Doc Brown's
GCE AS A2 A Level Chemistry
 Advanced
Level Chemistry Revision on Volumetric Titrations
(reedit)
GCE A
Level ASA2 IB Acidbase and other nonredox volumetric titration
quantitative calculation ANSWERS to PART 2
Questions 21 to onwards
PART 2 also includes some A Level gas volume and
gravimetric questions as well as more acidalkali and acidcarbonate
titrations and standardising hydrochloric acid calculations
Calculations
index  all academic levels
All my advanced level
preuniversity chemistry notes
The basics
of volumetric titrations and calculations
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brown (email if you think there is an error)
PART 1 Questions
PART 1
Question Answers
PART 2 Questions
Redox
Titration Q's
Qualitative
Analysis
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 Q21 to Q33
Most of the answers have been
rounded up or rounded down to three significant figures (3sf)
Q21 ANSWERS
(a)
mol HCl = 0.100 x
10.5/1000 = 0.00105
from the neutralisation
equation Ca(OH)_{2}(aq) + 2HCl(aq) ===> CaCl_{2}(aq) +
2H_{2}O(l)
2 mol HCl needed to
neutralise 1 mol Ca(OH)_{2}
therefore
mol Ca(OH)_{2}
= 0.00105/2
= 0.000525
(in 25 cm^{3} or 25/1000 = 0.025 dm^{3})
Concentration of calcium
hydroxide
= 0.000525/0.025
=
0.0210 mol dm^{3}
(b) (i)
Mr[Ca(OH)_{2}]
= 74, mass = mol x formula mass
mass Ca(OH)_{2} in 1
dm^{3} = 0.021 x 74 = 1.554 g
Therefore
concentration of
Ca(OH)_{2} = 1.55 g
dm^{3}
(ii)
Since 100 cm3 is 1/10th
of 1000 cm^{3} or 1 dm^{3}
The
concentration of Ca(OH)_{2}
= 0.155 g/100 cm^{3}
Q22 ANSWERS
NaOH + HCl ===> NaCl + H_{2}O
from equation mol NaOH = mol
HCl = 0.100 x 20.55/1000 = 0.002055
25.0 cm^{3} = 25/1000 = 0.025 dm^{3}
so,
molarity NaOH
=
0.002055/0.025
= 0.0822 mol
dm^{3}
Q23 ANSWERS
mol NaOH in titration =
0.1025 x 17.65/1000 = 0.001809125
from the equation 2NaOH + H_{2}SO_{4}
==> Na_{2}SO4 + 2H_{2}O
mol H_{2}SO_{4}
= mol NaOH/2
mol H_{2}SO_{4}
= 0.001809125/2 = 0.0009046 in 25.0/1000 = 0.0250 dm^{3}
molarity H_{2}SO_{4}
= 0.0009046/0.025
= 0.0362
mol dm^{3} (3sf)
Q24 ANSWERS
In the titration: mol NaOH =
0.1 x 22.5/1000 = 2.25 x 10^{3}
Each mol of citric acid
requires 3 mol of NaOH for complete neutralisation,
therefore, mol citric acid
in 25.0 cm^{3} of the diluted solution = 2.25 x 10^{3}/3 = 7.5 x 10^{4}
mol
Since only 1/10th of the
diluted solution was used in the titration ...
... the total moles of
citric acid in the original cordial is 10 x 7.5 x 10^{4} = 7.5
x 10^{3} mol
M_{r}(citric acid, C_{6}H_{8}O_{7})
= 192, mass = mol x formula mass,
so mass of citric acid = 192
x 7.5 x 10^{3} = 1.44 g in 25.0/1000 = 0.0250 dm^{3}
So the
concentration of
citric acid
in the original cordial is 1.44/0.025
=
57.6 g/dm^{3}
Q25 ANSWERS
Mass hydrated salt = 4.28,
mass anhydrous salt = 1.89, mass water driven off = 4.28  1.89 = 2.39
M_{r}(H_{2}O)
= 18, M_{r}(Na_{2}SO_{4}) = 142
mol H_{2}O = 2.39/18
= 0.1328, mol Na_{2}SO_{4} = 1.89/142 = 0.01331
molar ratio H_{2}O/Na_{2}SO_{4}
= 0.1328/0.01331 = 9.98 i.e. ~10, therefore
x =
10,
so, in this hydrated form of
sodium sulphate, the
formula is
Na_{2}SO_{4}.10H_{2}O
Although the ratio is not
precisely 10.0 (actually 0.02 x 100/9.98, about 0.2% off!),
its pretty close given that the masses were only quoted to three
significant figures and would be highly unlikely to be any other
integer!
Q26 ANSWERS
the equation will be M +
2HCl ==> MCl_{2} + H_{2}, from equation molar ratio M :
H_{2} of 1 : 1
therefore mol M = mol H_{2}
= 75/24000 = 3.125 x 10^{3}
mol M = mass M/A_{r}(M),
therefore A_{r} = mass M/mol M = 0.428/3.125 x 10^{3} =
136.96
A_{r} is
~137
which
corresponds to barium Ba
Q27 ANSWERS
Na_{2}CO_{3}
+ 2HCl ==> 2NaCl + H_{2}O + CO_{2}
M_{r}(Na_{2}CO_{3})
= 106, so mol Na_{2}CO_{3} titrated = 0.132/106 =
0.001245
from the molar equation,
mol HCl = 2 x mol Na_{2}CO_{3} = 0.00249
molarity HCl
= mol/volume in
dm3 = 0.00249/(24.8/1000) =
0.0996 mol dm^{3} (0.0996M)
Q28 ANSWERS
(a)
Mr(Na_{2}CO_{3})
= 106, mol Na_{2}CO_{3} = 1.30/106 = 0.01236 mol
volume = 250 cm^{3}
= 25.0/1000 = 0.025 dm^{3}
therefore molarity of Na_{2}CO_{3}
solution = 0.01236/0.025 =
0.04904 mol dm^{3} (4sf)
(b) mol Na_{2}CO_{3}
in each titration = 0.04904 x 25.0/1000 = 0.001226
from the molar equation
Na_{2}CO_{3} + 2HCl ==> 2NaCl + H_{2}O + CO_{2}
each mole of Na_{2}CO_{3}
needs two moles of HCl for complete neutralisation
therefore mol HCl =
0.001226 x 2 = 0.002452 in 24.35 cm^{3} (or 24.35/1000 =
0.02435 dm^{3})
therefore molarity HCl
=
0.002452/0.02435 = 0.1007
mol dm^{3} (0.101 M 3sf)
Q29 ANSWERS
From equation mol Na_{2}CO_{3}
= mol HCl/2, M_{r}(Na_{2}CO_{3}) = 106, mass
= mol x M_{r}
Analysis (i)
(a) mol HCl = 1.00 x
20.95/1000 = 0.02095, (b) mol Na_{2}CO_{3} =
0.02095/2 = 0.010475
(c) mass titrated based
on HCl titre = 0.010475 x 106 = 1.11035g, (d) % purity = 100 x
1.11035/1.113 =
99.76%
Analysis (ii)
(a) mol HCl
= 1.00 x
20.55/1000 = 0.02055, (b) mol Na_{2}CO_{3} =
0.02055/2 = 0.010275
(c) mass
based on HCl
titre = 0.010275 x 106 =
1.089g
(d) % purity = 100 x
1.08915/1.092 =
99.74%
Analysis (iii)
(a) mol HCl
= 1.00 x
21.90/1000 =
0.0219
(b) mol Na_{2}CO_{3} = 0.0219/2
= 0.01095
(c) mass based on HCl
titre = 0.01095 x 106 =
1.1607g
(d) % purity = 100 x 1.1607/1.166
=
99.54%
Most likely analysis result
All three values are
reasonably close together, so the best estimate would be to average
them
(99.76 + 99.73 +
99.55)/3 = 99.68%, 99.7%
(1dp, 3sf) the analysis method is not really accurate enough for 4sf
In this method the main
error is from the titration value, where an error of ≥0.05 cm^{3}
is likely.
For a titration of 21 cm^{3},
this equates to an error of 100 x 0.05/21 = 0.24%, so quoting beyond
3 significant figures or 1 decimal place is inappropriate i.e. the
analytical result is best quoted as
99.7^{+}/_{}0.2%
Q30 ANSWERS
Na_{2}CO_{3}
+ 2HCl ==> 2NaCl + H_{2}O + CO_{2}, 1 mol
hydrochloric acid relates to 0.5 moles of sodium carbonate
mol HCl used = 0.100 x
24.65/1000 = 0.002465, mol Na_{2}CO_{3} titrated =
0.002465/2 = 0.0012325
M_{r}(Na_{2}CO_{3})
= 106, so mass Na_{2}CO_{3}`titrated = 106 x 0.0012325 =
0.130645
Therefore mass of H_{2}O
in sample= 0.352  0.130645 = 0.221355 g
mole ratio Na_{2}CO_{3}
: H_{2}O is therefore 0.0012325 : 0.221355/18, giving 0.0012325
: 0.0122975
diving through by 0.0012325
gives a ratio of 1 : 9.98 (only a 0.2% error if x = 10)
Therefore the value of
x can be reliably deduced as
10,
since it would be expected to be an integer.
i.e. the formula of hydrated
sodium carbonate crystals ('washing soda') is Na_{2}CO_{3}.10H_{2}O
Q31 ANSWERS
(a)
mass CuSO_{4}.xH_{2}O
= 3.33, mass anhydrous CuSO_{4} = 2.13g
therefore mass of water
driven off = 3.33  2.12 = 1.21 g
% water of
crystallisation
= 100 x 1.21/3.33 =
36.3%
(63.7% is CuSO_{4})
(b)
M_{r}(H_{2}O)
= 18, M_{r}(CuSO_{4}) = 159.5
using the % composition
from (a) you can calculate a mole ration based on 100g of the
hydrated salt.
mol CuSO_{4} =
63.7/159.5 = 0.399, mol H_{2}O = 36.3/18 = 2.016
the mole ratio H_{2}O/CuSO_{4}
= 2.016/0.399 = 5.05
To within a 1% error
x = 5,
so the
formula of hydrated copper(II) sulfate is
CuSO_{4}.5H_{2}O
(Note: assuming 5 is the
answer, the error would be 100 x (5.05  5.0)/5 = 1.0%, which
would be quite acceptable.
Q32 ANSWERS in
preparation
(a)(i) On adding the acid to
the alkaline carbonate mixture solution, what are the colour changes for
the endpoints of titration (1) and (2)?
In titration (1) phenolphthalein
turns from pink to colourless (exactly when it becomes first
colourless!)
In titration (2) methyl orange
changes from yellow to the first orange colour.
Detailed theory on acidbase
titrations, pH curves and use of indicators
(a)(ii) Give the equations for
what happens in each titration.
In titration (1) only the Na_{2}CO_{3}
is halfneutralised to NaHCO_{3}
(i) Na_{2}CO_{3}
+ HCl ==> NaCl + NaHCO_{3}
In titration (2) all of the
carbonates are neutralised in two stages
(i) Na_{2}CO_{3}
+ HCl ==> NaCl + NaHCO_{3}
(ii) NaHCO_{3} + HCl
==> NaCl + H_{2}O + CO_{2}
(b)(i) Calculate the moles of
Na_{2}CO_{3} in the prepared solution and calculate its
molarity.
From titration (1) the Na_{2}CO_{3}
is halfneutralised by 9.80 cm^{3} of HCl.
From the equation (i) 1 mole of
HCl reacts with 1 mole of Na_{2}CO_{3}
mol = molarity x volume
Therefore mol HCl = mol Na_{2}CO_{3}
= 0.5000 x 9.80/1000 = 0.0049 mol
Since each aliquot titrated is
1/10th of the prepared solution, total mol Na_{2}CO_{3}
= 10 x 0.0049 = 0.049
250 cm^{3} = 0.25 dm^{3},
so molarity Na_{2}CO_{3} = 0.049/0.25 =
0.196 mol/dm^{3}
(b)(ii) Calculate the mass of
sodium carbonate in the solution and hence the percentage Na_{2}CO_{3}
in the original solid mixture.
mass Na_{2}CO_{3}
= M_{r}(Na_{2}CO_{3}) x mol = 106 x 0.049 =
5.194 g
% Na_{2}CO_{3}
= 5.194 x 100 / 7.357 =
70.6%
Extra calculations for further
practice and will partly help you to solve Q33
(c)(i) In the titrations, what
total volume of the HCl was used to neutralise the Na_{2}CO_{3}?
Since it took 9.80 cm^{3}
HCl to effect
Na_{2}CO_{3}
+ HCl ==> NaCl + NaHCO_{3}
on an equimolar basis, it takes
another 9.8 cm3 HCl to effect NaHCO_{3} + HCl
==> NaCl + H_{2}O + CO_{2}
Therefore a total volume of
19.60 cm^{3} HCl would be required to totally neutralise
the Na_{2}CO_{3}
(c)(ii) In titration (2) what
volume of HCl was used to neutralise the NaHCO_{3} in the
original mixture?
If 19.60 cm3 HCl was used to
neutralise the Na2CO3 in the original mixture, then,
the volume of HCl needed to
neutralise the original NaHCO_{3} must = 24.75  19.60 =
5.15 cm^{3} HCl
(c)(iii) Calculate the number
of moles of NaHCO_{3} in the 25.0 cm^{3} aliquot at the
start and the total moles in the prepared solution.
from ? one mole of HCl reacts
with 1 mole NaHCO_{3}, therefore
moles HCl = mol NaHCO_{3}
= 0.5000 x 5.15 / 1000 = 0.002575
therefore total moles = 0.002575
x 10 = 0.02575 mol NaHCO_{3} in the prepared
solution.
(c)(iv) Calculate the mass off
NaHCO_{3} in the prepared solution, and hence its % in the
original solid mixture.
mass = moles x formula mass, mass
= 0.02575 x 84 = 2.163 g
% NaHCO_{3} in mixture
= 2.163 x 100 / 7.357 =
29.4%
If your answer to (b)(ii) +
(c)(iv) do not add up to 100%, you have made an error
somewhere!
Q33 ANSWERS
(a) Explaining the titrations ...
There are three bases present, in
order of strength: (i) OH^{} > (ii) CO_{3}^{2}
> (iii) HCO_{3}^{}, so this gives the order
in which they will be neutralised, shown in equations (i) to (iii) below ...
Titration (1) using
phenolphthalein indicator corresponds to titration of NaOH plus Na_{2}CO_{3}
(but only half neutralised to NaHCO_{3})
The sequence of reaction is as follows as the pH
decreases on adding the acid ...
(i) NaOH + HCl ==> NaCl + H_{2}O
(normally a sharp inflection at pH 7 in the pH curve, but presence
of Na_{2}CO_{3} blurs this and the pH is much higher)
actual ionic equation: OH^{}_{(aq)}
+ H^{+}_{(aq)} ==> H_{2}O_{(l)}
(ii) Na_{2}CO_{3}
+ HCl ==> NaCl + NaHCO_{3} (complete at ~pH 89, clear
2nd inflection in the pH curve and detectable with an appropriate
indicator)
actual ionic equation: CO_{3}^{2}_{(aq)}
+ H^{+}_{(aq)} ==> HCO_{3}^{}_{(aq)}
The pK_{ind} of
phenolphthalein (9.3) can detect the endpoint at the end of the
reaction (ii).
In this titration you will NOT
observe effervescence at all, except if you overshoot the endpoint
by some margin, the colour change is pink to colourless.
Titration (2) using methyl
orange indicator corresponds to the titration of NaOH + all of Na_{2}CO_{3}
The sequence of reactions is as
follows as the pH decreases on adding the acid ...
(i) NaOH + HCl ==> NaCl + H_{2}O
(normally a sharp inflection at pH 7 in the pH curve, but presence
of Na_{2}CO_{3} blurs this and the pH is much higher)
actual ionic equation: OH^{}_{(aq)}
+ H^{+}_{(aq)} ==> H_{2}O_{(l)}
(ii) Na_{2}CO_{3}
+ HCl ==> NaCl + NaHCO_{3} (complete at ~pH 89, clear
2nd inflection in the pH curve and detectable with an appropriate
indicator)
actual ionic equation: CO_{3}^{2}_{(aq)}
+ H^{+}_{(aq)} ==> HCO_{3}^{}_{(aq)}
(iii) NaHCO_{3} + HCl
==> NaCl + H_{2}O + CO_{2} (complete at ~pH 4,
clear 3rd inflection in the pH curve and detectable with an
appropriate indicator, effectively you have a solution of 'carbonic
acid' at the end of the titration)
actual ionic equation: HCO_{3}^{}_{(aq)}
+ H^{+}_{(aq)} ==> H_{2}O_{(l)}
+ CO_{2(aq/g)}
Note that (ii) + (iii) add up to
(iv) Na_{2}CO_{3} + 2HCl ==> 2NaCl + H_{2}O
+ CO_{2}
AND reaction (iii) is the
difference between the two titrations.
The pK_{ind} of methyl
orange is 3.7 and matches the pH of endpoint of the reaction (iii)
when everything is completely neutralised, colour change is yellow
to first trace or orange.
In this titration you may observe effervescence in the final stages as carbon dioxide is
formed.
Note: You can do this titration
exercise as a double indicator single titration. The reason is
quite simple.
When titration (1) is complete, the
phenolphthalein has gone colourless {reactions (i) + (ii) completed},
so, at this point, you can add the methyl orange indicator and continue
the titration to give the extra HCl needed to complete the
neutralisation {to complete reaction (iii)}.
Detailed theory on acidbase
titrations, pH curves and use of indicators
(b) Calculating the molarity of the
sodium carbonate formed by solution X on absorbing CO_{2}
The difference between the titrations
is the reaction (iii) NaHCO_{3} + HCl ==> NaCl + H_{2}O + CO_{2}
From this we can calculate the
concentration of NaHCO_{3}, hence the original concentration of Na_{2}CO_{3}
titration (2)  (1) = 22.55 
12.45 = 10.10 cm^{3} of 1.000 mol/dm^{3} HCl
therefore: mol HCl reacting with
NaHCO_{3} = 1.000 x 10.10/1000 = 0.0101
From the titration equation: NaHCO_{3}
+ HCl ==> NaCl + H_{2}O + CO_{2}
mol NaHCO_{3} = mole HCl =
0.0101, {and 25 cm^{3} = 25/1000 = 0.025 dm^{3}, also required in (c)}
molarity = mol / volume in dm^{3},
{and 25 cm^{3} = 25/1000 = 0.025 dm^{3}, also required in (c)}
molarity NaHCO_{3} formed =
0.0101/0.025 = 0.404 mol/dm^{3}
but in reaction (ii) each mol of Na_{2}CO_{3}
gives 1 mol of NaHCO_{3}
so, the original molarity of Na_{2}CO_{3}
formed
= 0.404 mol/dm^{3}
(c) Calculating the molarity of the
unreacted sodium hydroxide
Since 10.10 cm^{3} of the HCl
was used to effect ...
(iii) NaHCO_{3} + HCl ==> NaCl + H_{2}O
+ CO_{2}
Prior to this, on an equimolar basis,
another 10.10 cm^{3} of the
HCl would be used to effect ...
(ii) Na_{2}CO_{3} + HCl ==> NaCl + NaHCO_{3}
Therefore a total of 20.20 cm^{3} of
the HCl was used to completely neutralise the Na_{2}CO_{3}
therefore 22.55  20.20 = 2.35 cm^{3}
of the HCl was used to neutralise the residual NaOH
mol HCl used = 1.000 x 2.35/1000 =
0.00235
since titration equation is HCl +
NaOH ==> NaCl + H_{2}O,
mol HCl = mol NaOH, so mol NaOH =
0.00235
molarity = mol / volume in dm^{3},
25 cm^{3} = 0.025 dm^{3}
molarity NaOH = 0.00235/0.025
= 0.094 mol/dm^{3}
(d) Calculating the volume of carbon
dioxide gas absorbed by the NaOH (molar gas volume = 24 dm^{3} at
RTP)
from calculation (b) molarity of Na_{2}CO_{3}
= 0.404 mol/dm^{3}
and from the equation for CO_{2}
absorbed: 2NaOH + CO_{2} ==> Na_{2}CO_{3} + H_{2}O
mol Na_{2}CO_{3} =
mol CO_{2} , so in 2 dm^{3} of the solution there is
the equivalent of 2 x 0.404 mol = 0.808 mol CO_{2}
so volume CO_{2}
absorbed = 0.808 x 24 = 19.392 = 19.4 dm^{3} (3sf,
1dp)
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