) Na_{2}CO_{3(aq)} + 2HCl_{(aq)} ==>
2NaCl_{(aq)} + H_{2}O_{(l)} + CO_{2(g)}
9a(ii)
20.0 cm^{3} of 1.0
mol dm^{3} hydrochloric acid contains 1.0 x 20.0/1000 = 0.02 mol HCl
From the equation, 0.020 mol
HCl reacts with 0.010 mol Na_{2}CO_{3}, M_{r}(Na_{2}CO_{3})
= 106
therefore mass Na_{2}CO_{3}
titrated = 0.01 x 106 = 1.06 g per aliquot,
since 250 cm^{3} is
^{1}/_{10}th of the aliquot, 10 x 1.06 = 10.6 g of
Na_{2}CO_{3} would be used to make up the solution.
Molarity of Na_{2}CO_{3(aq)}
= (10.6 g/106 g mol^{1})/0.25 dm^{3}
= 0.40 mol dm^{3}
9(b)(i) CH_{3}COOC_{6}H_{4}COOH
+ NaOH ==> CH_{3}COOC_{6}H_{4}COO^{}Na^{+}
+ H_{2}O
9b(ii)
23.0 cm^{3} of 0.100
mol dm^{3} NaOH contains 0.100 x 23.0/1000 = 0.0023 mol NaOH
From the equation, mol Aspirin
= mol NaOH, M_{r}(CH_{3}COOC_{6}H_{4}COOH)
= 180
so need Aspirin mass of 0.0023
x 180
= 0.414 g
9b(iii)
The last stage in the
synthesis of 2ethanoylhydroxybenzoic acid ('Aspirin') is made by esterifying
2hydroxybenzoic acid with ethanoic anhydride.
M_{r}(HOC_{6}H_{4}COOH)
= 138, is 42 units less than aspirin. In terms of this particular impurity the
%
aspirin will be overestimated
for the following reason. The
2hydroxybenzoic acid will also be titrated with the aspirin, and, with its
smaller molecular mass, it will need more alkali to neutralise than aspirin
per equivalent mass of material. This can result in >100% purity!!!!
9(c)(i) It is a ligand substitution/replacement
reaction.
[Ca(H_{2}O)_{6}]^{2+}_{(aq)}
+ EDTA^{4}_{(aq)} ==> [CaEDTA]^{2}_{(aq)} +
6H_{2}O_{(l)}
more simply
Ca^{2+}_{(aq)}
+ EDTA^{4}_{(aq)} ==> [CaEDTA]^{2}_{(aq)}
or [Ca(H_{2}O)_{6}]^{2+}_{(aq)}
+ H_{2}EDTA^{2}_{(aq)} ==> [CaEDTA]^{2}_{(aq)}
+ 2H^{+}_{(aq)} + 6H_{2}O_{(l)}
more simply Ca^{2+}_{(aq)} + H_{2}EDTA^{2}_{(aq)}
==> [CaEDTA]^{2}_{(aq)} + 2H^{+}_{(aq)}
9(c)(ii) M_{r}(CaCO_{3}) = 100, mol CaCO_{3}
= mol Ca^{2+} in solution = 0.250/100
= 0.00250 mol
since 250 cm^{3} =
0.25 dm^{3},
molarity Ca^{2+} = 0.0025/0.25
= 0.010 mol dm^{3}
9(c)(iii) mole CaCO_{3} =
mol Ca^{2+} = mol EDTA used in titration.
Therefore from c(ii) mol Ca^{2+}
= mol EDTA = 0.01 x 25.0/1000 = 0.00025 mol in 25.70 cm^{3} (0.0257
dm^{3}) EDTA solution,
so
molarity EDTA = 0.00025/0.0257
=
0.00973 mol
dm^{3} (3 sf)
9(c)(iv) M_{r}(apatite) =
(5 x 40) + 3 x (31 + 4 x 16) + (16 + 1) = 502
% Ca in apatite = 200 x
100/502
= 39.8%
9(c)(v) In the titration mol Ca^{2+}
= mol EDTA,
therefore mol Ca^{2+}
= 22.5 x 0.0200/1000 = 0.00045,
since ^{10}/_{250}
of the original solution was used in the titration,
the total mol of calcium in
the tooth solution = 0.00045 x 250/10 = 0.01125 mol Ca
total mass of Ca in tooth = 0.01125 x 40 =
0.45 g
% by mass Ca in the tooth
= 0.45 x 100/1.40
= 32.1 %
Q10
ANSWERS (a) Ag^{+}_{(aq)} + Cl^{}_{(aq)} ==>
AgCl_{(s)}
(sodium ions and
nitrate ions etc. are spectator ions)
(b) from equation: moles silver
nitrate (AgNO_{3}) =
moles chloride ion (Cl^{})
moles = molarity AgNO_{3} x volume of AgNO_{3} in dm^{3}
= 0.100 x 13.8/1000 = 1.38 x 10^{3}
mol Cl^{} (in 25.0 cm^{3} aliquot)
(c) moles in 1 dm^{3}
of diluted seawater = 1.38 x 10^{3} x 1000/25 = 0.0552 (scaling up
to 1000 cm^{3})
So molarity of chloride in
diluted seawater is 0.0552 mol dm^{3}
(d) Now in the titration 25.0 cm^{3}
of the 250 cm^{3} was used,
so the molarity of chloride ion in the
original seawater must be scaled up accordingly.
molarity of chloride ion in
seawater = 0.0552 x 250/25.0
= 0.552 mol dm^{3}
(e) M_{r}(NaCl) = 23 +
35.5 = 58.5
concentration of NaCl in g
dm^{3}
= molarity x formula mass = 0.552 x 58.5
= 32.3 g dm^{3}
Q11
ANSWERS (a)
moles = molarity AgNO_{3} x volume in dm^{3} = 0.100
x 19.7/1000 = 1.97 x 10^{3} mol Cl^{} ion
[AgNO_{3}:NaCl or Ag^{+ }: Cl^{}
is 1 : 1, see Q10(a)/(b)] f. mass NaCl = 23 + 35.5 = 58.5
(b) mass = mol x formula
mass = 1.97 x 10^{3} x 58.5 = 0.1152 g NaCl
(c) % purity = 0.1152 x
100/0.12
= 96.0 % in terms of NaCl (3sf)
Q12
ANSWERS (a)
mole Cl^{} = moles Ag^{+} [=AgNO_{3}, see Q10(a)/(b)]
mole Cl^{} = molarity
AgNO_{3} x vol AgNO_{3} = 0.100 x 21.2/1000 = 2.12 x 10^{3}
mol Cl^{}
(b)
Since calcium chloride is CaCl_{2},
mol CaCl_{2} = mole Cl^{}/2 = 2.12 x 10^{3}/2
=
1.06 x 10^{3} mol CaCl_{2}
(c) M_{r}(CaCl_{2})
= 40 + (2 x 35.5) = 111
mass = mol x f. mass =
1.06 x 10^{3} x 111
= 0.1177 g CaCl_{2}
(d) Since ^{1}/_{10}th
of original solution titrated, original mass of CaCl_{2} in mixture
= 10 x 0.1177 g = 1.177g CaCl_{2}
(1.78g 3sf)
(e) Therefore % = 1.177 x 100/5.0
= 23.5% CaCl_{2} (3 sf)
and
% NaNO_{3} =
100  23.5 = 76.5% (3 sf)
Q13
ANSWERS (a)
moles = mass/f. mass, f. mass Na_{2}CO_{3} = 106,
mol Na_{2}CO_{3} = 13.25/106 = 0.125
molarity = mol/vol. in dm^{3},
250 cm^{3} = 0.250 dm^{3},
molarity Na_{2}CO_{3}
= 0.125/0.250
= 0.50 mol dm^{3}
(b) Na_{2}CO_{3(aq)} + 2HCl_{(aq)} ==>
2NaCl_{(aq)} + H_{2}O_{(l)} + CO_{2(g)}
(c) mol = molarity x volume
mol Na_{2}CO_{3}
titrated = 0.5 x 25.0/1000 = 0.0125 mol Na_{2}CO_{3} (in
the 25 cm^{3} aliquot pipetted)
(d) from equation, mole ratio Na_{2}CO_{3}:HCl
is 1:2,
so mol HCl = 2 x mol Na_{2}CO_{3}
= 2 x 0.0125 = 0.025 mol HCl (in the 24.65 cm^{3} titre)
(e) molarity = mol/vol. in dm^{3},
dm^{3} = cm^{3}/1000, 24.65/1000 = 0.02465 dm^{3}
therefore molarity HCl
= 0.025/0.02465 = 1.014 mol dm^{3} (1.01 3sf)
Q14
ANSWERS (a) mol HCl = 1.00 x 25.3/1000
= 0.0253 mol
(b) reacting mole ratio, Na_{2}CO_{3}:HCl
is 1:2
so
mol Na_{2}CO_{3}
titrated = mol HCl/2 = 0.0253/2
= 0.01265 mol
(c) mass = mol x f. mass, so
mass Na_{2}CO_{3} = 0.01275 x 106
= 1.34 g
therefore
% purity of Na_{2}CO_{3}
= 1.34 x 100/1.35
= 99.3% (3sf)
Q15
ANSWERS (a) An appropriate quantity of the acid is weighed out, preferably
on a 4 sf electronic balance. It can be weighed into a conical flask by
difference i.e. weight acid added to flask = (weight of boat + acid)  (weight
of boat). The acid is dissolved in water, or aqueousethanol if not very
soluble in water. The solution is titrated with standard sodium hydroxide
solution using phenolphthalein indicator until the first permanent pink.
The burette should be rinsed with the sodium
hydroxide solution first. During the titration the flask should be rinsed around the inside to
ensure all the acid and alkali react, and dropwise addition close to the endpoint
to get it to the nearest drop  the first permanent pink colour.
The pK_{ind} for
phenolphthalein is 9.3, and its effective pH range is 8.3 to 10.0. The pH of a
solution of the sodium salt of the acid (from strong base + weak acid) is in
this region and so the equivalence point can be detected with this indicator.
(b) moles = molarity x volume
in dm^{3} (dm^{3} = cm^{3}/1000)
mol NaOH = 0.100 x 20.5/1000 =
0.00205 mol
(c) mol NaOH = mol RCOOH =
0.00205
because 1:1 mole ratio for a
monobasic acid: RCOOH + Na^{+}OH^{} ==> RCOO^{}Na^{+}
+ H_{2}O
(d) moles = mass (g)/M_{r},
so M_{r} = mass/mol = 0.279/0.00205 = 136.1
(e) The simplest aromatic
carboxylic acid is benzoic acid C_{6}H_{5}COOH, M_{r} =
122
136122 = 14, which suggests an
'extra' CH_{2} (i.e. CH_{3} attached to the benzene ring
instead of a H), so, since the COOH is attached to the ring, there are three
possible positional/chain isomers of CH_{3}C_{6}H_{4}COOH
(M_{r} = 136)
2, 3 or 4methylbenzoic acid.
Q16
ANSWERS (a)
moles = molarity x volume in dm^{3} (dm^{3} =
cm^{3}/1000)
mol NaOH = 0.0995 x
19.85/1000 = 0.001975 mol
(b) The titration reaction for
complete neutralisation is:
R(COOH)_{2} + 2Na^{+}OH^{}
==> R(COO^{}Na^{+})_{2} + 2H_{2}O
this 1 : 2 reaction mole ratio
means that mol dibasic acid = mol NaOH/2 = 0.001975/2 = 0.0009875
(c) moles = mass (g)/M_{r},
so M_{r} = mass/mol = 0.103/0.0009875 = 104.3 (approx. 104
3sf)
(d) Since a dibasic acid, and 2
x COOH = 2 x 45 = 90 mass units, the remaining 14 units could be CH_{2},
and so the structure is likely to be
HOOCCH_{2}COOH,
propanedioic
acid (malonic acid), M_{r} = 144
Q17
ANSWERS (a) mol NaOH = 0.100 x 19.25/1000 =
0.001925
(b) mol NaOH = mol acid =
0.001925, mol ratio 1:1, C_{6}H_{5}COOH + Na^{+}OH^{} ==>
C_{6}H_{5}COO^{}Na^{+}
+ H_{2}O
M_{r} (C_{6}H_{5}COOH) =122, mol = mass (g)/M_{r}
or mass =
mol x M_{r}
so mass acid = 0.001925
x 122 = 0.2349 g
(c) % purity = actual mass of acid
titrated x 100 / mass of original sample
% purity = 0.2349 x
100/0.236 = 99.5% (3 sf)
Q18
ANSWERS (a) M_{r} (C_{6}H_{5}COOH) =122, mol acid = mass (g)/M_{r} = 0.250/122 =
0.002049 mol
(b) mol alkali = 0.002049
mol, since mol acid, 1:1 mole ratio in reaction (see Q17(b)).
(c) Since 0.002049 mol NaOH in
22.5 cm^{3}, so scaling up
molarity NaOH = 0.002049
x 1000/22.5 = 0.0911 mol dm^{3} (3 sf)
Q19 ANSWERS (a) The ratio of 2M/0.1M is 20, so need to do 20 fold dilution.
So 25.0 cm^{3}
of 2M diluted to 500 cm^{3} gives an approximately 0.1 mol
dm^{3} solution. You can do this with a measuring cylinder
and beaker.
In practice you
could make up 12.5 cm^{3} of the 2M acid up to 250 cm^{3}.
You could do this with a burette and a 250 cm^{3} standard
volumetric flask, but standardisation of the acid is still required.
(b) The solubility of calcium
hydroxide is low, so it would give a very inaccurate tiny titration
value with relatively concentrated acid. For any accurate work e.g. to
3sf, standardisation of reagents is required.
(c) 1 x 50.0 cm^{3}
pipette or 2 x 25.0 cm^{3} pipette would be the most convenient
(or accurate burette?).
Phenolphthalein is used for strong basestrong acid titrations.
(d) Ca(OH)_{2(aq)} + 2HCl_{(aq)} ==>
CaCl_{2(aq)} + 2H_{2}O_{(l)}
(e) from equation,
mole ratio Ca(OH)_{2} : HCl is 1:2
and since moles
solute = molarity x volume in dm^{3}
mol HCl used in
titration = 0.1005 x 15.22/1000 = 0.001530 mol HCl
therefore mol Ca(OH)_{2}
= 0.001558/2 = 0.000765 mol Ca(OH)_{2}
(f) Scaling up from mol
Ca(OH)_{2} in 50 cm^{3} to 1 dm^{3} (1000 cm^{3})
molarity Ca(OH)_{2}
= 0.00765 x 1000/50 = 0.1558 = 0.153 mol dm^{3} (to
3sf)
(g) Mr[Ca(OH)_{2}]
=74, so solubility in g dm^{3} = 0.153 x 74 = 1.13 g dm^{3}
(3 sf)
Since density of
water is ~1.0 g cm^{3}, the solubility of Ca(OH)_{2}
is about 0.113 g/100 cm^{3} H_{2}O
Q20 ANSWERS (a)
+ CO_{2} === heat/pressure/NaOH ==>
(sodium salts status
ignored)
(b)
+
== reflux ==>
+
(c) 2hydroxybenzoic acid M_{r}(C_{7}H_{6}O_{3})
= 138.1, aspirin M_{r}(C_{9}H_{8}O_{4})
= 180.3
(d) Aspirin is not very soluble in pure water and is
much more soluble in ethanol solvent.
(e) The titre/mass data: Note: If you use average mass/average titre you get
55.73 cm^{3} 0.1000M NaOH/g
Note that Q9b explains why
0.350.40g is a convenient mass to weigh out for an individual
titration.
mass of
aspirin (g) 
titration/cm^{3} of 0.1M
NaOH 
titre/mass 
0.3591 
20.05 
55.83 
0.3532 
19.65 
55.63 
0.3686 
20.60 
55.89 
0.3583 
19.90 
55.54 
0.3635 
20.25 
55.71 
av titre/mass =
55.72 cm^{3}/g 
(f) (i) 55.72 cm^{3}
(ii)
+ NaOH ==>
+ H_{2}O
(iii) From the titration
equation, 1 mole of aspirin = 1 mole of NaOH
therefore mol aspirin =
mol NaOH = 55.72 x 0.1000 / 1000 = 0.005572 mol (mol = molarity x
volume in dm^{3})
(iv) mass = Mr(aspirin)
x 0.005572 = 180.3 x 0.005572 = 1.00463g
(v) % purity = 100 x g
mass titrated/ 1g = 100.5% (1dp, 4sf)
(g) You need to reread the opening paragraph of the question,
particularly the last line. The problem with this titration is that you also
titrate the 2hydroxybenzoic acid impurity. This has a smaller molecular mass
than aspirin, therefore for equivalent masses, the 2hydroxybenzoic acid will
give more moles to react with the NaOH and so give a false, and over estimated
value of moles of aspirin, hence a % purity of >100.
(h) (i) The moles of acid = 0.005572 for 1g,
since M_{r} = mol/mass, M_{r}(av) = 1/0.005572 = 179.5
(ii) M_{r}(av) =
179.5 = {x X 138.1(C_{7}H_{6}O_{3})
+ (100x) X 180.3(C_{9}H_{8}O_{4})}
/ 100
where x = the % of
2hydroxybenzoic acid, and multiplying through by 100 gives ...
17950 = 138.1x + 18030 
180.3x
which on rearranging gives
...
42.2x = 80, x = 80/42.2
= 1.9
therefore the aspirin
contained 1.9% impurity of 2hydroxybenzoic acid.
(i) (i) % error = ~ 0.1 x 100
/20 = 0.5% (this would be good for any student!)
(Other sources of error are ignored for the purpose of this calculation)
(ii) The error range
would be 179.5 ±0.5% of 179.5, 179.5±0.9, giving a range of 178.6 to
180.4.
(iii) Using an M_{r}(av)
of 178.6 gives 4.0% of 2hydroxybenzoic acid, 180.4 gives 0.24% !
(iv) This means, for a not
unreasonable error range based on the titration, that a wide range of
values of the % impurity in the aspirin, including the possibility of a
ve percentage. One must conclude that this is not a very accurate
method for small percentages of this particular acidic impurity. It
seems at first that the 0.5% error seems ok, but it isn't in reality,
the error in the method is comparable to the actual % of the impurity!
(v) For low % impurity
you need, if possible, a method that directly, as well as accurately
measures the impurity. In this case you can use colorimetry, e.g.
measure the absorbance of the colour produced by reacting the
2hydroxybenzoic acid with iron(III) chloride solution. This gives a
purple colour and is a test for the phenol group. (I haven't written
this up yet!, but I have described the
use of colorimetry in
determining the formula of a transition metal complex which outlines
the essential principles.)
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