EDTA complexometric titrations

EDTA Titrations

(CHELOMETRIC TITRATIONS)
(Complexometric Titrations)


I. CONCEPT OF CHELATION.

A. Coordination bonding

- between metal ion (M) and a ligand (L). e- in the bond are
donated by L.


M :L
1. Examples of ligands.

Cl-, Br-, I-, F-, NH3, ....

Monodentate ligand: through 1 atom coordination

Multidentate ligand (chelating L): through >1 atoms coordination

2. Stepwise formation constants.

Cu2+ + NH3 = Cu(NH3)2+; K1 = 1 x 104

Cu(NH3)2+ + NH3 = Cu(NH3)22+; K2 = 2 x 103

Cu(NH3)2+ + NH3 = Cu(NH3)32+; K3 = 5 x 102

Cu(NH3)2+ + NH3 = Cu(NH3)42+; K4 = 1 x 102
3. Overall formation constant.

Cu2+ + 4 NH3 = Cu(NH3)42+

B4 = K1K2K3K4 = 1 x 1012


· B4 is very large.

· Since this corresponds to a series of stepwise rxns, the rxn is
not quantitative.

· More favorable for application to quantitative analysis if all
ligands were attached to the same molecule.
2. Structure of complex (chelate) between Ca2+ and EDTA.

EDTA: ETHYLENEDIAMINETETRAACETICACID
"H4Y"
H4Y = 4H+ + Y4-

Ca2+ + Y4- = CaY2-;

Formation constant: [MY n - 4 ]
Mn+ + Y4- = MYn-4; Kf =
[M n + ][Y 4- ]
Sometimes Kf is written as KMY

3. Table of Kf for metal-EDTA complexes.
Table 13-2 (page 264):

Note: Kf corresponds to rxn between the metal ion and the
fully deprotonated EDTA (Y4-).

We must learn how to predict the consequence of solution pH
on the net ("conditional") formation constant (KMY'). or Kf'
C. Conditional Formation Constants.

1. Stepwise acid dissociation constants for H4Y.

H4Y = H+ + H3Y-; Ka1 = 1.0x10-2

H3Y- = H+ + H2Y2-; Ka2 = 2.1x10-3

H2Y2- = H+ + HY3-; Ka3 = 7.8x10-7

HY3- = H+ + Y4-; Ka4 = 6.8x10-11

EDTA purchased for laboratory work usually is Na2H2Y,
which has a higher solubility than H4Y.

If considering the protonation capability of amines, EDTA
is a hexaprotic acid.
2. Definitions.

a. Formation constant:
[MY n -4 ]
M n + + Y 4 - = MY; Kf =
[M n + ][Y 4- ]
b. Conditional constant:
[MY n-4 ]
M n+ + Y' = MY; K =
f
[M n+ ][Y' ]
[Y'] is analytical concn of EDTA, i.e., [EDTA].

[Y'] = [H4Y] + [H3Y-] + [H2Y2-] + [HY3-] + [Y4-]

Substituting each component with the Ki expression:

[Y'] = [H 4 Y] { 1 + }
K 1 K 1 K 2 K 1 K 2 K3 K 1 K 2 K3 K 4
+
+ + 2+ + 3
+
[H ] [H ] [H ] [H + ]4
c. Fractional concns for each species.
[Y 4- ]
Y 4- =
[Y ]
K a1 K a 2 K a 3 K a 4
Y 4- =
[H + ]4 + K a1 [H + ]3 + K a1 K a 2 [H + ]2 + K a1 K a 2 K a 3 [H + ] + K a1 K a 2 K a 3 K a 4


D the denominator
[HY 3- ] K a1 K a 2 K a 3 [H + ]
HY3- = =
[Y ] D
[H 2 Y 2- ] K a1 K a 2 [H + ]2
H Y2 - = =
2 [Y ] D
[H 3Y - ] K a1 [H + ]3
H Y- = =
3 [Y ] D
[H 4 Y] [H + ]4
H 4Y = =
[Y ] D

3. Calculation of Kf from values of Kf' and pH.

[Y4-] = Y4- [Y']

[MY n-4 ] [MY n-4 ]
Kf = =
[M ][Y ] [M n+ ] Y4- [Y]
n+ 4-




Rearranging
[MY n-4 ]
K f Y 4- = K'f =
[M n+ ][Y]


Want a minimum value of 1x108 FOR Kf'

Adjust the pH to obtain a satisfactory value of Kf'.
4. Calculation of minimum pH which will give K'CaY =1x108.

For Ca2+ + Y4- = CaY2-,
KCaY = 5x1010


Since we want,
K'CaY = K CaY aY 4- = 1x10 8


8
aY 4 - = K'CaY = 1x10 = 2x10-3
K CaY 5x1010


And log(Y4-) = -2.7

From the plot of log(Y4-) vs. pH, we find this value
corresponds to a pH of ca. 7.5
Example: Calculate the minimum value of pH which will give
K'MgY = 1x108
(Ca2+ and Mg2+ are primary sources of water "hardness")

K'MgY = 5x108

1x108
Y 4- = =2x10-1
5x108

log( Y 4 - )=0.7
Minimum pH is ~10. This can be achieved with NH4OH.
III. INDICATORS FOR CHELOMETRIC TITRATIONS.
A. Common metal ion indicators.

B. Eriochrome Black T (EBT) Indicator (In)
used in Lab#7 for determination of Ca2+ + Mg2+ by EDTA titration.
1. Structure of EBT




2. Acid-base properties of EBT.

pH 6.3 pH 11.6
H2In- HIn2- In3-

(red) (blue) (orange)
KMgIn > KCaIn




3. Reaction of EBT with metals.

Color of MgIn is red.


Mg2+ + HIn2- MgIn- + H+

(blue) (red)
4. Process of titration

Titrate the Ca2+ first because KCaY>KMgY.

Then titrate the Mg2+ to the end point of EBT.

What if there is no Mg2+ in the sample?
The end point would be very poor (gradual)

Some Mg2+ is added to the indicator reagent to ensure the
presence of Mg2+.
IV. MASKING OF INTERFERING METAL IONS.

A. Reason: KMY values for most metal ions are larger than
for Ca2+ and Mg2+.



B. Cyanide as a masking agent.
Fe3+, Zn2+, Ni2+, Co2+ + CN- M(CN)xm-x
2Cu2+ + 6CN- 2Cu(CN)2- + (CN)2
(Cu2+ is reduced to Cu+.)


Note: cyanogen, (CN)2 is a toxic gas
C. Triethanol amine for masking Al3+.

Al3+ + TEA Al(TEA)3+


CH2CH2OH

TEA = :N - CH2CH2OH


CH2CH2OH
V. BACK TITRATIONS


First,
Addition of excess EDTA titrant to allow for slow rxns
to proceed to completion or to obtain a more easily observed
end point.

Then,
The excess EDTA is then titrated with a standard solution of
metal ion.
Titration Curves

The ion Mn+ (100.0 mL of 0.05 M metal ionbuffered to pH 9.00) was
titrated with 0.05 M EDTA

What is the equivalence volume, Ve in millimeters?

Calculate the concentration of free metal ion at V= ½ Ve

What fraction of (�y4- ) of free EDTA is in the form of Y4- at pH
9.00?

The formation constant (Kf) is 10^12.00. Calculate the conditional
formation constant. (Kf)1 (=�y4- Kf)

Calculate the concentration of free metal ion at V= Ve