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Complexation Titration
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Forming Complexes
Most metals ions react with electron-pairdonors to form coordination compounds or
complexes.
The donor species (ligand) must have atleast one pair of unshared electrons
available for bond formation.
Cu(H2O)42+, Cu(NH3)42+, Cu(NH2CH2COO)2
CuCl42-
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Ligands
A ligands is a neutral molecule or ion having alone pair that can be used to form a bond to a
metal ion. Chelating agents: unidentate, bidentate,
tridentate, tetradentate,pentadentate,
hexadentate
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Gramicidin A antibiotic ion channel
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Nonact informs a complex with the K + ion; the
coordination occurs through the 8 O atoms.
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Chelation in Biochemistry
Chelating ligands can form
complex ions with metals
through multiple ligands.This is important in many
areas, especially
biochemistry.
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Box 13-1 Chelation Therapy & Thalassemia
A successful drug for iron excretion
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The ions of alkali metals cam form complexes with crown
ether and cryptand
D. J. Cram, C. J. Pedersen and J.-M. Lehn Nobel prize inChemistry in 1987
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Crown Ether Complex of Potassium
Potassium permanganate is dissolved in
benzene by addition of 18-crown-6.
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Coordination Compounds
colored & paramagnetic (often)consists of a complex ion
(1) Coordination compounds are neutral
species in which a small number ofmolecules or ions surround a central
metal atom or ion.
ex.[Co(NH3)5Cl]Cl2
complex ion : [Co(NH3)5Cl]2+
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Coordination Compounds
coordinate covalent bond
Complex ion = metal cation + ligands
e acceptor e donor
center (one) surrounding (
2 )transion metal
Lewis acid Lewis base
[ Co(NH3)5Cl ]Cl2
H2O , NH3 , :Cl
..
.... ..
..
ionic force
counter ionscentral metal ligands
complex ion
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20.3 Coordination Compounds
(2) Coordination number :
The # of donor atoms surrounding the central
metal
The most common : 4 or6
(3) Ligands :
A neutral molecule or ion having a line pair that
can be used to from a bond to a metal ion.
monodentate : H2O , NH3bidentate : en , ox
polydentate : EDTA
Chelating agents
http://localhost/var/www/apps/conversion/tmp/scratch_10/f20.6.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_10/f20.7-8.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_10/f20.7-8.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_10/f20.6.ppt8/12/2019 1 Complexation Titration
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13-1 Metal-Chelate Complexes
EDTA forms strong 1:1 complexes with most metal ions
As a metal-binding agent
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Useful chelating agents
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13-2 EDTA(ethylenediaminetetraacetic acid, a hexadentate)
(1) The most widely used chelating agent in titration
(2) Forms strong 1:1 complexes regardless of the charge onthe cation
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Acidic Properties of EDTA
C CN N
CH2-COOH
CH2-COOH
HOOC-H2C
HOOC-H2C
H H
HH
C CN N
CH2-COOH
CH2-COO-
-OOC-H2C
HOOC-H2C
H H
HH
H++
H
H4Y
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C CN NCH2-COOH
CH2-COO-
-OOC-H2C
-OOC-H2C
H H
HH
H++
H
C CN N
CH2-COO-
CH2-COO-
-OOC-H2C
-OOC-H2C
H H
HH
H++H
H3Y-K1=1.0210
-2
H2Y-2K2=2.1410
-3
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C CN N
CH2-COO-
CH2-COO-
-
OOC-H2C
-OOC-H2C
H H
HH
H+
C CN N
CH2-COO-
CH2-COO-
-OOC-H2C
-OOC-H2C
H H
HH
HY-3K3=6.9210
-7
Y-4K4=5.510
-11
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The Nature of EDTA Complexes
with Metal Ions
The reagent combines with metal ions in a 1:1 ratioregardless of the charge on the cation.
M+n+Y-4 MY+(n-4)
]][[
][4
)4(
YM
MY
K n
n
MY
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(1) Multidentate chelating agents form strongercomplexes (Kf) with metal ions than bidentateor monodentate ligands.
(2) Neutral EDTA is a tetrabasicacid
(3) Metal-EDTA complex is unstable at both low pH& high pH.
At low pH H+& M n+
At high pH OH-& EDTA
For EDTA
][Y][M
][MY
KMYYM 4
4
f
44
n
nnn
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Pb2+as example:
At pH 10, tartrate is present
to prevent Pb(OH)2
Pb-tartrate complex must be
less stable than Pb-EDTA
(4) Auxiliary complexing agents:
prevent metal ions from
precipitating.
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13-3 Metal Ion Indicators
Metal ion indicator: a compound whose color changes whenit binds to a metal ion.
For an useful indicator, it must bind metal less strongly thanEDTA does.the indicator must release its metal to EDTA
Example: MgIn + EDTAMgEDTA + In Indicator is pH dependent.
If metal blockthe indicator, use back titration.
NH3
NH3
[H3N:Cu:NH3]2+Cu2+ + 4:NH3
pale bule deep bule
:
:
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Demonstration 13-1 Metal Ion IndicatorColor Changes
P.294
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COLOR PLATE 8
Titration of Mg2+by EDTA, Using Eriochrome Black T Indicator
(a) Before (left), near (center), and after (right) equivalence point.
(b) Same titration with methyl red added as inert dye to alter
colors.
Demonstration 13-1 Metal Ion Indicator Color Changes
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Indicator for EDTA Titrations
Ertichrome Black T
H2O+H2In-
HIn-2
+H3O+
H2O+HIn-2
In-3
+H3O+
O2N
SO3-
OH
N
N
OH
K1=510-7
K2=2.810-12
red blue
blue orange
MIn-+HY
-3 HIn-2
+MY-2
red blue
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At the end point:
MgIn + EDTA MgEDTA + In-
(red) (colourless) (colourless) (Blue)
Before Titration:
Mg2+ + In- MgIn(colourless) (blue) (red)
During Titration: Before the end point
Mg2+ + EDTA MgEDTA
(free Mg2+
ions) (Solution red due to MgIn complex)
Compounds changing colour when binding
to metal ion.
Kffor Metal-In-< Kffor Metal-EDTA.
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Requirements for Indicator
Metal-indicator complex must be less stablethan the metal-EDTA complex.
Binding between metal and indicator must notbe too weak. It has to avoid EDTA replacing at
the beginning of the titration.
In general, the metal-indicator complex shouldbe 10 to 100 timesless stable than the metal-
titrant complex.
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Eriochrome Black T is blue, but turns red in the presence of metals
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13-4 EDTA Titration Techniquesare useful for the determination of [metal]
Direct titration Titrate with EDTA Buffered to an appropriate pH Color distinct indicator Auxiliary complexing agent
Back titration Excess EDTA, & titrate with metal ion For analyte
ppt in the absence of EDTA :
Ex: (Al3+-EDTA)at pH 7, indicator Calmagite) backtitration with Zn2+react slowly with EDTA
block the indicator
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Displacement titration No satisfactory indicator Ex1: Hg2++ MgY2-HgY2-+ Mg2+ KfHgY
2- >MgY2-
Ex2: 2Ag++ Ni(CN)42-2Ag(CN)2+ Ni2+ , Ni2+is titrated with EDTA
Indirect titration Determine [Anion] that precipitate metal ions: CO3
2-, CrO42-S2-SO4
2-
Ex: SO42-+ Ba2+BaSO4(s) at pH 1
filter BaSO4(s) and boil with excess EDTA at pH 10
Ba(EDTA)2-and excess EDTA is back titration with Mg2+
Masking
Masking prevents one element from interfering in the analysis ofanother element.
Ex: Al3++ Mg2++ F-AlF63++ Mg2+then only Mg2+can be react with
EDTA masking Al3+with F- Masking agent: CN- , F- (using with pH control to avoid HCN & HF)
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In general, the metal-indicator complex should
be 10 to 100 times less stable than the metal-
titrant complex
Expt:
The formation constants of the EDTA complexes
of Ca2+
and Mg2+
are too close to differentiatebetween them in an EDTA titration, so they will
titrate together. Ca2+can actually be titrated in the
presence of Mg2+by raising the pH to 12 with
strong alkali; Mg(OH)2precipitates and does nottitrate.
E ilib i C l l ti
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Equilibrium Calculations
Involving EDTA
EDTA titrations are always performed insolutions that are buffered to a known pH to
avoid interferences by other cations or toensure satisfactory indicator behavior.
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Conditional Formation Constant
.applicableisfor whichpHat theonlyconstantaisK
][
][
][
][
]][[
][
][][][][][
][][][][
4
'
MY
)4(
4
'
4
)4(
4
)4(
43
2
2
34
4321321
2
21
3
1
4
43214
Tn
n
MYMY
T
n
n
MYn
n
MY
T
CM
MY
KK
CM
MYK
YM
MYK
YHYHYHHYYC
KKKKHKKKHKKHKH
KKKK
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D
HKKK
DHKK
D
HK
D
H
KKKKHKKKHKKHKHD
][
][
][
][
][][][][
3213
2
212
3
11
4
0
4321321
2
21
3
1
4
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13-5 The pH-dependent Metal-EDTA
Equilibrium
Since the anion Y4-is the ligand species in complexformation, the complexation equilibria are affected
markedly by the pH
Fraction Composition of EDTA Solutions
.....KKK
][H
KK
][H
K
][H1
1
][Y
[EDTA]
KKKKK][HKKKKK][HKKKK][HKKK][HKK][HK][H
KKKKKK
][EDTAY[EDTA]
][Y
][Y][HY]Y[H]Y[HY][H]Y[H]Y[H
][Y
654
3
65
2
6Y
4
5432154321
2
4321
3
321
4
21
5
1
6
654321
Y
Y
4
4
Y
4322345
26
4
Y
4
4
44
4
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15D-3 Equilibrium: pH dependent M-Y
YH4
4
a4a3a2a1a3a2a1
2
a2a1
3
a1
4
a4a3a2a14
a4a3a2a1
4
a4a3a2
3
a4a3
2
a444
YH
4
4
CY
KKKK][HKKK][HKK][HK][H
KKKK
KKKK
][H
KKK
][H
KK
][H
K
][H1
1
][Y
C
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Species EDTA as a function of pH
Y4- complexes with metal ions and so the complexation equilibria are very pH dependent
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Fig. 9.1. Fraction of EDTA species as a function of pH.
Y complexes with metal ions, and so the complexation equilibria are very pH dependent.
Only the strongest complexes form in acid solution, e.g., HgY2-; CaY2-forms in alkaline solution.
Gary Christian,
Analytical Chemistry,
6th Ed. (Wiley)
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C l l ti f th C ti
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Calculation of the Cation
Concentration in EDTA Solutions
Calculate 4
Calculate conditional formation constants K
Ni+2+Y-4 NiY -2
184
)4(
102.4]][[
][
YNi
NiYK n
n
MY
0.015M-xx x
Calculate equilibrium [Ni+2] at pH=3 and pH=8
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pH 4 pH 4
2 3.710-14 8 5.410-3
3 2.510-11
9 5.210-2
4 3.610-9 10 3.510-1
5 3.510-7 11 8.510-1
6 2.210-5 12 9.810-1
7 4.810-4
Value for 4 for EDTA at selected pH value
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)(101.8][Ni1027.2015.0
1027.2102.4104.5K8pH
)(102.1][Ni1005.1015.0
1005.1102.4105.2K3pH
KK
CY][H]Y[H]Y[H][HY][Y][Ni
10216
2
16183'
NiY
528
2
81811'
NiY
NiY4
'
NiY
T43
2
2
342
Mxx
x
Mxx
x
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C diti l F ti C t t
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Conditional Formation Constant
Most of the EDTA is not Y4-
belowpH=pK6=10.37. The species HY3-, H2Y
2-, and soon, predominate at lower pH.
It is convenient to express the fraction of freeEDTA in the form Y4-
P.300
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The number Ktf =Y4- ,Kfis called the conditional formationconstantor the effective formation constant.
P.300
(1) We can use Kfto calculate the equilibrium concentrationsof the different species at a given pH.
(2) Kf : HgY-2PbY-2CaY-2KfpHKf
pH
pH9.0Kf
EDTA(pH9.0)
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Complex-Formation Titrations
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Complex-Formation TitrationsFormation Constants for EDTA Complexes
Cation KMY Log KMY Cation KMY Log KMYAg+ 2.1 x 107 7.32 Cu2+ 6.3 x 1018 18.80
Mg2+ 4.9 x 108 8.69 Zn2+ 3.2 x 1016 16.50
Ca2+
5.0 x 1010
10.70 Cd2+
2.9 x 1016
16.46Sr2+ 4.3 x 108 8.63 Hg2+ 6.3 x 1021 21.80
Ba2+ 5.8 x 107 7.76 Pb2+ 1.1 x 1018 18.04
Mn2+
6.2 x 1013
13.79 Al3+
1.3 x 1016
16.13Fe2+ 2.1 x 1014 14.33 Fe3+ 1.3 x 1025 25.1
Co2+ 2.0 x 1016 16.31 V3+ 7.9 x 1025 25.9
Ni2+ 4.2 x 1018 18.62 Th4+ 1.6 x 1023 23.2
K = conditional formation constant = K
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Fig. 9.2. Effect of pH on Kf values for EDTA chelates.
Kf= conditional formation constant = Kf4.
It is used at a fixed pH for equilibrium calculations (but varies with pH since 4does).
Gary Christian,
Analytical Chemistry,
6th Ed. (Wiley)
As the pH increases the equilibrium shifts to the right
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Fig. 9.3. Titration curves for 100 mL 0.1 M Ca2+
versus 0.1 M Na2EDTA at pH 7 and 10.
As the pH increases, the equilibrium shifts to the right.
Gary Christian,
Analytical Chemistry,
6th Ed. (Wiley)
E l
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Example atp . 277
pH affects the titration of Ca2+with EDTA
K,
f
is smaller at lower pH.
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470
Figure 17-7Influence of pH on the
titration of 0.0100 M
Ca2+with 0.0100 M
EDTA. Note that the
end point becomes lesssharp as the pH
decreases because the
complex formation
reaction is lesscomplete under these
circumstances.
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471
Figure 17-8
Titration curves for 50.0 mL of 0.0100 M solutions of
various cations at pH 6.0.
The points represent the pH at which the conditional formation
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Fig. 9.4. Minimum pH for effective titrations
of various metal ions with EDTA.
constant, Kf', for each metal is 106, needed for a sharp end point.
Gary Christian,
Analytical Chemistry,
6th Ed. (Wiley)
The Effect of Other Complexing
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472
Figure 17-10
Influence of ammoniaconcentration on the end
point for the titration of 50.0
mL of 0.00500 M Zn2+.
Solutions are buffered to pH
9.00. The shaded region
shows the transition range
for Eriochrome Black T.
Note that ammonia
decreases the change in
pZn in the equivalence-
point region.
p gAgents on EDTA Titration Curves
Formation constants are the equilibrium
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Formation constants are the equilibrium
constants for complex ion formation one
ligand at a time.
The overall or cumulative formation of the complex ion at
any given step in this process is denoted as beta, .For step 2: the cumulative equation is the sum ofequations 1 and 2 in the formation process.
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12.5 Auxiliary Complexing Agents
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12.5 Auxiliary Complexing Agents
Metal-ligand Equilibria; Fractions of metal ion in
solution: General formula to find fraction of metal ion
in solution with all of its complexed ions:
Problem: Ammonia complexes of zinc: Zn2+and NH3form four complexes. If the concentration of free,
unprotonated NH3is 0.10 M, find the fraction of zinc in
the form Zn2+.
Problem: Find the fraction of thallium present whenthallium complexes with 0.20 M CN. Thallium forms
four complexes with cyanide ion: log K1= 13.21, log K2
= 13.29, log K3= 8.67, log K4= 7.44.
13 6 EDTA Tit ti C
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13-6 EDTA Titration Curves
The end point break depends upon1) [Mn+]
2) [L1]
3) [pH] selectivity
4) Kf
The smaller Kf, the more alkaline the solution must
be to obtain a kfof 106.
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12-3 EDTA Titration Curves
The titration curve is a graph of pMversus the volume of added EDTA.
The right side figure illustrates forreaction of 50.0 mL of 0.050 0 M
Mn+with 0.050 0 M EDTA,assuming Kf
= 1.15 1016,
where the concentration of free
Mn+decreases as the titration
proceeds.
There are three regions in anEDTA titration curve:
(a) Before the equivalence point.
(b) At the equivalence point.
c After the e uivalence oint.
EDTA Titration Curve
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EDTA Titration CurveRegion 1
Excess Mn+left after each addition
of EDTA. Conc. of free metal
equal to conc. of unreacted Mn+.
Region 2
Equivalence point:[Mn+] = [EDTA]
Some free Mn+generated by
MYn-4 Mn++ EDTA
Region 3Excess EDTA. Virtually all metal
in MYn-4form.
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Since at equivalence point,there is exactly as muchEDTA as metal in the
solution, we can treat thesolution as if it had beenmade by dissolving pureMYn4. Some free Mn+isgenerated by the slightdissociation of MYn4:
The Ca2+end point is moredistinct than the Sr2+end
point because theconditional formationconstant, for CaY2is greaterthan that of SrY2
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470
Figure 17-6
EDTA titration curves for
50.0 mL of 0.00500 MCa2+(KCaY=1.7510
10) and
Mg2+(KMgY=1.72108) at pH
10.0. Note that because of the
larger formation constant, the
reaction of calcium ion withEDTA is more complete, and a
larger change occurs in the
equivalence-point region. The
shaded areas show the
transition range for the
indicator Eriochrome Black T.
The titration rxn:
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Mn++ EDTAMYn-4
Kf= 4Kf
Three regions:(1) Before equivalence
point : excess Mn+(2) At equivalence point
[Mn+]= [EDTA]
(3) After equivalence point :
excess EDTA
Example at p.302
f C 2 2
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Titration of Ca+2and Mg+2
The K of EDTA of Ca+2and Mg+2are tooclose to differentiate between them in an
EDTA titration.
Generally, they will titrate together. This titration is used to determine totalhardness of water.
Ti i f C 2
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Titration of Ca+2
EB-T cannot be used to indicate the directtitration of Ca+2in the absence of Mg+2with
EDTA.
The indicator forms too weak a complex withCa+2to give a sharp end point.
Example: calculate the pZn of solutions
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Example: calculate the pZn of solutions
prepared by adding 20.0, 25.0, and 30.0 mL of
0.0100M EDTA to 50.0 mL of 0.00500M Zn2+
.Assume that both the Zn2+and EDTA solutions
are 0.100M in NH3and 0.175M NH4Cl to provide
a constant pH of 9.0
pH9.0(NH3+NH4Cl)0.0100M
EDTA50.00mL 0.00500M Zn2+pZn
EDTA(a) 20.0 mL (b)25.0 mL (c) 30.0 mL
KZnY:pHNH3
K"ZnY= M4KZnY =[ZnY2-]
CTCM
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476
Figure 17-11
Structure andmolecular model of
Eriochrome Black
T. The compound
contains a sulfonicacid group that
completely
dissociates in
water and two
phenolic groupsthat only partially
dissociate.
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478
Figure 17-12
Structural formulaand molecular
model of
Calmagite. Note the
similarity toEriochrome Black T
(see Figure 17-11).
Example 17 5
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Example 17-5
Determine the transition ranges for Eriochrome
Black T in titrations of Mg2+and Ca2+at pH 10.0,given that (a) the second acid dissociation
constant for the indicator is
(b) The formation constant for MgIn-is
(c) Ca2+ Kf= 2.5x105
H2O + HIn2- In3- + H3O+ K2= 2.8 x 10-12
Mg2+ + In3- MgIn- Kf= 1.0 x 107
pH10.0EBTCa2+
Mg2+EBT
R l ti
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Resolution
A small measured amount of Mg+2is added tothe Ca+2solution.
Ca+2gives more stable K than Mg+2.
A correction is made for the amount of EDTAused for titration of the Mg+2.
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At the end point:MgIn + EDTA MgEDTA + In-
(red) (colourless) (colourless) (Blue)
Before Titration:
Ca+2+EDTA+Mg2++In- MgIn+CaEDTA
During Titration: Before the end point Ca2+ + EDTA CaEDTA
Kffor Mg+2-EDTA < Kffor Ca
+2-EDTA.
EDTA tit ti l l ti
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EDTA titration calculations
Problem 12-7: Consider the titration of 25.0 mL of 0.0200M MnSO4with 0.0100 M EDTA in a solution buffered to
pH 8.00. Calculate pMn2+at the following volumes of
added EDTA and sketch the titration curve: (a) 0 ml, (b)
20.0 ml, (c) 40.0 ml, (d) 49.0 ml, (e) 49.9 ml, (f) 50.0 ml,
(g) 50.1 ml, (h) 55.0 ml, (i) 60.0 ml.
Solution: from Table 12.2, we get log(Kf) = 13.89. At pH= 8.00, (Y4-) = 4.2 x10-3.
The above calculation will be carried out based on
whether the titration is before or after the equivalence
point. Here Ve*0.0100 M = 25.0*0.0200 M; Ve = 50.0 ml
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(a) to (e) are before the equivalence point and will follow the samecalculation equation:
[Mn2+] * (0.025 + V) = 0.025 *0.020.01*V
(f) this is the equivalence pointKf = 4.2 x10-3x 1013.89= [MnY2-]/([Mn2+][EDTA])
= (0.02*0.025/0.075)/ ([Mn2+])2
[Mn2+] =8.179 x10-7.
pMn2+=
Calculations (g) to (i) are beyond the equivalence point, where[EDTA] can be obtained from the following
[EDTA]*(0.025 + V) = 0.01*V0.02*0.025
then Kf = 4.2 x10-3x 1013.89= [MnY2-]/([Mn2+][EDTA])
A beaker containing 50.0 mL of 0.300 M Ca2+ at pH 9 is titrated
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g pwith 0.150 M EDTA.The pCa at the equivalence point is:(a) 4.97.
(b) 5.13.(c) 5.84.
For Ca2+ Kf= 1010.65 ; = 0.041 at pH = 9
2
' 9
21.8 10f
CaYK
Ca EDTA
22
2
9 9
0.1
1.8 10 1.8 10
CaYCa
2 67.45 10Ca
pCa= 5.13
At the equivalent point, all Ca2+ions have formed complex
and free Ca2+come from the dissociation of complex:
CaY2- Ca2+ + EDTA
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"50 Mg2+ 0.050M " EDTA 0.050M . pH .10.0
Mg2+ ? K=2.02 x 108
EDTA ,+Mg2 :. Y4- - Mg2+ 1:1 MgY2-.
[MgY2-]50.0ml x 0.05M
100ml= 0.025M
Mg2+ + Y4- MgY2-
K =[MgY2-]
[Mg2+] [Y4-]= 2.02 x 108
0.025M
[Mg2+]= 2.02 x 108
[Mg2+][Mg2+] = 1.1 x 10-5M
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OCH3
OH
+HN
CO2-
CO2-
CH3
NH
+
-O2C
-O2C
SO3-
Xylenol Orange
M-In + EDTA M-EDTA + In
' '
M-In+ EDTA M-EDTA + In
OH
-
O3S
O2N
N N
OH
Eriochrome black T
pK1=6.3
pK1=11.6
+Zn2 EDTA
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EDTA 0.01M
Zn-EDTA
n(EDTA)=n(Zn2+)
Erio T
EDTA 0.01M
Zn2+, 20ml
Buffer pH=10
Erio T-Zn
Zn2+ + EDTA Zn-EDTA
+Ca2 EDTA
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Ca2+ + EDTA Ca-EDTA
(x mole)Ca
2+
+(y mole)
EDTA-Mg
(y mole)Ca-EDTA+ (y mole)Mg2++ (x-ymole)Ca2+
x mole> y mole
K(EDTA-Mg)= 4.9 x 108
K(EDTA-Ca)= 5.0 x 1010
EDTA
(y mole)Ca-EDTA+ (y mole)Mg-EDTA+ (x-ymole)Ca-EDTA
mole EDTA= C x V =(y mole)+(x-ymole)= x mole
Mg-EDTA
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1:1 Mg:EDTA
EDTA+Mg2 +Mg2EDTA
g
+Ca2 EDTA C 2 M EDTA EDTA C EDTA C EDTA M EDTA
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EDTA 0.01M
Ca-EDTA, Mg-EDTA
n(EDTA)=n(Ca2+)
Erio T
EDTA 0.01M
Ca2+, 20ml
~1ml Mg-EDTA
Buffer pH=10
Erio T-Ca
Ca2+ + Mg-EDTA+EDTA Ca-EDTA+ Ca-EDTA+ Mg-EDTA
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First
2nd
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3rd
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