Acid Base Titrations in Aqueous Solvents
Introduction:
All kind of titrations in various samples are performed today in process analysers and
laboratories, by far the most of them are acid base titrations. Acids and bases play an important
role in many processes and are controlled at a regularly basis. In most cases titration in water is
quite adequate to determine the acid or base concentration. Generally for the titration of acids a
strong base like sodium or potassium hydroxide is used. For the titration of bases a very strong
acid like hydrochloric acid is used, the hydrochloric acid can be replaced by nitric acid or
sulphuric acid if the sample contains substances reacting with chloride such as silver ions.
Titration is an absolute method so requires no calibration, on top of that multiparameter
determinations are possible: with one acid as a titrant several bases can be determined. Titration
is a quite versatile technique where it concerns the concentration range that can be analysed.
Variations in sample size (between 0.1 and 100 ml), in concentration of the titrant (between 0.01
M and 1 M) and that of the burette size (between 2 ml and 50 ml) can be made. Most
determinations in this monograph are potentiometric titrations with a glass electrode as indication
system. In section 9 an example is given with an antimony electrode and section 10 on a
photometric indication.
1) Classification of Acids according to strength:
An acid dissolved in water gives the next reaction: HA + H2O ÙH3O + + ADepending on the strength of an acid the equilibrium lies to the right or to the left. A very strong
acid, like hydrochloric acid completely dissociates in water. A weak acid, like acetic acid only
partly dissociates and most of the acetic acid is present in the form of the non dissociated acid.
For the classification of the acid strength the dissociation constant Ka is used.
[ H 3 O + ][ A − ]
[ HA ]
Example: Calculation of the part of the acid that has not been dissociated.
An Acid HA with Ka=10-4 is dissolved in water at a concentration of 1 mol / l
Reaction: HA + H2O Ù H3O + + ASo [H3O+] is equal to [A- ]
The concentration is 1 mol / l in total, so [HA] + [A-] = 1 mol / l
K a =
Ka =
[ H 3 O + ][ A − ]
[ HA ]
[ A − ]2
→ 10 − 4 =
[ HA ]
The greater part of the total concentration HA is in non dissociated form HA so [HA] ~ 1 mol/l
and [A- ] 2 = 10-4 and [A-] =10-2
So about 1 % of the acid has been dissociated and 99% is not.
Definition
pKa = -log Ka
For easy operations normally the unit pKa is used for the acid strength.
pKa means the -log of the Ka value.
Example:
Ka of Acetic Acid is:
1.75 * 10 -5
So log Ka = - 4.76 and the pKa = 4.76
-1-
No. 4 Acid Base Titrations in Aqueous Solvents
2a) Classification of Acids according to pKa :
Some examples of acids, besides the listed ones many more of them can be found in literature.
Very Strong Acids
pKa < 0
Strong Acids
0 < pKa< 4.5
Weak Acids
HClO4
HI
HCl
H2SO4
HNO3
pKa
-9
-9
-6
-3
-1
HSO4H3PO4
HF
Formic Acid
1.9
2.1
3.1
3.7
Acetic Acid
H2PO4HCN
4.7
7.2
9.1
HPO4-2
Phenol
12.3
10
4.5 < pKa <9.5
Very Weak Acids
pKa >9.5
The very strong acids all completely dissociate in water, the differences in pKa value in this group
are only noticeable in non-aqueous solvents such as acetone and alcohols. In water all the very
strong act the same. The figure below presents titration curves of acids with different pK values
using the degree of conversion λ versus pH.
λ = 0 at start of the titration and λ = 1 in the Endpoint
Fig.1
Acids with pK numbers from 2 - 10
12.00
pK
10.00
10
9
8.00
8
pH
7
6
6.00
5
4
4.00
2.00
0.00
3
2
0.10
0.20
0.30
0.40
λ
0.50
-2-
0.60
0.70
0.80
0.90
1.00
1.10
1.20
No. 4 Acid Base Titrations in Aqueous Solvents
As can be seen from the curves:
-the potential jump gets smaller with higher pK values
-acids can be titrated with a reasonable accuracy till a pK of 8
-very weak acids show no inflection point at all !
-for the weaker acids (in the range of pK 3.5 and 10.5) at the factor λ = 0.5
the pK values correspond to the pH values.
2b) Classification of Bases according to pKb :
A base dissolved in water gives the next reaction:
B + H2O Ù OH - + HB +
Strong bases are for instance NaOH, KOH, and tetra methyl ammonium hydroxide.
Analogue to acids a strong base reacts completely and a weak base partly dissociates.
The strength of a base is given by the Kb value, following the formulae:
Kb =
[OH − ][ HB + ]
[B]
For easy operations the next equation is used: Definition
pKb = -log Kb
Some examples of bases
Weak bases
Ammonia
Pyridine
Aniline
4.5 < pKb < 9.5
4.7
8.8
9.4
Very weak bases pKb > 9.5
Caffeine
Acetamid
Urea
13.4
13.5
13.8
Equation: pKa+pKb=pKw
Water dissociates for a very small part in [H3O+] and [OH-] , following the reaction:
2 H2O Ù H3O+ + OH- , the K of this reaction Kw = 10-14 and the pKw = 14
Ammonia reacts with water:
NH3 + H20 Ù NH4+ + OH+
The (NH4) reacts with water
NH4+ + H20 Ù NH3 + H3O+
Because the ratio between [H3O+] and [OH-] is known by the pKw, the pKa and pKb can easily
be calculated if one of both is known because
pKa+pKb = pKw =14.0
pKb = 4.7
So Ammonia (NH3) is a weak base with
pKa = 9.3
Therefore the Ammonium (NH4) + is a weak acid with
In literature many tables of the pKa of acids and pKb of bases can be found, sometimes the
constant is represented just as pK and the reader has to find out himself if is the stated pK is the
pKa or the pKb. Unfortunately some tables do not mention the pKb of bases but just record the
pKa of the corresponding acid. Because the sum of pKa and pKb is 14, calculation of pKb of the
corresponding base is easy.
3a) Polybasic Acids:
Some acids react with water in steps; a well-known example is Phosphoric Acid (Fig. 2a)
During the titration H3PO4 reacts with hydroxide in three steps.
pKa1 = 2.1
H3PO4 + OH- Î H2PO4- + H2O with
pKa2 = 7.2
H2PO4- + OH- Î HPO4-2 + H2O with
PO4-3 + H2O with
pKa3 = 12.3
HPO4-2 + OH- Î
A titration could theoretically result in three endpoints. The last EP however cannot be detected
because of the high pKa number (fig 2a).
-3-
No. 4 Acid Base Titrations in Aqueous Solvents
Fig. 2a
Fig. 2b
Na3PO4 with 0.2 M HCl
H3PO4 with 0.2 M NaOH
12
12
11
11
10
10
EP2
EP1
9
9
8
8
7
7 pH
pH
6
6
5
5
EP1
EP2
4
4
3
3
2
2
0
1
2
3
4
5
6
V [ml]
7
8
0
9
1
2
3
4
5
6
V [ml]
7
8
9
10
3b Polyacid bases:
The Na3PO4 sodium salt can be titrated with HCl ( fig. 2b) , so again a titration in 3 steps that
ends in H3PO4. The pKb values can be calculated from the pKa numbers of phosphoric acid,
using the expression pKa + pKb = pKw=14. The corresponding pKb numbers are respectively
1.7, 6.8 and 11.9. Again the last EP cannot be detected because the base is too weak.
4 Separation of a Mixture of Acids first idea
Mixtures of two acids can be determined in one titration if there is a significant difference in pKa
value. The strongest acid is titrated first, resulting in the first endpoint, the second endpoint can
represent the weaker acid. If the difference in pK is too small no separate endpoints will be found,
only the sum of the acids can be determined. In this section some examples are given
4a) HCl and HNO3
The determination of hydrochloric acid and nitric acid is showed in fig. 3. The pKa numbers are
respectively -6 for the hydrochloric and - 1 for the nitric acid. As mentioned earlier these
negative values have no meaning in water and both acids act similar. The result is that no
seperation occurs. The sum of both acids of coarse can easily be determined. In this particular
case the HCl, containing chloride can be determined by a titration with AgNO3. The difference
between the results in the titration with NaOH and the result in the titration with AgNO3
represents the HNO3.
Fig.3
HCl
(pKa –6)
and
HNO3 (pKa –1)
H C l + H N O 3 w ith 0 .2 M N aO H
10
9
EP1
8
7
-No separation
-Sum can be
determined
pH 6
5
4
3
2
1
0 .0
2 .5
-4-
5 .0
V [m l]
7 .5
1 0 .0
1 2 .5
No. 4 Acid Base Titrations in Aqueous Solvents
4b) HCl and Phosphoric Acid
Solutions of hydrochloric acid are dissociated for 100 %, solutions of phosphoric acid are
dissociated for about 25 %. These values as such, indicate that a titration of a mix of both acids
will give problems in detecting the first endpoint. And indeed no separation occurs.
The first endpoint detected in the mix represents the sum in the HCl and first endpoint of H3PO4.
Fortunately the endpoint from the reaction H2PO4- + OH- Î HPO4-2 + H2O can be used for the
determination of the phosphoric acid. Because the same consumption of hydroxide is used for
the reaction H3PO4 + OH- Î H2PO4- + H2O the determination of HCl is possible too, just by
subtracting the difference of the detected endpoints from the first endpoint.
Fig. 4
HCl
H3PO4
H2PO4-
( pKa –6 )
( pKa 2.1)
( pKa 7.2)
Titrant 0.2 M NaOH
12
A)
11
10
9
-No separation H3PO4 and
HCl
H 3PO 3
B) HCl
C) HCl+ H 3PO 3
8
pH7
6
-Separation H3PO4 and H2PO4-
5
4
A
3
B
C
2
0.
1.
2.
3.
4.
6.
V [ml]
7.
8.
9.
10.
12.
4c) Hydrochloric and Acetic Acid
Both acids can be determined, the separation even occurs if the ratios hydrochloric acid and
acetic acid are varied from 1-10 to 10-1. In figure 5 curves are shown of the ratios 1-10, 1-1 and
10-1. The endpoint in this mixture of hydrochloric acid is about pH 3.5. HCl alone gives an
endpoint at about pH 7 (see fig. 4). The inflection of the hydrochloric acid in the mix is much
smaller because of the presence of the inflection of the acetic acid.
Fig.5
HCl
(pKa
Acetic Acid (pKa
H C l + A cetic Acid with 0.2 M N aO H
-6)
4.7)
11
10
9
-Separation HCl and
Acetic acid
8
A) HCl -Acetic Acid 1 - 10
B)
´´
1 - 1
C)
´´
10 - 1
EP 2
EP2
EP 2
7
pH 6
5
A
4
B
EP1
E P1
EP1
3
C
2
0
1
2
3
-5-
4
5
6
7
V [ml]
8
9
10
11
12
No. 4 Acid Base Titrations in Aqueous Solvents
4d) Hydrochloric Acid + Formic Acid
The titration results in a poor inflection for the hydrochloric acid. Accuracy will be far from optimal,
furthermore if the performance of the electrode declines false endpoints will be found. To check
inflections, the differentiated ERC curve as represented in fig.6b can be viewed. Large potential
breaks in the normal curve, like the second endpoint, result in sharp peaks and the top reaches
high ERC values above 70. The first endpoint is a poor inflection: the top reaches a value of 7.
Into account has to be taken that this peak is on a sloping baseline, after correction the ERC
value drops to 3. This figure is definitely too low for quantitative measurements. Discrete values
of 5 are the minimum, and advised is 10 for a reasonable accuracy. So for mixed acids of this
composition only the total of acids can be determined.
Fig. 6a
HCl (pKa -6)
Fig. 6b
HC l+Formic Acid
HC l + Formic Acid
12
80
11
Formic Acid
(pKa 3.7)
70
10
60
9
EP2
8
-Too poor
separation
7
6
50
pH
EP1
ERC
40
EP2
30
5
4
20
EP1
3
10
2
0
0
1
2
3
4
5
6 7
V [ml]
8
9
10 11
12
0
1
2
3
4
5
6
7
V [ml]
8
9
10
11
4e) Phosphoric acid and Acetic Acid
The EP2 from the acetic acid is the smallest inflection. EP3 and EP1 have larger inflections, it is
advised to use these endpoints for the calculation of the acetic acid. The phosphoric acid can be
calculated from the EP1. The difference between EP3 and EP1 represents the sum of the
consumption of the H2PO4- and the acetic acid. As for phosphoric acid solutions the consumption
of the H3PO4 and the H2PO4- consumption are basically the same, the difference is the acetic
acid.
H3PO4
H2PO4Acetic
Acid
pKa
2.1
7.2
Fig.7 a
Phosphoric+ Acetic Acid with
0.2 M NaOH
1
1
EP3
40
8
7 pH
Poor
seperation
60
50
1
9
4.7
Fig.7 b
Phosphoric + Acetic Acid
30 ERC
EP1
EP2
EP3
EP2
6
20
5
4
EP1
10
3
2
0
0
5
1
V [ml]
1
2
-6-
2
0
5
1
V [ml]
1
2
2
No. 4 Acid Base Titrations in Aqueous Solvents
4 f) H3PO4 + Formic acid:
No separation of the H3PO4 and the formic acid, the first detected endpoint represents the sum of
both acids. Because the H2PO4- step is well separated, calculation of the phosphoric as well as
the formic is possible. Note in the differentiated curve the small ‘hill’ around 5 ml. The peak height
is so low that the endpoint from the H3PO4 is not detected.
Fig. 8a
H3PO4
H2PO4Formic
Acid
pKa
2.1
7.2
Fig. 8b
H3PO4+Formic Acid
12
60
55
11
50
10
EP2
45
9
3.7
40
8
35
pH7
EP1
30ERC
6
No separation
H3PO4 and
Formic Acid
H3PO4+Formic Acid
EP2
25
EP1
20
5
15
4
10
3
5
2
0.0
2.5
5.0
0
V [ml]
7.5
10.0
12.5
15.0
17.5
0.0
2.5
5.0
7.5
V [ml]
10.0
12.5
15.0
17.5
5 Separation of a mixed bases
Analogue to acids, the simultaneous determination of bases can be done if there is a significant
difference in pKb of the bases. Is the difference too small then only the sum of both bases can be
determined. The stronger base, having the lowest pKb value is titrated first the one with the
highest pKb is titrated last.
5a Amino ethoxy ethanol and hydroxyl amine.
Amino ethoxy ethanol pKb: 5.4
Hydroxylamine:
pKb: 8.0
The amines are well separated, so good quantitative determinations are possible. See the
differentiated curves in fig. 9b.
Fig. 9 a
Amines with 0.1 M H2SO4
A) Amino
Ethanol
A:
Amino Ethoxy
ethoxy ethanol
B) Hydroxylamine
B : Hydroxyl amine
A
C) Mix
C : Mix
1
1
.
Fig. 9b
9
8
A
7
B
pH
6
Amines with 0.1 M H2SO4
A
70
60
50
C
60
ERC
B
30
5
B
4
C
C
C
20
3
10
2
0
0.
2.
5.
7.
10.
V [ml]
12.
15.
17.
-7-
0.
2.
5.
7.
10.
V [ml]
12.
15.
No. 4 Acid Base Titrations in Aqueous Solvents
5b) Carbonate, Bicarbonate and Hydroxide
The Carbonic acid H2CO3 can dissociate in two steps, first in the formation of Bicarbonate (HCO3) with a pKa of 6.3 and the second to Carbonate (CO3–2) with pKa 9.4. The H2CO3 is not stable in
water and falls apart in CO2 and H2O. The carbonate and bicarbonate are stable bases and can
be titrated with acid. Using the expression pKa + pKb = pKw =14 the pKb can be calculated.
So the pKb of Carbonate is 4.6 and the pKb of bicarbonate is 7.7
In samples with Carbonate and Hydroxide in about equal concentrations 3 EPs exist
(Curve1 fig.10)
EP1: NaOH + HCl Î NaCl + H2O
EP2: Na2CO3 + HCl Î NaHCO 3 + NaCl
EP3: NaHCO3+ HCl Î CO2 + H2O +NaCl
The first EP cannot be used for quantitative measurements, but all concentrations can be
calculated from the last two remaining endpoints because the consumption for the Na2 CO3 is
equal to the consumption of NaHCO3 .
With low concentrations carbonate (curve2 in fig.10) two endpoints are detected.
EP1: sum of NaOH + Na2CO3 and EP2: NaHCO 3
Fig. 10
H yd ro xid e a n d C a rb on a te w ith 0.2 M H C l
13
2
12
=
-
1) High CO3 and high OH
2) Low CO3= and high OH-
EP1
11
1
10
EP1
9
EP2
8
pH 7
EP2
6
5
EP3
4
3
2
1
0 .0
0 .7
1 .4
2 .1
2 .8
3 .5
V [m l]
4 .2
4 .9
5 .6
6 .3
7 .0
6) Rules for seperation of acids and bases based on the pK
For acids rules of thumb
The negative values for the pKa of very strong acids have a meaning only in organic solvents.
For the rule of thumb in water, use for all of the very strong acids in these calculations the value
pK = 0 .As can be seen in the foregoing examples successful seperation based on the difference
of pK values is achieved easier with weaker acids.
Strong acids: Difference in pKa at least pKa > 4
Weak acids: Difference in pKa at least pKa > 2
For bases rules of thumb:
Strong bases: Difference in pKb at least pKb >4
Weak bases: Difference in pKb at least pKb >2
-8-
No. 4 Acid Base Titrations in Aqueous Solvents
7) Titration of very weak acids
For Ammonia (NH3) the pKb is 4.7, using the formulae pKa+pKb=14 the pKa for the
corresponding acid the Ammonium ion (NH4)+ is 9.3. The (NH4)+ ion can be titrated with sodium
hydroxide, because the pKa value is above 8 the accuracy will be limited. As a result the curve
shows a small inflection. The results of different quantities of a standard solution of ammonium
chloride result in a relative standard deviation of 0.5 % . Note that the endpoint itself varies with
the concentration.
Fig. 11 Different concentrations (NH4)+
N H 4 C l w ith 0 .2 M N a O H
11
EP1
EP1
EP1
EP1
4
6
EP1
EP1
10
pH
9
8
0
2
V [m l]
8
10
12
Standard solution of ammoniumchloride, 1-10 ml of standard + 50 ml water
Titrant 10 ml burette 0.2 M NaOH, DET ( MPD = 4, drift 0.1 pH /min)
No.
ml of sample
EP in ml
EP in pH
NH4Cl in
g/l
1
1
1.039
10.40
11.12
2
2
2.052
10.56
10.98
3
4
4.135
10.73
11.06
4
6
6.161
10.81
10.99
5
8
8.214
10.88
10.99
6
10
10.273
10.88
11.02
Average
10.71
11.03
Relative Standard deviation
0.5 %
8) Examples of the formation of a second compound
Sometimes chemistry can help to solve a titration problem, some acids or bases form on addition
of an auxiliary solution a compound that can be titrated easier.
8a) Ammonia and Formaldehyde
An example is the reaction of ammonia with formaldehyde, adding an excess of formaldehyde
converts ammonia to the hexamethylene tetramine.
Reaction: 4 NH3 + 6 HCOHÎ (CH2) 6 N4 + 6 H2O
The hexamethylene tetramine ( pKb = 9.1) is a much weaker base then ammonia (pKb = 4.7) .
So for the titration of ammonia with acid, the addition of formaldehyde would not improve the
titration. But if we add formaldehyde to ammonium chloride, the weak acid ammonium (pKa 9.3)
is converted to a to the stronger hexamethylene tetramine ion (pKa 4.9). This improves the
curves remarkable. To check the improvement of the formaldehyde addition some titrations were
performed with NaOH 0.2 M and standard samples of HCl, NH4Cl and a mix of both (with the
-9-
No. 4 Acid Base Titrations in Aqueous Solvents
addition of formaldehyde 5 ml 37 % pre neutralised to pH 7). When the determination of the
ammonium is done in two determinations, the overshoot of titrant in the HCl titration has to be
added to the consumption in the ammonium titration.
1 NH4Cl
2 HCl
3 mix in one titration
4) HCl + NH4Cl mix in two titrations, titration of the HCl
5) HCl + NH4Cl mix in two titrations, second titration
of the NH4Cl after addition of formaldehyde
Fig. 12 a
Fig. 12b
HCl, NH4Cl,HCl+NH4Cl
HCl, 5 ml Formaldehyde 37 %, NH4Cl
12
10
EP1
11
EP2
9
10
2
9
1
3
7
EP1
7
pH
pH
8
EP1
8
EP1
6
5
EP1
6
5
5
4
4
3
3
4
2
2
0.0
2.5
5.0
7.5
10.0
12.5
-0.0
0.5
1.0
1.5
2.0
2.5
3.0
V [ml]
3.5
4.0
4.5
5.0
5.5
6.0
6.5
V [ml]
No formaldehyde addition
Sample
HCl det. NH4Cl det.
6.169 ml
1 NH4Cl
5.483 ml
2
5.473 ml
6.189 ml
3 HCl + NH4Cl
Formaldehyde addition
Sample
HCl (4)
NH4Cl (5).
HCl + NH4Cl
5.472 ml
6.177 ml
HCl + NH4Cl
5.471 ml
6.160 ml
HCl + NH4Cl
5.473 ml
6.149 ml
average
5.472 ml
6.162 ml
Relative
0.02 %
0.23 %
Stand. Dev
Comparing the standard deviations it is obvious that the deviation of the NH4Cl is relatively high
compared to the HCl deviation. This example follows the rule that deviations will increase with
increasing pK value.
8b) Boric acid and Mannitol
Fig.13
On the addition of the sugar
Mannitol the Boric Acid with
a pKa of 9.2 can be converted
to the boric ester.
Because this ester has a pK
of 6 titration results will improve
remarkably.
Boric acid
Boric Acid and Mannitol
(1)
(2)
EP1
10.92
9.36
EP1
1
pH
7.80
6.24
2
4.68
0
1
-10-
2
3
4
5
6
V [ml]
7
8
9
10
11
12
No. 4 Acid Base Titrations in Aqueous Solvents
9) Hydrogen Fluoride with an Antimony Electrode
For potentiometric acid base titrations the glass electrode is used almost exclusively as
indication system. As the glass is attacked by fluoride solutions below the pH of 4, the
the alternative of an Antimony Electrode can be used here.
Fig. 14
H F w ith 0 .2 M
N a O H
( A n t im o n y e le c t r o d e )
1 0
9
8
p H
E P 1
7
6
5
4
0
1
2
3
V
[m l]
4
5
6
7
10) Acid Base Titrations using Color Indicators.
One of the oldest forms of titration is the addition of color indicators for Endpoint indication.
A color indicator is an organic dye that has different colors depending on the pH of the solution.
An example is methyl orange, above pH 4.4 the color is yellow and below pH 3.1 the color is red.
The region between pH 3.1 and 4.4 is called, the pH interval, between these two pH values the
color change is from red to orange to yellow. For an acid having its endpoint at about pH 3.7
methylorange can be used. The titration can be performed visually, the addition of titrant is
continued until the original red color changes to orange.
Fig. 15
.
yellow
pH 5
EP
3
red
ml
pH indicator
Methylorange
Methyl red
p- Nitrophenol
Phenolphthalein
pH interval
3.1 - 4.4
4.2 - 6.3
5.6 - 7.6
8.2 - 9.8
Color change
red - yellow
red - yellow
colorless- yellow
colorless -red violet
-11-
No. 4 Acid Base Titrations in Aqueous Solvents
Thymolphtalein
9.3 - 10.5
colorless -blue
The disadvantage of visual indication is when the titration curve has not a clear inflection, the
color change of the indicator is to gradually then. Adding more indicator to the solution will
introduce a second problem because the indicator itself is also an acid or base and so is
consuming acid or base. When two acids or bases are present in the sample, for each
determination a different indicator has to be used.
In fig. 16a Sodium carbonate is titrated a fotometric indication using the indicators
phenolphthalein (PP) and methylorange (MO.) is used . At the beginning of the titration the
solution is red/violet and due to the color of the PP and the transmission is low. Adding acid will
lower the pH and the PP will become colorless. At this stage the resulting solution gets yellow
due to the yellow color of the MO and this results in a high Transmission. Continuing the titration
the MO will change its color from yellow to red and the Transmission will drop again. The raise
and fall in the transmission results in the detection of an endpoint.
As a comparison fig. 16 b gives the determination with a glass electrode for detection of the
endpoint.
Fig. 16 a
Fig. 16b
Na2CO3 with glass electrode,
Na2CO3, fotometric. 540 nm
phenolphthaleine + methylorange addition
100
11
95
EP1
10
90
9
80
EP1
8
85
EP2
%T
7
pH
6
75
5
70
EP2
4
65
3
60
2
55
0
1
2
3
4
V [ml]
5
6
7
8
0
1
2
3
4
5
6
7
V [ml]
8
9
10
11
Jaap Guijt
20 February 2004
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12