20) Complexometric Titration

Billones Lecture Notes

UNIVERSITY OF THE PHILIPPINES MANILA

The Health Sciences Center

Complex-Formation TitrationsCOMPLEX-FORMATION REACTIONS

H

H

N

H

2

O

O

H

Cu

O

N

H

2

N

H

2

O

O

O

C

H

2

C

H

2

Cu2+ +

..

....

+ 2H+

N

C

H

2

C

H

2

N

C

H

2

C

H

2

C

H

2

C

H

2

O

O

H

O

O

H

O

O

H

O

O

H

.. ..

M

N

O

O

N

O

O

O

O

Mn+ +

Copper(II) ion forms a neutral complex with glycine, Cu(NH3CH2COO)2.

Most metal ions form stable complexes with EDTA.




Complexometric titrimetry are titrimetric methods based on complex formation.

It is based upon a particular class of coordination compounds called chelates.

A chelate is produced when a metal ion coordinates with two (or more) donor groups of a single ligand to form a five- or six-membered heterocyclic ring.

Multidentate ligands, particularly those having four or six donor groups, have two advantages over the unidentate ones.

1) react more completely with cations and thus provide sharper end points

2) react with metal ions in a single-step process




pM

0

10

20

0EP

1:1

2:14:1

Volume of Titrant

Curves for Complex Formation Titrations

Much sharper end point is obtained with a reaction that takes place in a single step

Multidentate ligands are ordinarily preferred for complexometric titrations.




TITRATIONS WITH AMINOCARBOXYLIC ACIDS

Ethylenediaminetetraacetic acid (EDTA)

EDTA is the most widely used complexometric titrant.

The EDTA molecule has six potential sites (i.e. hexadentate) for bonding a metal ion; the four -COOH groups and the two -NH2 groups.

EDTA species:

H4Y, H3Y-, H2Y2-, HY3-, and Y4-.

α

0

0.5

1.0

0pH

148642 10 12

α0α1

α2 α3 α4

H4YH2Y2-

H3Y-

HY3- Y4-




It is apparent that the H2Y2- predominates in moderately acidic media (pH 3 to 6).

Only at a pH values greater than 10 does Y4- become a major component of solutions.

COMPLEXES OF EDTA AND METAL IONS

EDTA reagent combines with metal ions in a 1:1 ratio regardless of the charge on the cation.

Ag+ + Y4- AgY3-

Al3+ + Y4- AlY-

1 : 1 M : EDTA

Billones Lecture NotesThe Health Sciences Center


N

+

C

H

2

C

H

2

N

+

C

O

O

-

C

O

O

H

H

H

H

O

O

C

-

O

O

C

N

+

C

H

2

C

H

2

N

+

C

O

O

-

C

O

O

H

H

H

-

O

O

C

-

O

O

C

N

+

C

H

2

C

H

2

N

+

C

O

O

-

C

O

O

-

H

H

-

O

O

C

-

O

O

C

N

+

C

H

2

C

H

2

N

C

O

O

-

C

O

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-

H

-

O

O

C

-

O

O

C

N

C

H

2

C

H

2

N

C

O

O

-

C

O

O

-

-

O

O

C

-

O

O

C

.. ..

.. ..

.. ..

Structure of H4Y and its dissociation products

H4Y

H3Y-

H2Y-2

HY-3

Y-4




NaOH

The dissociation constants for the acidic constants for the acidic groups in EDTA are

K1 = 1.02 x 10-2

K2 = 2.14 x 10-3

K3 = 6.92 x 10-7

K4 = 5.50 x 10-11

Formation constants KMY for common EDTA complexesThe constant refers to the equilibrium involving the deprotonated species Y4- with the metal ion:

Mn+ + Y4- MY(n-4) KMY = [MY(n-4)+] [Mn+][Y4-]




Cation KMY logKMY Cation KMY logKMY

Ag+ 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 16.46

Sr2+ 4.3 x 108 8.63 Hg2+ 6.3 x 1021 21.80

Ba2+ 5.8 x107 7.76 Pb2+ 1.1 x 1018 18.04

Mn2+ 6.2 x 1013 13.79 Al3+ 1.3 x 1016 16.13

Fe2+ 2.1 x 1014 14.33 Fe3+ 1.3 x 1025 25.1

Co2+ 2.0 x 1016 16.31 V3+ 7.9 x1025 25.9

Ni2+ 4.2 x 1018 18.62 Th4+ 1.6 x 1023 23.2




Equilibrium Calculations Involving EDTA

A titration curve for the reaction of a cation Mn+ with EDTA consists of a plot of pM versus reagent volume.

Values for pM are readily computed in the early stage of a titration by assuming that the [Mn+] is equal to cM.

Calculating [Mn+] in a buffered solution containing EDTA is a relatively straightforward procedure provided the pH is known.

The α4 for H4Y would be defined as

α4 = [Y4-] cT

[Y4-] = α4cT

where cT is the total molar concentration of uncomplexed EDTA:

cT = [Y4-] + [HY3-] + [H2Y2-] + [H3Y-] + [H4Y]




Conditional Formation Constants, K’ (pH dependent K)

Conditional or effective formation constants are pH dependent equilibrium constants that apply at a single pH only.

To obtain K’ for the equilibrium shown in the equation, we substitute α4cT for [Y4-] in the formation-constant expression:

Mn+ + Y4- MY(n-4)

KMY = [MY(n-4)+] [Mn+]α4cT

KMY = [MY(n-4)+] [Mn+][Y4-]

but [Y4-] = α4cT




Combining the two constants yields a new constant K’MY

K’MY = [MY(n-4)+] = α4KMY [Mn+]cT

K’MY = α4KMY

K’MY describes equilibrium relationships only at the pH for which α4 is applicable.

Computation of α4 Values for EDTA Solutions

α4 = ___K1K2K3K4_____________ __ [H+]4 + K1[H+]3 + K1K2[H+]2 + K1K2K3[H+] + K1K2K3K4

α4 = K1K2K3K4 D

or

αHY3- = K1K2K3[H+] αH2Y2- = K1K2[H+]2 D D




αH3Y- = K1[H+]3 αH4Y = [H+]4 D D

where K1, K2, K3 and K4 are the four dissociation constants for H4Y and D is the denominator.

α4 at Selected pH ValuespH α42.0 3.7 x 10-14

3.0 2.5 x 10-11

4.0 3.6 x 10-9

5.0 3.5 x 10-7

6.0 2.2 x 10-5

7.0 4.8 x 10-4

8.0 5.4 x 10-3

9.0 5.2 x 10-2

10.0 3.5 x 10-1

11.0 8.5 x 10-1

12.0 9.8 x `10-1

Only about 4 x 10-12 percent of the EDTA exists as Y4- at pH 2.00.




ExampleCalculate α4 and the mole percent of Y4- in a solution of EDTA that is buffered to pH 10.20.

[H+] = antilog (-10.20) = 6.31 x 10-11

From the values for the dissociation constants for H4Y, we obtain

K1 = 1.02 x 10-2 K1K2K3 = 1.51 x 10-11

K1K2 = 2.18 x 10-5 K1K2K3K4 = 8.31 x 10-22

Numerical values for the several terms in the denominator:

[H+]4 = (6.31 x 10-11)4 = 1.58 x 10-41

K1[H+]3 = (1.02 x 10-2)(6.31 x 10-11)3 = 2.56 x 10-33

K1K2[H+]2 = (2.18 x 10-5)(6.31 x 10-11)2 = 8.68 x 10-26




K1K2K3[H+] = (1.51 x 10-11)(6.31 x 10-11) = 9.53 x 10-22

K1K2K3K4 = 8.31 x 10-22

D = 1.78 x 10-21

significant configuration

The equation α4 = K1K2K3K4 becomes D

α4 = K1K2K3K4 = 8.31 x 10-22 = 0.466 D 1.78 x 10-21

mol % Y4- = 0.47 x 100% = 47%

Only the last two terms in the denominator contribute significantly to the sum D at pH 10.20.

At low pH values, in contrast, only the first two or three terms are important




Calculation of the Cation Concentration in EDTA SolutionsCalculate the equilibrium concentration of Ni2+ in a solution with an analytical NiY2- concentration of 0.0150 M at a pH (a) 3.0 and (b) 8.0.

Ni2+ + Y4- NiY2-

KMY = [NiY2-] = 4.2 x 1018

[Ni2+][Y4-]

The equilibrium concentration of NiY2- is equal to the analytical concentration of the complex minus the concentration lost by dissociation.

NiY2- ⇋ Ni2+ + Y4-

Kinstab = 1 very small KMY




[NiY2-] = 0.0150 – [Ni2+] ≈ 0.0150 because Kinstab is small

Since the complex is the only source of both Ni2+ and the EDTA species,

[Ni2+] = cT = [Y4-] + [HY3-] + [H2Y2-] + [H3Y-] + [H4Y]

Substitution of this equality gives

K’MY = α4KMY = [NiY2-] = [NiY2-] [Ni2+]cT [Ni2+]2

(a) α4 is 2.5 x 10-11 at pH 3.0. Substitution of this value and the concentration of NiY2- into the equation for K’MY gives

K’MY = 0.0150 = α4KMY = 2.5 x 10-11 x 4.2 x 1018 = 1.05 x 108

[Ni2+]2




[Ni2+] = sqrt(0.0150/1.05 x 108) = 1.2 x 10-5 M

[Ni2+] << 0.0150, as assumed

b) At pH 8.0, the conditional constant is much larger. Thus,

K’MY = 5.4 x 10-3 x 4.2 x 1018 = 2.27 x 1016

Substitution into the equation for K’MY followed by rearrangement gives

[Ni2+] = sqrt(0.0150 / 2.27 x 1016) = 8.1 x 10-10 M

large K’MY, small 1/K’MY , low [ion]

ExampleCalculate the concentration of Ni2+ in a solution that is prepared by mixing 50.0 mL of 0.0300 M Ni2+ with 50.0 mL of 0.500 M EDTA. The mixture was buffered to a pH of 3.00.




Here, the solution has an excess of EDTA, and the analytical concentration of the complex is determined by the amount of Ni2+ (limiting reagent) originally present. Thus,

cNiY2- = 50.0 x 0.03000 = 0.0150 M 100

cEDTA = 50.0 x 0.0500 – 50 x 0.0300 = 0.0100M 100

Again assume that [Ni2+] << [NiY2-] so that

[NiY2-] = 0.0150 – [Ni2+] ≈ 0.0150 M

At this point, the total concentration of uncomplexed EDTA is given by its analytical molarity:

cT = 0.0100 M




Substitution into the equation gives

K’MY = [NiY2-] = 0.0150 = α4KMY [Ni2+] cT [Ni2+] 0.0100

= 2.5 x 10-11 x 4.2 x 1018 = 1.05 x 108

[Ni2+] = 0.0150 = 1.4 x 10-8

0.0100 x 1.05 x 108

EDTA Titration Curves

Example

Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.00500 M Ca2+ with 0.0100 M EDTA in a solution buffered to a constant pH of 10.0.




Calculation of Conditional Formation Constant

K’CaY = [CaY2-] = α4KCaY

[Ca2+] cT

From Table, α4 = 0.35 and KCaY = 5.0 x 1010

K’CaY = 0.35 x (5.0 x 1010) = 1.75 x 1010

Substitution gives

Preequivalence Point Values for pCaBefore the EP is reached, the [Ca2+] is equal to the sum of the contributions from the untitrated excess of the cation and from the dissociation of the complex, which is equal to cT, and assumed to be small. For example, after addition of 10.00 mL reagent:

[Ca2+] = 50.0 x 0.00500 – 10.0 x 0.0100 + cT ≈ 2.50 x 10-3 60.0




pCa = -log (2.50 x 10-3) = 2.60Other pre-EP data are derived in this way.

The Equivalence-Point pCaCompute first the analytical concentration of CaY2-.

cCaY2- = 50.0 x 0.00500 = 3.33 x 10-3

50.0 + 25.0

The only source of Ca2+ ions is the dissociation of the CaY2- complex.

At EP, [Ca2+] = cT [CaY2-] = 3.33 x 10-3 – [Ca2+] ≈ 3.33 x 10-3 M

Substituting into the conditional formation constant expression gives

K’MY = [CaY2-] = 0.00333 = 1.75 x 1010

[Ca2+]cT [Ca2+]2




[Ca2+] = sqrt(0.00333 / 1.75 x 1010) = 4.36 x 10-7 M

pCa = -log (4.36 x 10-7) = 6.36

Postequivalence-Point pCaBeyond the equivalence point, analytical concentrations of CaY2-

and EDTA are obtained directly from the stoichiometric data.

For example, after the addition of 35.0 mL of reagent,

cCaY2- = 50.0 x 0.00500 = 2.94 x 10-3 M 50.0 + 35.0 the complex is diluted

cEDTA = 35.0 x 0.0100 – 50.0 x 0.00500 = 1.18 x 10-3

85.0As an approximation, we can write

[CaY2-] = 2.94 x 10-3 – [Ca2+] ≈ 2.94 x 10-3

from ionization




cT = 1.18 x 10-3 + [Ca2+] ≈ 1.18 x 10-3 Mfrom ionization

Substitution into the K’CaY expression gives

K’CaY = 2.94 x 10-3 = 1.75 x 1010

[Ca2+] x 1.18 x 10-3

[Ca2+] = 2.94 x 10-3 = 1.42 x 10-10

1.18 x 10-3 x 1.75 x 1010

pCa = -log (1.42 x 10-10) = 9.85

The approximation that [Ca2+] is small is clearly valid.

Curve A (red) in the following figure is a plot of the data for the Ca titration in the example.Curve B is the titration curve for a solution of magnesium ion under identical conditions.




pM

2

10

0

EP

Volume of 0.0100 M EDTA

EDTA titration curves for 5.0 mL of 0.00500 M Ca2+ (K’ for CaY2- = 1.75 x 1010) and Mg2+ (K’ for MgY2- = 1.72 x 108) at pH 10.0.

The shaded areas show the transition range for Eriochrome Black T (EBT).

pCa

EP25.0 mL 35.0 mL

pMg

4

6

8

CaIn- + HY3- ⇋ HIn2- + CaY2-MgIn- + HY3- ⇋ HIn2- + MgY2-

redred

blueblue




Titration curves for Ca+2 solutions buffered to various pH levels

Recall that α4, and hence K’CaY, becomes smaller as the pH decreases.

Influence of pH on the titration 50.0 mL of 0.0100 M Ca2+ with 0.0100 M EDTA

pCa

2

10

0

pH 12

4

6

8

pH 10

pH 8

pH 6

Volume of 0.0100 M EDTA50.00 mL

• Smaller ΔpM at e.p• End points for EDTA

titrations become less sharp as pH decreases because t h e c o m p l e x formation reaction is less complete under these circumstances.

• Lower pH, smaller α4

• Smaller α4, smaller K’MY




Cations with larger formation constants provide good end points even in acidic media.

Titration curves for 50.0 mL of 0.0100 M cation solutions at pH 6.0

pM

8

20

0

10

12

16

Volume of 0.0100 M EDTA50.00 mL

4

KCaY2- = 5.0 x 1012

KFeY2- = 2.1 x 1014

KZnY2- = 3.2 x 1014

KHgY2- = 6.3 x 1021

KFeY- = 1.3 x 1025

• smaller KMY, smaller ΔpM




8

10

12

14

16

18

20

22

24

26

0 2 4 6 8 10 12

pH

log

KM

YMinimum permissible pH for a satisfactory end point in the titration

Fe3+

In3+ Th4+

Sc3+Hg2+

Ga3+

La3+VO2+

Cu2+

Ni2+

Pb2+,Y3+Sm3+

Zn2+

Cd2+

Al3+Co2+

La3+Fe2+

Mn2+

Ca2+

Mg2+, Sr2+

moderately acidic environment is satisfactory for many divalent heavy-metal cations

Consider a mixture of Ca2+ and Zn2+.

Titration w/ EDTA at pH 10 (both Ca2+ and Zn2+)

Titration w/ EDTA at pH 6 (only Zn2+ will titrate)

mmol Zn2+ = mL EDTA x M EDTA @ pH 6

mmol Ca2+ = (mL x M EDTA @ pH 10) – mmol Zn




THE EFFECT OF OTHER COMPLEXING AGENTS ON EDTA TITRATION CURVES

Many cations form hydroxide precipitates when the pH is raised to the level required for their successful titration with EDTA.

When this problem is encountered, an auxiliary complexing agent is needed to keep the cation in solution.

For example, zinc(II) is ordinarily titrated in a medium that has fairly high concentrations of ammonia and ammonium chloride.

These species buffer the solution to a pH that ensures complete reaction between cation and titrant

In addition, ammonia forms ammine complexes with zinc(II) and prevents formation of the sparingly soluble zinc hydroxide.

Zn(NH3)42+ + HY3- ⇋ ZnY2- + 3NH3 + NH4+

The solution also contains such other zinc/ammonia species as Zn(NH3)32+, Zn(NH3)22+, and Zn(NH3)2+.




Complexation of a cation by an auxiliary-complexing reagent causes preequivalence pM values to be larger than in a comparable solution with no such reagent.

pZn

4

12

0Volume of 0.0100 M EDTA

cNH3 = 0.100

6

8

10

14Influence of ammonia concentration on the end point for the titration of 50.0 mL of 0.00500 M Zn2+.

25.0 mL

cNH3 = 0.010




Titration Curves When a Complexing Agent is PresentA quantitative description of the effects of an auxiliary complexing agent (ACA) can be derived by a procedure similar to that used to determine the influence of pH on EDTA titration curves.

A quantity αM is defined that is analogous to α4:

αM = [Mn+] ; [Mn+] = αMcM cM

where cM is the sum of the concentrations of species containing the metal ion exclusive of that combined with EDTA.

cM = [Zn2+] + [Zn(NH3)2+] + [Zn(NH3)22+] + [Zn(NH3)32+] + [Zn(NH3)42+]

The value of αM can be expressed readily in terms of the ammonia concentration and the formation constants for the various ammine complexes.




K1 = [Zn(NH3)2+] [Zn2+][NH3]

[Zn(NH3)2+] = K1[Zn2+][NH3]

Similarly, it is readily shown that

[Zn(NH3)22+] = K2[Zn2+][NH3]2

[Zn(NH3)32+] = K3[Zn2+][NH3]3

[Zn(NH3)42+] = K4[Zn2+][NH3]4

Substitution of these expressions into the cM equation gives

cM = [Zn2+](1 + K1[NH3] + K2[NH3]2 + K3[NH3]3 + K4[NH3]4)

Substituting this expression for cM (here, [Mn+] = [Zn2+]), leads to

αM = 1 1 + K1 [NH3] + K2[NH3]2 + K3[NH3]3 + K4[NH3]4




Finally, a conditional constant for the equilibrium between EDTA and zinc(II) in an ammonia/ammonium chloride buffer is

K”ZnY = [ZnY2-] = α4αMKZnY

cMcT

where K”ZnY is a new conditional constant that applies at a single pH as well as a single concentration of ammonia.

ExampleCalculate the pZn for solutions prepared by adding 20.0, 25.0, and 30.0 mL of 0.0100 M EDTA to 50.0 mL of 0.00500 M Zn2+. Assume that both the Zn2+ and EDTA solutions are 0.100 M in NH3 and 0.175 M in NH4Cl to provide a constant pH of 9.0.

The logarithms of the stepwise formation constants for the four zinc complexes with ammonia are 2.21, 2.29, 2.36, and 2.03. Thus,




K1 = antilog 2.21 = 1.62 x 102

K2 = antilog (2.21 + 2.29) = 3.16 x 104

K3 = antilog (2.21 + 2.29 + 2.36) = 7.24 x 106

K4 = antilog (2.21 + 2.29 + 2.36 + 2.03) = 7.76 x 108

Calculation of Conditional Constant, K”

A value of the αM can be obtained by assuming that the equilibrium molar and the analytical concentrations of ammonia are essentially the same; thus, for [NH3] = 0.100,

αM = 1 = 1.17 x 10-5

1 + 16 + 316 + 7.24 x 103 + 7.76 x 104

K”ZnY = α4 x αM x KZnY

= 5.2 x 10-2 x 1.17 x 10-5 x 3.2 x 1016

K”ZnY = 1.9 x 1010




Calculation of pZn after Addition of 20.0 mL of EDTA

At this point, only part of the zinc has been complexed by EDTA.

The remainder is present as Zn2+ and the four ammine complexes.

cM = 50.0 x 0.00500 – 20.0 x 0.0100 = 7.14 x 10-4 M 70.0

Substitution of this value gives

[Zn2+] = cMαM = (7.14 x 10-4)(1.17 x 10-5) = 8.35 x 10-9 M

pZn = 8.08


At the equivalence point, the analytical concentration of ZnY2- iscZnY2- = 50.0 x 0.00500 = 3.33 x 10-3 M 50.0 + 25.0




The sum of the concentrations of the various zinc species not combined with EDTA equals the sum of the concentrations of the uncomplexed EDTA species:

cM = cT

and

[ZnY2-] = 3.33 x 10-3 – cM ≈ 3.33 x 10-3 M

Substituting, we have

[ZnY2-] = K”ZnY

cM2

3.33 x 10-3 = 1.9 x 1010

cM2

cM = 4.18 x 10-7 M

Employing [Zn2+] = cMαM , we obtain




Zn2+] = cMαM = (4.18 x 10-7)(1.17 x 10-5) = 4. 90 x 10-12

pZn = 11.31


The solution now contains an excess of EDTA; thus,

cEDTA = cT = EDTA – sample VT

= 6.25 x 10-4 M

and since essentially all of the original Zn2+ is now complexed,

cZnY2- = [ZnY2-] = 50.0 x 0.00500 = 3.12 x 10-3 M 80.0

Rearranging K”ZnY = [ZnY2-] = α4αMKZnY gives, cMcT




cM = [ZnY2-] = 3.12 x 10-3 = 2.63 x 10-10 M cTK”ZnY (6.25 x 10-4)(1.9 x 1010)

From the equation [Mn+] = αMcM,

[Zn2+] = cMαM = (2.63 x 10-10)(1.17 x 10-5) = 3.07 x 10-15

pZn = 14.51

INDICATORS FOR EDTA TITRATION

Eriochrome Black T (EBT) is a typical metal-ion indicator that is used in the titration of several common cations.Its behavior as a weak acid is described by the equations:

H2O + H2In- ⇋ HIn2- + H3O+ K1 = 5 x 10-7 Red blue

H2O + HIn2- ⇋ In3- + H3O+ K2 = 2.8 x 10-12

blue orange




The metal complexes of Eriochrome Black T are generally red as in H2In-.

Thus, for metal ion detection, it is necessary to adjust the pH to 7 or above so that the blue form of the species, HIn2-, predominates in the absence of a metal ion.

Until the equivalence point in a titration, the indicator complexes the excess metal ion, so the solution is red.

When the EDTA becomes present in slight excess, the solution turns blue as a consequence of the reaction

MIn- + HY3- ⇋ HIn2- + MY2-

red blue

TITRATION METHODS EMPLOYING EDTA

Direct Titration




Methods Based on Indicators for the Analyte Ion

Mn+ + EDTA ⇋ M(EDTA)

color 1 color 2

Over 40 elements can be determined by direct titration with EDTA using metal ion indicators.

Methods Based on Indicators for an Added Metal IonA small amount of a cation is introduced that forms an EDTA complex that is less stable than the analyte complex and for which a good indicator exists.

MgY-2 + Ca+2 ⇋ CaY-2 + Mg+2

(less stable)

Mg+2 + EBT ⇋ Mg(EBT)red

Mg(EBT) + Y4- ⇋ MgY-2 + EBTred blue




Back-Titration MethodsBack-titration is useful for the determination of the cations that form stable EDTA complexes and for which a satisfactory indicator is not available.

The method is also useful for cations that react only slowly with EDTA.

A measured excess of standard EDTA solution is added to the analyte solution.

After the reaction is judged complete, the excess EDTA is back-titrated with a standard magnesium or zinc ion solution to an Eriochrome Black T or Calmagite end point.

M + Y(xs) ⇋ MY + Y(unreacted)

Y(unreacted) + Mg ➝ MgY (should be less stable than MY)

Displacement MethodsAn unmeasured excess of a solution containing the magnesium or zinc complex of EDTA is introduced into the analyte solution.




If the analyte forms a more stable complex than that of magnesium or zinc, the following displacement reaction occurs:

MgY2- + M2+ ➝ MY2- + Mg2+

The liberated cation is then titrated with the standard EDTA.analyte

Interference from a particular cation can sometimes be eliminated by adding a suitable masking agent, an auxiliary ligand that preferentially forms highly stable complexes with the potential interference.

For example, cyanide ion is often employed as a masking agent to permit the titration of magnesium and calcium ions in the presence of ions such as cadmium, cobalt, copper, nickel, zinc, and palladium.

All of the latter form sufficiently stable cyanide complexes to prevent reaction with EDTA.




Illustration Showing How Masking and Demasking Agents Can Be Used to Improve The Selectivity of EDTA Titrations

Lead, magnesium, and zinc can be determined on a single sample by two titrations with standard EDTA and one titration with standard Mg2+. The sample is first treated with an excess of NaCN, which masks Zn2+ and prevents it from reacting with EDTA.

Zn2+ + 4CN- ➝ Zn(CN)42-

Zn is masked, no reaction with EDTA

1. The Pb2+ and Mg2+ are then titrated with standard EDTA.

After the equivalence point has been reached, a solution of the complexing agent BAL (2-3-dimercapto-1-propanol, CH2SHCHSHCH2OH), which we will formulate R(SH)2, is added to the solution.

This bidentate ligand reacts selectively to form a complex with Pb2+ that is much more stable than PbY2-:

PbY2- + 2R(SH)2 ➝ Pb(RS2)22- + Y4-




2. Y4- is related to Pb2+. It is determined by titration with Mg2+

The liberated Y4- is then titrated with a standard solution of Mg2+. Finally, the zinc is demasked by adding formaldehyde.

Zn(CN)42- + 4HCHO + H2O ➝ Zn2+ + 4HOCH2CN + 4OH-

3. Zn2+ is titrated with EDTA.

The liberated Zn2+ is then titrated with the standard EDTA solution.

ExampleSuppose the initial titration of Mg2+ and Pb2+ required 42.22 mL of 0.02064 M EDTA. Titration of the Y4- liberated by the BAL consumed 19.35 mL of 0.007657 M Mg2+. Finally, after addition of formaldehyde, the liberated Zn2+ was titrated with 28.63 mL of the EDTA. Calculate the percentages of the three elements in a 0.4085 g sample.




The initial titration reveals the number of millimoles of Pb2+ and Mg2+ present.

no. mmol (Pb2+ + Mg2+) = 42.22 x 0.02064 = 0.87142

The second titration gives the number of millimoles of Pb2+. Thus,

no. mmol Pb2+ = 19.35 x 0.007657 = 0.14816

no. mmol Mg2+ = 0.87142 – 0.14816 = 0.72326

Finally, from the third titration we obtain

no. mmol Zn2+ = 28.63 x 0.02064 = 0.59092

To obtain the percentage, we write

0.14816 mmol Pb x 0.2072 g Pb / mmol x 100% = 7.515% Pb 0.4085 g sample




0.72326 mmol Mg x 0.024305 g Mg / mmol x 100% = 4.303% Mg 0.4085 g sample

0.59092 mmol Zn x 0.06539 g Zn / mmol x 100% = 9.459% Zn 0.4085 g sample




A 1.3174-g sample containing the chloride salts of magnesium, mercury(II), and zinc was dissolved in 250.0 mL of solution. A 50.00-mL aliquot was treated with 10 mL of an NH3/NH4Cl buffer at pH 10 followed by 25.00 mL of 0.05331 M EDTA. After a few minutes of mixing, the excess EDTA was back-titrated with 11.43 mL of 0.01816 M MgCl2. A second 50.00-mL aliquot was made basic and treated with excess NaCN, complexing both the mercury and the zinc. The magnesium in this sample required 16.83 mL of 0.005583 M EDTA for titration. The solution remaining at the end of this titration was treated with excess formaldehyde, which reacts with the free CN- and with Zn(CN)42-:

CN- + HCHO +H2O H2C(OH)(CN) + OH-

The liberated Zn+2 required 28.47 mL of the EDTA for titration. Calculate the percentage of each metal in the sample.

Exercise




Solution

mmol xs EDTA = 25.00 mL x 0.05331 M = 1.3328 mmol

mmol unreacted EDTA = 11.43 mL x 0.01816 M = 0.20757 mmol

mmol M+2 = 1.3328 mmol - 0.20757 mmol = 1.12523 mmol

ABANGAN ...

20) Complexometric Titration

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