Types of chemistry
Although any type of chemical reaction may be used for titrimetric
analysis, the most often used fall under the categories of:
 Bronsted acid - base: HA + B ↔ HB+ + A Complex formation: M(aq) + nL(aq) ↔ MLn(aq)
 Oxidation – reduction: Ox + Red ↔ Red’ + Ox’
 Precipitation: M(aq) + nL(aq) ↔ MLn(s)
Lewis acid-base chemistry is often involved in precipitation and
complex formation chemistry.
Complexation reactions
Metal Complexes
• Lewis acids are electron pair acceptors.
• Coordination complexes: metal compounds formed by Lewis acidbase interactions.
• Complexes: Have a metal ion (can be zero oxidation state) bonded
to a number of ligands. Complex ions can be charged. Example,
[Ag(NH3)2]+.
• Ligands are Lewis bases.
• Square brackets enclose the metal ion and ligands.
• Chelate or chelon effect: More stable complexes are formed with
chelating agents than the equivalent number of monodentate
ligands. This is due to entropy (randomness) of the reaction – the
more molecules, the lower the entropy and vice-versa. The
interaction from all the different sites together is quite strong.
Ligands with More than One Donor Atom
[Ni(H2O)6]2+(aq) + 6NH3
[Ni(NH3)6]2+(aq) + 6H2O(l)
Kf = 4  108
[Ni(H2O)6]2+(aq) + 3en
[Ni(en)3]2+(aq) + 6H2O(l)
Kf = 2  1018
• Sequestering agents are chelating agents that are used to
remove unwanted metal ions.
• In medicine sequestering agents are used to selectively
remove toxic metal ions (e.g. Hg2+ and Pb2+) while
leaving biologically important metals.
• One
very
important
chelating
agent
is
ethylenediaminetetraacetate (EDTA).
• EDTA occupies 6 coordination sites, for example
[CoEDTA]- is an octahedral Co3+ complex.
• Both N atoms (blue) and O atoms (red) coordinate to the
metal.
• EDTA is used in consumer products to complex the metal
ions which catalyze decomposition reactions.
Widely used chelator: (1) Direct titration
(2) Indirect determination through
a sequence of reactions
EDTA
* It forms 1:1 complexes with most metals.
• (Not with Group 1A metals – Na, K, Li)
* Forms stable water soluble complexes.
* High formation constants.
• A primary standard material – a highly purified
compound that serves as a reference material.
•Highlighted, acidic protons
lost upon metal complexation.
CH2CO2H
HO2CH2C
HNCH2CH2NH
pK2 = 1.5
CH2CO2H
HO2CH2C
H6Y2+
pK1 = 0.0
pK3 = 2.0
Hydroxyl
protons
pK4 = 2.66
pK5 = 6.6
pK6 = 10.24
Ammonium
protons
Fraction of EDTA in the form Y 44
Y 
4
H Y  H Y

Y 

2
6

Y 
 H Y   H Y  H Y  HY  Y 
5


4
2
3
3
4
2
4
Y 4
EDTA
[EDTA] : Total concentration of all “free” uncomplexed
EDTA species in solution.
 4Y  
K1 K 2 K 3 K 4 K 5 K 6
H   H  K  H  K K  H  K K K  H  K K K K  H K K K K K  K K K K K K 
 6
 5
 4
1
 3
1
2
 2
1
2
3

1
2
3
4
1
2
3
4
5
1
2
3
4
5
Note that only the fully ionised , -4 – charged anion binds to metal ions
6
Fractional Composition Diagram for EDTA
At this range Y4- predominates,
thus titrations are routinely done
in buffered solutions near or
above pH 10.
Formation Constant or Stability Constant:
Equilibrium constant for the reaction of metal with
a ligand.
M n   Y 4  MY n 4

Y 4

Y 
4
Therefore,
EDTA

MY 

M  EDTA
n4
Kf
[ MY n  4 ]
and K f  [ M n ][Y 4 ]
n
Y 4
Y    EDTA
4
Y 4
Conditional Formation Constant
Fixing the pH by buffering:
Then  Y
4
is constant.
Thus, conditional formation constant:
n4
[
MY
]
'
K f   Y 4 K f 
[ M n  ][ EDTA]
Consider EDTA complex formation as if the
uncomplexed EDTA is in one form.
At any fixed pH, find  Y
4
and evaluate Kf’
Effective titration: *Reaction must go to completion.
*Large Kf
*Analyte and titrant essentially
completely reacted at the
equivalence point and:
n(Metal) = n(Titrant)
*Metals with higher Kf values can
be titrated at lower pH
*pH and thus Kf’ dependent
Effect of pH on EDTA Titration of Ca 2+
Less distinct
end point
EDTA Titration Curves
Titration reaction:
For large Kf’:
M
n
 EDTA  MY
n4
K 'f   Y 4 K f
Reaction complete at each point in
the titration.
Titration curve: Plot pM (= -log[M]) vs. volume EDTA
EDTA Titration Curve
Region 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 3
Excess EDTA. Virtually all metal
in MYn-4 form.
Example
Consider the titration of 25.0 mL of 0.020 M MnSO4 with
0.010 M EDTA in a solution buffered at pH 8.00.
Calculate pMn2+ at the following volumes of added
EDTA and sketch the titration curve:
0 mL
20.0 mL
40.0 mL
49.0 mL
49.9 mL
50.0 mL
50.1 mL
55.0 mL
60.0 mL
Mn2+ + EDTA  MnY2K 'f   Y 4 K f  (5.6 x10 3 )(7.4 x1013 )  4.2 x1011
End point volume = 50.0 mL
Region 1
1. 0.0 mL EDTA:
0.020 M Mn2+:
p Mn2+ = -log[Mn2+] = -log(0.020) = 1.70
2. 20.0 mL EDTA:
Initial Mn2+ volume
 50.0  20.0 
 25.0 
[ Mn ]  
0.020 

50.0


 45.0 
2
Fraction
Remaining
Original Dilution
Mn2+ conc. Factor
 [Mn2+] = 0.00671 M

Total volume
of solution
pMn2+ = -log[Mn2+] = 2.18
Use same method to calculate pMn2+ for any EDTA volume
before equivalence point (= 50.0 mL EDTA)
Region 2
At the Equivalence Point:
[Mn2+] = [EDTA]
50.0 mL EDTA
virtually all metal is in MnY2- form.
Assume negligible dissociation, then:
[ MnY
2
 25.0 
]  (0.020 M )

 75.0 
Initial
Mn2+ conc.
Dilution
Factor
[MnY2-] = 6.67 x 10 –3 M
Initial Mn2+ volume
Total volume
of solution
Region 2 (continued)
At the Equivalence Point:
Mn2+ + EDTA  MnY2Initial conc.
-
-
0.00667
Final conc.
x
x
0.00667 - x
[ MnY 2 ]
'
11

K

4
.
2
x
10
f
and
2
[ Mn ][ EDTA]


x = 3.98 x 10–8 M
pMn2+ = -log[Mn2+] = 7.40
0.00667  x
'
11

K

4
.
2
x
10
f
2
x
Region 3
After the equivalence point: All Mn2+ in the MnY2- form
& there is excess EDTA.
55.0 mL EDTA:
 5.0 
[EDTA]  (0.010)

 80.0 
Original
Dilution
EDTA Conc.
Factor
 [EDTA] = 6.25 x 10–4 M
Excess EDTA
volume
Total volume
of solution
Initial Mn2+ volume
[ MnY
2
 25.0 
]  (0.020 M )

 80.0 
Initial
Mn2+ conc.

Dilution
Factor
Total volume of
solution
[MnY2-] = 6.25 x 10–3 M
0.000625
[ MnY 2 ]
'
11
'
11

K

4
.
2
x
10
 K f  4.2 x10 and
f
2
2
[ Mn ]0.00625
[ Mn ][ EDTA]

[Mn2+] = 2.31 x 10–14 M

pMn2+ = -log[Mn2+] = 13.62
Manganese Ion EDTA Titration
12.00
10.00
pM
8.00
6.00
4.00
2.00
0.00
0.0
10.0
20.0
30.0
40.0
Volume of 0.010 M EDTA Soln (mL)
50.0
60.0
70.0
EDTA Titration Curves for Ca 2+ and Sr 2+
(Buffered at pH 10)
*Ca2+ end point more distinct.
*Lower pH, Kf ’ decreases, &
End point less distinct.
*We cannot raise pH
arbitrarily:
Metal hydroxides might
precipitate.
Auxiliary Complexing Agents
*Ligand strongly binds to metal & prevents hydroxide
precipitation at high pH.
*Auxiliary ligand binds less than EDTA binding to metal.
*NH3 normally used: NH3 fixes pH and
complexes metal species
*Tartrate, citrate, or triethanolamine may be used.
Auxiliary Complexing Agents
Metal – Ligand Equilibria
M + L ⇌ ML
[ ML ]
1 
[ M ][ L]
M + 2L ⇌ ML2
[ ML2 ]
2 
[ M ][ L]2
i = overall or cumulative formation constant
*Fraction of uncomplexed metal ion, M:
M
CM is total concentration of all forms of metal
M = M, ML, and ML2.
[M ]

CM
Auxiliary Complexing Agents
CM = [M] + [ML] + [ML2]
[ML]  1[M ][ L] and
Therefore,
Mass balance expression
[ ML2 ]   2 [ M ][ L]2
C M  [ M ]  1[M ][ L]   2 [M ][ L]2

C M  [ M ] 1   1 [ L]   2 [ L] 2
 M
[M ]

CM
M
[M ]

[ M ]1   1 [ L]   2 [ L] 2 

M 
1
1   1 [ L]   2 [ L] 2
Example
Consider the titration of 50.0 mL of 0.00100 M Zn2+
with 0.00100 M EDTA at pH10 in the presence of
0.10 M NH3. (This is the concentration of NH3. There is
Also NH4+ in the solution.) Find pZn2+ after addition of
20.0, 50.0, and 60,0 mL of EDTA.
Note: We always assume that EDTA is a much stronger
complexing agent than NH3.
Kf for EDTA > Kf for NH3.
Solution
Zn2+ - NH3 complexes:
Zn(NH3)2+, Zn(NH3)22+, Zn(NH3)32+, and Zn(NH3)42+
 Zn 
2
1
1   1 [ L ]   2 [ L ] 2   3 [ L ]3   4 [ L ] 4
1 = 1.51 x 102, 2 = 2.69 x 104, 3 = 5.50 x 106, and 4 = 5.01 x 108
[L] = [NH3] = 0.10 M

 Zn  1.8x10
5
2
Very little free Zn2+ in the presence of 0.10 M NH3.
 Most Zn2+ complexed by NH3
At pH 10,  Y 4   0.36
K   Zn 2 K   Zn 2  Y 4 K f
''
f
'
f
= (1.8 x10-5) (0.36) (1016.50)
= 2.05 x 1011
1. Addition of 20.0 mL EDTA sol’n:
C Zn 2 
 50.0  20.0 
 50.0 

0.00100 

50.0


 70.0 
= 4.3 x 10-4 M
[Zn 2 ]   M CZn 2  (1.8x10 5 )(4.3x10 4 )  7.7 x10 9 M

pZn2+ = -log[Zn2+] = 8.11
2. Equivalence point: Addition of 50.0 mL EDTA:
 50.0 
[ ZnY ]  (0.00100 M )

 100.0 
2
= 5.00 x 10-4 M
5.00 x10 4  x
''
11

K

2
.
05
x
10
f
x2
 X = C Zn
2
=4.9 x 10-8 M
[Zn 2 ]   M CZn 2  (1.8x10 5 )(4.9 x10 8 )  8.9 x10 13 M

pZn2+ = -log[Zn2+] = 12.05
3. After the equivalence point: 60.0 mL EDTA
 10.0 
[EDTA]  (0.00100)

 110.0 
= 9.1 x 10-5 M
 50.0 
[ ZnY ]  (0.00100 M )

 110.0 
= 4.5 x 10-4 M
2
[ ZnY 2 ]
'
16.50
16

K


K

(
0
.
36
)(
10
)

1
.
1
x
10
f
f
2
Y 4
[ Zn ][ EDTA]

[Zn2+] = 4.3 x 10–16 M

pMn2+= 15.36
Note: Past equivalence point problem independent on
presence of NH3. Both [EDTA] and [ZnY2-] known.
EDTA Titrations at Different Concentrations of
Auxiliary Complexing Reagent (NH3).
Small  pZn near
equivalence point.
Significant  pZn
Near equiv. Point.
(More distinct end
point)
Metal Ion Indicators
Compounds changing colour when binding to metal ion.
Kf for Metal-In < Kf for Metal-EDTA.
Before Titration:
•
Mg2+
+
(colourless)
In

(blue)
MgIn
(red)
During Titration: Before the end point
•
Mg2+
+
EDTA

MgEDTA
(free Mg2+ ions) (Solution red due to MgIn complex)
At the end point:
3. MgIn
+
EDTA
(red)
(colourless)

MgEDTA
(colourless)
+
In
(Blue)
EDTA Titration Techniques
1. Direct Titration
*Buffer analyte to pH where Kf’ for MYn-2 is large,
*and M-In colour distinct from free In colour.
*Auxiliary complexing agent may be used.
2. Back Titration
*Known excess std EDTA added.
*Excess EDTA then titrated with a std sol’n of a second
metal ion.
*Note: Std metal ion for back titration must not displace
analyte from MYn-2 complex.
2. Back Titration: When to apply it
*Analyte precipitates in the absence of EDTA.
*Analyte reacts too slowly with EDTA.
*Analyte blocks indicator
3. Displacement Titration
*Metal ions with no satisfactory indicator.
*Analyte treated with excess Mg(EDTA)2Mn+ + MgYn-2

MYn-4
* Kf’ for MYn-2 > Kf’ for MgYn-2
+
Mg2+
4. Indirect Titration
*Anions analysed: CO32-, CrO42-, S2-, and SO42-.
Precipitate SO42- with excess Ba2+ at pH 1.
*BaSO4(s) washed & boiled with excess EDTA at pH 10.
BaSO4(s) + EDTA(aq) 
BaY2-(aq)
+ SO42-(aq)
Excess EDTA back titrated:EDTA(aq) + Mg2+MgY2-(aq)
Alternatively: *Precipitate SO42- with excess
Ba2+ at pH 1.
*Filter & wash precipitate.
*Treat excess metal ion in filtrate with EDTA.
5. Masking
*Masking Agent: Protects some component of analyte
from reacting with EDTA.
*F- masks Hg2+, Fe3+, Ti4+, and Be2+.
*CN- masks Cd2+, Zn2+, Hg2+, Co2+, Cu+, Ag+, Ni2+, Pd2+,
Pt2+, Hg2+, Fe2+, and Fe3+,
but not Mg2+, Ca2+, Mn2+, Pb2+.
*Triethanolamine: Al3+, Fe3+, and Mn2+.
*2,3-dimercapto-1-propanol: Bi3+, Cd2+, Cu2+, Hg2+,
and Pb2+.
*Demasking: Releasing masking agent from analyte.
OH
M CN 
nm
m
Mn+

 mH 2 CO  mH  mH2C
Metal-Cyanide Formaldehyde
Complex
CN
*Oxidation with H2O2 releases Cu2+ from
Cu+-Thiourea complex.
*Thus, analyte selectivity:
1. pH control
2. Masking
3. Demasking