Doc Brown's
GCE AS A2 A Level Chemistry
- Advanced
Level Chemistry Revision on Volumetric Titrations
GCE A
Level AS-A2 IB Chemistry Volumetric Analysis
Acid-base and other non-redox volumetric titration
quantitative calculation questions
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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These pages are designed to give you problem solving practice in volumetric
titration calculation questions - most involve some kind of volumetric analysis
where you titrate exact volumes of solutions, an accurately weighed mass.
PART 1
of A Level an chemistry volumetric titration analysis worksheet of structured questions: Worked out titration
questions - Q1-8 and
Q13-14 &
19 based on
acid-base titrations (acid-alkali, oxide, hydroxide, carbonate
and hydrogencarbonate)
and
Q15-18 based on alkali (NaOH)-organic acid titrations e.g.
standardising sodium hydroxide solution or like Q20 about aspirin analysis.
Q9 includes useful exemplars for coursework on how much to use in titrations including
EDTA,
Q10-12 are on silver nitrate-chloride ion titrations, further
Q's will be added in the future.
Appendix 1. information on EDTA
structure and function of EDTA in
titrations.
PART 1
Question Answers *
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 GCSE STARTER CALCULATIONS

The non-redox
titration Questions 1 to 20
I've tried to quote the data to the appropriate
significant figures and associated 'trailing zeros'.
Q1
A solution of sodium hydroxide contained 0.250 mol dm-3.
Using phenolphthalein indicator, titration of 25.0 cm3 of this solution required 22.5 cm3 of a hydrochloric acid solution for complete neutralisation.
(a) write the equation for the titration reaction.
(b) what apparatus would you use to measure out (i) the sodium hydroxide solution? (ii) the hydrochloric acid solution?
(c) what would you rinse your apparatus out with before doing the titration ?
(d) what is the indicator colour change at the end-point?
(e) calculate the moles of sodium hydroxide neutralised.
(f) calculate the moles of hydrochloric acid neutralised.
(g) calculate the concentration of the hydrochloric acid in mol/dm3 (molarity).
Q2
A solution made from pure barium hydroxide contained 2.74 g in exactly 100 cm3 of water.
Using phenolphthalein indicator, titration of 20.0 cm3 of this solution required 18.7 cm3 of a hydrochloric acid solution for complete neutralisation.
[atomic masses: Ba = 137, O = 16, H = 1)
(a) write the equation for the titration reaction.
(b) calculate the molarity of the barium hydroxide solution.
(c) calculate the moles of barium hydroxide neutralised.
(d) calculate the moles of hydrochloric acid neutralised.
(e) calculate the molarity of the hydrochloric acid.
Q3
4.90g of pure sulphuric acid was dissolved in water, the resulting total volume
was 200 cm3.
20.7 cm3 of this solution was found on titration, to completely neutralise 10.0 cm3 of a sodium hydroxide solution.
[atomic masses: S = 32, O = 16, H = 1)
(a) write the equation for the titration reaction.
(b) calculate the molarity of the sulphuric acid solution.
(c) calculate the moles of sulphuric acid neutralised.
(d) calculate the moles of sodium hydroxide neutralised.
(e) calculate the concentration of the sodium hydroxide in mol dm-3 (molarity).
Q4
100 cm3 of
a magnesium hydroxide solution required 4.5 cm3 of sulphuric acid (of concentration 0.100 mol dm-3) for complete neutralisation.
[atomic masses: Mg = 24.3, O = 16, H = 1)
(a) give the equation for the neutralisation reaction.
(b) calculate the moles of sulphuric acid neutralised.
(c) calculate the moles of magnesium hydroxide neutralised.
(d) calculate the concentration of the magnesium
hydroxide in mol dm-3 (molarity).
(e) calculate the concentration of the magnesium
hydroxide in g cm-3.
Q5
Magnesium oxide is not very soluble in water, and is difficult to titrate directly.
Its purity can be determined by
use of a 'back titration' method.
4.06 g of impure magnesium oxide was
completely dissolved in 100 cm3 of hydrochloric acid, of concentration 2.00 mol dm-3 (in excess).
The excess acid required 19.7 cm3 of sodium hydroxide (0.200 mol dm-3) for neutralisation
using phenolphthalein indicator and the end-point is the first permanent
pink colour.
This 2nd titration is called a 'back-titration', and is used to determine the unreacted acid.
[atomic masses: Mg = 24.3, O = 16)
(a) (i) Why do you have to use excess
acid and employ a back titration?
(ii) write equations for the two neutralisation reactions.
(b) calculate the moles of hydrochloric acid added to the magnesium oxide.
(c) calculate the moles of excess hydrochloric acid titrated.
(d) calculate the moles of hydrochloric acid reacting with the magnesium oxide.
(e) calculate the moles and mass of magnesium oxide that reacted with the initial hydrochloric acid.
(f) hence the % purity of the magnesium oxide.
(g) what compounds could be present in the magnesium oxide that could lead to a false value of its purity ? explain.
doc b future note:
(i) for an insoluble
carbonate you might need to use methyl orange/screened methyl orange
indicator because of dissolved carbon dioxide?
(ii) work out another
problem like this where 25cm3 aliquots are titrated, more
efficient and accurate than a one off titration.
Q6
2.00 dm3 of concentrated hydrochloric acid (10.0 M) was spilt onto a laboratory floor.
It can be neutralised with limestone powder.
[atomic masses: Ca = 40, C = 12, O = 16)
(a) give the equation for the reaction between limestone and hydrochloric acid.
(b) how many moles of hydrochloric acid was spilt?
(c) how many moles of calcium carbonate will neutralise the acid?
(d) what minimum mass of limestone powder is needed to neutralise the acid?
(e) If 1000 dm3 of sulphuric acid, of concentration 2.00 mol dm-3, leaked from a tank,
calculate the minimum mass of magnesium oxide required to neutralise it.
Q7
A 50.0 cm3 sample of sulphuric acid was diluted to 1.00 dm3.
A sample of the diluted sulphuric acid was analysed by titrating with aqueous sodium hydroxide.
In the titration, 25.0 cm3 of 1.00 mol dm-3 aqueous sodium hydroxide required 20.0 cm3 of the diluted sulphuric acid for neutralisation.
(a) give the equation for the full neutralisation of sulphuric acid by sodium hydroxide.
(b) calculate how many moles of sodium hydroxide were used in the titration?
(c) calculate the concentration of the diluted acid.
(d) calculate the concentration of the original concentrated sulphuric acid solution.
Q8
A sample of sodium hydrogencarbonate was tested for purity
using the following method. 0.400g of the solid was dissolved in 100 cm3 of
water and titrated with 0.200 mol dm-3 hydrochloric acid using methyl
orange indicator.
23.75 cm3 of acid was required for complete neutralisation. [Ar's:
Na = 23, H = 1, C = 12, O = 16]
(a) Write the equation for the
titration reaction.
(b) Calculate the moles of acid
used in the titration and the moles of sodium hydrogencarbonate titrated.
(c) Calculate the mass of sodium
hydrogen carbonate titrated and hence the purity of the sample.
(d) If 0.400g of another group 1
hydrogencarbonate in its pure state, was titrated with the same acid and it
took 20.00 cm3 to neutralise it, calculate ...
(i) moles of acid needed for
neutralisation and moles of hydrogen carbonate titrated
(ii) the formula mass of the
hydrogencarbonate
(iii) by working out the atomic mass
of M, suggest the identity of M in the group 1 hydrogencarbonate formula
MHCO3
Q9
This question involves theoretical calculations to do with 'how much to weigh
out' for titrations and a common requirement to show development in coursework
projects.
They involve reagents such as pure anhydrous sodium carbonate,
standardised hydrochloric acid and EDTA titrations
(theory).
Atomic masses: O = 16, H =
1, Na = 23, C = 12, Ca = 40, P = 31.0
9(a)(i) Write out the equation,
complete with state symbols for the reaction between hydrochloric acid and
sodium carbonate.
(ii) A pipetted 25.0 cm3
aliquot of a solution of sodium carbonate is to be titrated with an
approximately 1.0 mol dm-3
hydrochloric acid to be standardised.
What mass of dried anhydrous
sodium carbonate must be dissolved in 250 cm3 of deionised water, so that a 25.0 cm3
aliquot of the carbonate solution will give a 20.0 cm3 titration
with the hydrochloric acid?
What is the molarity of the
sodium carbonate solution, assuming 100% purity.
9(b)(i) The simplified molecular
structure of 2-ethanoylhydroxybenzoic acid ('Aspirin') is CH3COOC6H4COOH.
Give the equation of its reaction
with sodium hydroxide.
(ii) A sample of aspirin was to
be analysed for purity by titrating it with standardised 0.100 mol dm-3
sodium hydroxide using phenolphthalein indicator. Assuming 100% purity and
access to a 4 decimal place electronic balance, calculate the mass of Aspirin
that should be weighed out to give a titration of 23.0 cm3 of the
alkali.
(iii) The main contaminant is
likely to be unreacted 2-hydroxybenzoic acid. Why is this likely to be an
impurity? and how will this affect the % purity you calculate i.e. why and how
will the % purity be in error?
9(c)
Introduction to Q9(c)
EDTA can be used to estimate the
concentration of many metal ions in solution with a volumetric
titration. An indicator is used which forms a weak complex
with the metal ion. When all the free metal ions have been
titrated with an EDTA solution and hence more strongly complexed,
the indicator is displaced from its weak metal complex and a new
colour is observed at the end-point.
Pure calcium carbonate can be
used to make a standard calcium ion solution to practice a complexometric
titration of calcium ions with EDTA or determine the molarity of the EDTA
reagent.
See
Appendix 1.
for theoretical information on EDTA structure and function in titrations
(advisable to read).
(i) Give a simple equation to
show the chelation reaction between hydrated calcium ions and the EDTA anion
at pH10 and what sort of reaction is it?
(ii) To make a standard calcium
ion solution 0.250 g of A.R. calcium carbonate was dissolved in a little
dilute hydrochloric acid and made up to 250 cm3 in a calibrated volumetric flask.
Calculate the molarity of the calcium ion in this
solution.
(iii) Approximately 1.0g of the
solid disodium dihydrate salt of EDTA was dissolved in 250 cm3 of water in a
volumetric flask. 25.0 cm3 of this was pipetted into a conical flask
and ~1 cm3 of a conc. ammonia/ammonium chloride pH10 buffer was
added. After adding a few drops of Eriochrome Black T indicator, the EDTA solution
was titrated with the standard calcium ion solution (from part ii) until the reddish tinge turns to blue at
the endpoint. If 25.7 cm3 of the EDTA solution was required to
reach the equivalence point, what was the molarity of the EDTA?
(iv) In human teeth,
approximately 96% of the outer enamel and 70% of the inner dentine are
composed of the apatite mineral, calcium hydroxy phosphate.
If the simplest empirical formula is
Ca5(PO4)3OH
calculate the % calcium in the
apatite mineral.
(v) A dried 1.40g human tooth
was dissolved in a small quantity of hot conc. nitric acid. A drop of methyl
orange indicator was added followed by drops of 6M sodium hydroxide until
the indicator turned orange to neutralise the solution. The solution was
then made up to 250 cm3 in a volumetric flask. 10.0 cm3 of this
solution was pipetted into a conical flask and ~1 cm3 of a conc.
ammonia/ammonium chloride pH10 buffer was added.
This solution was then titrated
with 0.0200 mol dm-3 EDTA using Eriochrome Black T indicator. The
indicator turned blue after 22.5 cm3 of EDTA was added. Calculate
the average % by mass of calcium throughout the tooth.
Q10 25.0 cm3
of seawater was diluted to 250 cm3
in a graduated volumetric flask.
A 25.0 cm3 aliquot of the diluted
seawater was pipetted into a conical flask and a few drops of potassium
chromate(VI) indicator solution was added.
On titration with 0.100 mol dm-3
silver nitrate solution, 13.8 cm3 was required to precipitate all the
chloride ion.
[Atomic masses: Na = 23, Cl = 35.5]
(a) Give the ionic equation for the
reaction of silver nitrate and chloride ion.
(b) Calculate the moles of chloride
ion in the titrated 25.0 cm3 aliquot.
(c) Calculate the molarity of
chloride ion in the diluted seawater.
(d) Calculate the molarity of
chloride ion in the original seawater.
(e) Assuming that for every chloride ion there is a sodium ion,
what is the theoretical concentration of sodium chloride salt in g dm-3
in
seawater?
Q11 0.12 g of rock salt was dissolved in water and titrated with 0.100 mol dm-3
silver nitrate, and potassium chromate(VI) indicator, until the first permanent brown precipitate of silver chromate is
seen.
19.7 cm3 was required to titrate all the chloride ion.
[Atomic
masses: Na = 23, Cl = 35.5]
(a) How many moles of chloride ion
was titrated?
(b) What mass of sodium chloride
was titrated?
(b) What was the % purity of the
rock salt in terms of sodium chloride?
Q12 5.00 g of a solid mixture of anhydrous calcium chloride(CaCl2) and sodium nitrate
(NaNO3) was dissolved in 250 cm3 of
deionised water in a graduated volumetric flask.
A 25.0 cm3 aliquot of
the solution was pipetted into a conical flask and a few drops of potassium
chromate(VI) indicator solution was added.
On titration with 0.1 mol dm-3
silver nitrate solution, until the first permanent brown precipitate of silver
chromate was formed, 21.2 cm3 was required to precipitate all the
chloride ion. [Atomic masses: Ca = 40, Cl = 35.5]
(a) Calculate the moles of chloride
ion titrated.
(b) Calculate the equivalent moles
of calcium chloride titrated.
(c) Calculate the equivalent mass
of calcium chloride titrated.
(d) Calculate the total mass of
calcium chloride in the original 5.0 g of the mixture.
(e) The % of calcium chloride and
sodium nitrate in the original mixture.
Q13 A bulk solution of hydrochloric acid
was standardised using pure anhydrous sodium carbonate (Na2CO3,
a primary standard).
13.25 g of sodium carbonate was dissolved in about 150.0 cm3
of deionised water in a beaker.
The solution was then transferred, with
appropriate washings, into a graduated flask, and the volume of water made up to
250 cm3, and thoroughly shaken (with stopper on!) to ensure complete
mixing.
25.0 cm3 of the sodium
carbonate solution was pipetted into a conical flask and screened methyl orange
indicator added.
The aliquot required 24.65 cm3 of a hydrochloric
acid solution, of unknown molarity, to completely neutralise it. [atomic masses:
Na = 23, C = 12, O = 16]
(a) Calculate the molarity of the
prepared sodium carbonate solution.
(b) Write out the equation between
sodium carbonate and hydrochloric acid, including state symbols.
(c) How many moles of sodium
carbonate were titrated?
(d) How many moles of hydrochloric
acid were used in the titration?
(e) What is the molarity of the
hydrochloric acid?
Q14 For this question,
the relevant formula mass
and equation are in the answers to Q13.
A 1.35g sample of impure sodium carbonate was
titrated with standardised 1.00 mol dm-3 hydrochloric acid with methyl
orange indicator.
If 25.3 cm3 of acid was required for complete
neutralisation calculate the following:
(a) the moles of acid used in the
titration,
(b) the moles of sodium carbonate
titrated,
(c) the mass of sodium carbonate
titrated and hence its % purity.
Q15 (a) Describe a procedure that can used to determine the molecular mass
of an organic acid by titration with standardised sodium hydroxide solution.
Indicate any points of the procedure that help obtain an accurate result and
explain your choice of indicator.
0.279g of an organic monobasic
aromatic carboxylic acid, containing only the elements C, H and O, was dissolved
in aqueous ethanol. A few drops of phenolphthalein indicator were added and the
mixture titrated with 0.100 mol dm-3 sodium hydroxide solution.
It
took 20.5 cm3 of the alkali to obtain the first permanent pink. [at.
masses: C = 12, H = 1 and O = 16]
(b) How many moles of sodium
hydroxide were used in the titration?
(c) How many moles of the organic
acid were titrated? and explain your reasoning.
(d) Calculate the molecular mass of
the acid.
(e) Suggest possible structures of
the acid with your reasoning.
Q16 Using the method outlined in
the answer to
Q15(a), 0.103g of a dibasic/diprotic
non-aromatic carboxylic acid required 19.85 cm3 of a standardised
sodium hydroxide solution for complete neutralisation.
If the concentration of
the alkali was 0.0995 mol dm-3. [atomic masses: C = 12, H = 1 and O =
16]
Calculate ...
(a) moles of sodium hydroxide used
in the titration,
(b) moles of dibasic/diprotic acid
titrated giving your reasoning,
(c) the molecular mass of the acid,
(d) a possible structure of the
acid.
Q17 The % purity of an organic acid can be
determined by the procedure outlined in the answer to
Q15(a).
0.236g of benzoic acid required
19.25 cm3 of 0.100 mol dm-3 sodium hydroxide for complete
neutralisation.
Calculate ...
(a) moles of sodium hydroxide used
in titration,
(b) moles and mass of benzoic acid
titrated [at. masses: C = 12, H = 1 and O = 16]
(c) % purity of benzoic acid
from this assay titration.
Q18 Using the method described in
the answer to
Q15(a),
sodium hydroxide solution can be standardised. 0.250 g of very pure benzoic acid
(C6H5COOH)
was titrated with a solution of sodium hydroxide of unknown molarity.
If 22.5
cm3 of the alkali was required for neutralisation, calculate ...
(a) moles of acid titrated [at.
masses: C = 12, H = 1 and O = 16],
(b) mol alkali used in titration,
(c) the molarity of the alkali.
Q19 The solubility of calcium hydroxide in
water can be measured reasonably accurately to 3sf by titrating the saturated
solution with standard hydrochloric acid.
(a) If the standard
hydrochloric acid is made by diluting '2M' bench acid, what volume of the
'2M' acid is required to make up 250 or 500 cm3 of approximately 0.1
mol dm-3 hydrochloric acid and how might you do it?
(b) Why must the 2M acid be
diluted and why must the diluted acid be standardised?
In the calculation below assume
the molarity of the standardised hydrochloric acid is 0.1005 mol dm-3.
(for standardisation method see Q13)
At 25oC, a few grams
of solid calcium hydroxide was shaken with about 400 cm3 of
deionised water, and then filtered. 50.0 cm3 samples of the
'limewater' gave an average titration of 15.22 cm3 of 0.1005 mol
dm-3 hydrochloric acid using phenolphthalein indicator.
(c) If the acid is in the
burette, how would you measure out the calcium hydroxide solution? and why
is phenolphthalein indicator used?
(d) Give the equation for
calcium hydroxide reacting with hydrochloric acid.
(e) What is the reacting mole
ratio of Ca(OH)2 : HCl and hence calculate the moles of them
involved in the titration.
(f) Calculate the molarity of
the solution in terms of mol Ca(OH)2 dm-3.
(g) What is the approximate
solubility of calcium hydroxide in g Ca(OH)2 per 100g water?
Q20
ASPIRIN ASSAY ANALYSIS This question follows on in some respects from Q9b
which I'd forgotten I'd already written, apologies for some repetition!
2-ethanoylhydroxybenzoic acid (acetylsalicylic acid),
known commercially as aspirin, can be analysed by titration with standard sodium
hydroxide solution when a sample of it is dissolved aqueous alcohol (a mixture
of ethanol and water) and using phenolphthalein indicator (pKind = 9.3, useful
range pH 8.3-10).
In the pharmaceutical industry, aspirin is manufactured by
reacting 2-hydroxybenzoic acid
(salicylic acid) with ethanoic anhydride.
Prior to this reaction,
2-hydroxybenzoic acid is manufactured by reacting carbon dioxide with phenol,
the mixture is heated under pressure sodium
hydroxide in the so called Kolbe Reaction.
Aspirin, therefore, always contains a small
percentage of 2-hydroxybenzoic acid as an impurity!
(a) Give the equation for the Kolbe synthesis
of 2-hydroxybenzoic acid.
(b) Give the equation for the
formation of aspirin from 2-hydroxybenzoic acid.
(c) Give the molecular formulae and
calculate the molecular masses of 2-hydroxybenzoic acid and aspirin.
Accurate relative atomic
masses: Ar(C) = 12.01, Ar(H) = 1.01, Ar(O)
= 16.00
(d) Why must ethanol be added to
the water prior to doing the titration?
(e) Five samples of aspirin were
titrated with commercially purchased precisely 0.1000 mol dm-3 (0.1000M) sodium hydroxide solution and
the results are given below.
The titration values were
recorded to the nearest 0.05 cm3, which is reasonable of a
burette calibrated in 0.1 cm3 increments.
mass of
aspirin (g) |
titration/cm3 of 0.1M
NaOH |
titre/mass |
0.3591 |
20.05 |
? |
0.3532 |
19.65 |
? |
0.3686 |
20.60 |
? |
0.3583 |
19.90 |
? |
0.3635 |
20.25 |
? |
av titre/mass = ? cm3/g |
In each case calculate the
titre/mass and work out its average value for the five titrations.
(f) Assuming that only aspirin was
titrated (though not true), from the average titre/mass figure calculate the
'theoretical' % purity of the aspirin by the following sequence:
(i) What volume of 0.1000 M NaOH is
equivalent to 1.000 g of aspirin?
(ii) Give the reaction equation
for the titration.
(iii) How many moles of aspirin
can be titrated by your answer to (i)
(iv) from (iii) calculate the
theoretical mass of aspirin titrated.
(v) From (iv) calculate the
theoretical % purity of the aspirin!
(g) Why is the theoretical % purity
based on this titration method always likely to be over 100%?,
ignoring any titration errors - which does not necessarily explain why via this
method of analysis you will always tend to get >100%, especially if you do the
titration very accurately!
(h) Assuming that 2-hydroxybenzoic
acid is titrated with NaOH on a 1 : 1 molar basis, calculate the % of this
impurity in the aspirin by the follow sequence:
(i) From your answer to
(f)(iii) calculate an average molecular mass
(ii) From the average molecular
mass, and a little bit of algebra, using x as the % of the 2-hydroxybenzoic
acid impurity, calculate the value of x.
(i) Suppose for the sake of
argument, there was an error of 0.1 cm3 on the titration value which
is likely to be the biggest source of error. Obviously there are errors
associated with the NaOH molarity, the weighing, burette reading.
(i) What is the approximate %
error on the titration value?
(ii) What error range of values
for Mr(av) would this give?
(iii) Using the minimum and
maximum values from (ii), recalculate the % of 2-hydroxybenzoic acid in the
aspirin using the method indicated in (h) and quote the range of possible
values.
(iv) Comment on the results of
your calculations, a bit worrying for some coursework projects! yes?
(v) In principle, what must an
alternative method be capable of doing? Can you suggest an appropriate
method - and forget acid-alkali titrations!
ANSWERS
Appendix 1. Information on EDTA structure and function
EDTA is an acronym abbreviation for the old name
EthyleneDiamineTetraAcetic acid and is used in
equations.
It is a hexadentate ligand i.e.
it can donate 6 electron pairs to form 6 dative-covalent bonds and binds
strongly with many metal cations Mn+ where n is usually 2 or 3.
Multi-dentate ligands are called
chelating* agents, because the two ligand bonds become part of a five membered
ring system. (*The chelate is from the Greek word meaning a crab's claw)
The solid EDTA 'off the
shelf' used in analysis is usually the disodium dihydrate salt, which has
the structure
(Na+-OOCCH2)(HOOCCH2)NCH2CH2N(CH2COOH)(CH2COO-Na+).2H2O (Mr = 372.2)
The full unionised structure
is (HOOCCH2)2NCH2CH2N(CH2COOH)2 which we could abbreviate to H4EDTA
since theoretically four hydrogens from the four carboxylic acid groups are
ionisable.
With increasing pH (L ==>
R) the following structures becoming more likely,
H4EDTA(1) ==>
H3EDTA-(2) ==> H2EDTA2-(3)
==> HEDTA3-(4) ==> EDTA4-
( for those interested pKa(1) = 2.0, pKa(2)
= 2.66, pKa(3) = 6.16, pKa(4) = 10.24 )
of which H2EDTA2-
is the most prominent chelating species in solutions of pH10 in titrating
calcium ions though the complex is actually formed by the combination of a
metal ion and the EDTA4- ion.
The theory behind the
titration of calcium ions with EDTA reagent is a bit complicated and
the titration should be carried out in the presence of magnesium ions, usually
included in the EDTA volumetric reagent, but if not, they must be in the mixture
being titrated. This may seem to prelude an
incorrect titration for calcium since magnesium ions reacting with EDTA, but it doesn't (see explanation later).
Hydrated/hexa-aqua metal ions like
aqueous calcium and magnesium ions (M2+) give the following reaction with
EDTA reagent,
(eq 1) [M(H2O)6]2+(aq)
+ H2EDTA2-(aq) ==> [MEDTA]2-(aq,
colourless)
+ 2H+(aq) + 6H2O(l)
which, for theoretical
explanation of the titration and subsequent calculations, is best simplified to
(eq 2) M2+(aq)
+ H2EDTA2-(aq) ==> [MEDTA]2-(aq,
colourless)
+ 2H+(aq) (M = Mg, Ca)
and sometimes just shown as (eq
3) M2+(aq)
+ EDTA4-(aq) ==> [MEDTA]2-(aq)
Both calcium and magnesium EDTA
complexes are strongly
formed i.e. virtually 100% to the right BUT the Kstab for the
formation of the EDTA-calcium ion complex is greater than that for the EDTA-magnesium
ion complex i.e. the calcium ion complex is more stable and calcium ions
will displace magnesium ions from their EDTA complex.
The indicators used e.g.
Eriochrome Black T (represented in a 'free' anionic form as HIn2-)
weakly complexes with ions such as the magnesium ion.
(eq 4) Mg2+(aq, colourless)
+ HIn2-(aq,
blue)
[MgIn]-(aq, red) + H+(aq) (M = Mg, Ca)
Both metal ions form a weak complex
with the indicator at the start of the titration, but the indicator is displaced
by the stronger binding EDTA (ligand displacement reactions), but much more slowly from the calcium complex than the magnesium
complex i.e.
(eq 5) [MgIn]-(aq,
red) + H2EDTA2-(aq) =
fast => [MgEDTA]2-(aq, colourless)
+ HIn2-(aq, blue)
+ H+(aq)
(eq 6) [CaIn]-(aq,
red) + H2EDTA2-(aq) =
slow => [CaEDTA]2-(aq, colourless)
+ HIn2-(aq, blue)
+ H+(aq)
and this means without the
presence of magnesium ions, the end-point is sluggish giving an inaccurate one with
just the calcium ions present, because (eq 6) is too slow.
Therefore the order of complex ion
stability is [CaEDTA]2-(aq) > [MgEDTA]2-(aq)
> [MgIn]-(aq) and this order of stability is
crucial to the success of the titration as the ensuing argument will show.
In the EDTA solution are the
Mg-EDTA complex ion plus excess uncomplexed EDTA ions. As the EDTA reagent is
run into the calcium ion solution the calcium ion-EDTA complex is formed by
reactions (eq 7) or (eq 8).
(eq 7) Ca2+(aq)
+ H2EDTA2-(aq) ==> [CaEDTA]2-(aq)
+ 2H+(aq)
(eq 8) Ca2+(aq)
+ [MgEDTA]2-(aq) ==> [CaEDTA]2-(aq) + Mg2+(aq)
In (eq 8) the magnesium ion
is displaced from its EDTA complex on a 1 : 1 molar basis by the calcium ion and
then the free magnesium ions form a red complex with the blue indicator (eq 4)
below. This continues as long as there are still Ca2+ ions to titrate
and magnesium ions to be displaced i.e. no blue colour is seen yet.
(eq 4) Mg2+(aq, colourless)
+ HIn2-(aq,
blue)
[MgIn]-(aq, red) + 2H+(aq)
Continued addition of EDTA
eventually converts all the 'free' calcium ions into their EDTA complex ion via
(eq 7 or 8), BUT, the 1st drop of excess EDTA after the calcium ions are
all complexed then releases the blue form of the indicator via the fast
reaction (eq 5),
(eq 5) [MgIn]-(aq,
red) + H2EDTA2-(aq)
==> [MgEDTA]2-(aq, colourless)
+ HIn2-(aq, blue)
+ H+(aq)
so giving the sharp end point
from red to blue at pH10.
In the case of analysing a
mixture of calcium and magnesium ions in the same mixture, one method is to
analyse for Ca2+ as above (VCa). Obviously, you do
NOT add magnesium ions to the EDTA or the mixture being titrated if you wish to
estimate the total (Mg2+ + Ca2+). You add a known excess
of standardised EDTA solution (Vexcess) and then back titrate
with another standardized M2+ ion solution (Vback)
and the end point is blue to red.
Therefore:
(Vexcess) - (Vback) = (VCa+Mg)
and (VMg) = (VCa+Mg) - (VCa)

The full molecular
structure of the chelating anion derived from EDTA.
See Transition
metals Appendix 2.
Introduction to complexes
& ligands
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