**
** **
**
**
Doc Brown's
GCE AS A2 Level Chemistry**
**
Advanced
Level Chemistry Revision on Titrations**
**GCE A
Level AS-A2 IB Chemistry Volumetric Analysis: Acid-base and other non-redox volumetric titration
quantitative calculation **
**
answers to PART 1 Questions 1 to 20**
**PART 1
Questions *
PART 2 Questions**
**PART 2 Question Answers * Redox
Titration Q's**
**
Qualitative
Analysis **
**
If you find these useful or
spot a silly error please
EMAIL
query? comment**
**
**
ILLUSTRATIONS OF ACID-ALKALI
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!
**
**
**
**
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 2-ethanoylhydroxybenzoic acid ('Aspirin') is made by esterifying
2-hydroxybenzoic 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
2-hydroxybenzoic 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 aqueous-ethanol 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 drop-wise addition close to the end-point
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
136-122 = 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 __4-methylbenzoic 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**
HOOC-CH**_{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 base-strong 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) 2-hydroxybenzoic 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
0.35-0.40g 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** = **__1.00463__g
**(v) **% purity = 100 x g
mass titrated/ 1g** = **__100.5%__ (1dp, 4sf)
**(g) **You need to re-read the opening paragraph of the question,
particularly the last line. The problem with this titration is that you also
titrate the 2-hydroxybenzoic acid impurity. This has a smaller molecular mass
than aspirin, therefore for equivalent masses, the 2-hydroxybenzoic 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})
+ (100-x) X 180.3(C_{9}H_{8}O_{4})}
/ 100
where x = the % of
2-hydroxybenzoic 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 2-hydroxybenzoic 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 2-hydroxybenzoic 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
2-hydroxybenzoic 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.)
PART 1
Questions *
PART 2 Questions *
PART 2 Question Answers * Redox
Titration Q's **
**
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!
ILLUSTRATIONS OF ACID-ALKALI
TITRATIONS and SIMPLE STARTER CALCULATIONS
**
**
Advanced Level
ORGANIC CHEMISTRY REVISION
STUDY NOTES
Advanced Level INORGANIC CHEMISTRY REVISION
STUDY NOTES
Advanced Level PHYSICAL-THEORETICAL REVISION
STUDY NOTES
**
** |