Tài liệu Chapter 12: EDTA Titrations docx

>» The central atom, usually a metal cation, accepts the pair of electrons from the "donor molecule or ion" to form a coordinate covalent bond.. > When a bidentate or higher number of d

Chapter 12:

EDTA Titrations

Review of Textbook for Chem115/116: Chapters

17 & 23: Complex (coordinate compound)

> Coordinate covalent bonds: a bond formed when

both electrons of the bond are donated by one atom

Ag’ + 2(:NH,) @ [H,N: Ag :NH,]*

Electron configuration of Ag [Kr]4d!°5s!5P°

Ag* [Kr]4d!°5s° 5P°

Sp hybrid orbitals: accommodate 2 pairs of electrons

> Complex (Coordinate compound): a compound

consisting either of complex ions and other ions of

opposite charge or of neutral complex species.

| Basic Concepts and Terms

A Complex-Formation Reactions:

>some elements (mostly metal ions) can form coordination

compounds with molecules/ions which have a "spare pair" of electrons

>» The central atom, (usually) a metal cation, accepts the pair of

electrons from the "donor molecule or ion" to form a coordinate covalent bond

>The empty orbitals of the cation and the orbital occupied by the bonding pair of electrons on the donor ion or molecule (ligand) form a new molecular orbital

>The number of coordinate covalent bonds that a cation tends to

form with individual ligands (or functional groups on ligands) is

known as its coordination number

>» Typical values for coordination numbers are 2, 4 and 6

>» The compounds formed can be neutral, positively or negatively charged depending on the charge of the reacting species.

> When a ligand has a single complexing or donor group in its structure, it

is said to be unidentate,

> when there are two groups, it is bidentate,

> When there are three groups, it is tridentate ligand, etc

> When a bidentate (or higher number of donor groups present in the

ligand) forms a coordinate covalent compound with a metal cation, we call the resulting compound a metal chelate

Structure of a Metal/EDTA Chelate

1.1 < 1015

1.3 x 1016 1.3 xX 1075 7.9 x 1035

1.6 X 10%

log Kary |

18.80 -

16.50 | 16,46 21.80 18.04

16.13 25.1 25.9

23.2

10.0 20.0 30.0 40.0 50.0 60.0 Volume of 0.0100 M EDTA, mL

Any Questions?

II Titrations with Ethylenediaminetetraacetic Acid (EDTA)

two ammo groups on the ethylene group

EDTA is a hexadentate ligand and can coordinate up to Six positions on the central metal atom The structure

of a compound with six coordination sites is octahedral

EDTA Disassociation

EDTA Dissociation Constants:

HY + H0< => HY + HOt Kai = 1.02x10”

Hs Y + H0< => HY? + HOt Kya = 2.14x10°

H:Y” + H;0< => HY? + HOt Ka3 = 6.92x10”

HY? + H;0< = > Y'“+ H0? Kas = 5.50x10"1!

For our purposes, only the completely dissociated EDTA ion will be

able to form a chelate with the metal ion Therefore, the number

metal ions per unit volume solution chelated by EDTA will be pH-

dependent

mg VY“ K,K,K,K,

" C, [HƑ+K[H'Ƒ+K,K,[H'ƒ+K,K,K,[H']+K,K;K;K,

> Therefore, only EDTA solutions with pH's above 10 will [Y*’] as

a major fraction

> In addition, EDTA always forms 1:1 molar complexes with metal

ions

lil Equilibrium Calculations Involving EDTA

A Titration Curve: M™ + Y“ <=> MYP”-“”

A titration curve for the reaction of a cation with EDTA:

a plot of pM = (- log[M™]) on the y-axis, versus volume (mL) of titrant (EDTA solution) on the x-axis

Note in Table 14-1, that the Kyy values are in the range of 10’ to 10”

These reactions proceed very far to the right!

Since the fraction of [Y*] is dependent upon pH, it is necessary to take the pH of the solution into account when determining pM from the

equilibrium constant expression

We can determine the fraction of EDTA”, au, as a function of pH (see Table 14-2),

at pH = 10, a,= 0.35; the mole percent of [Y"] is 35%

at pH = 11, a, = 0.85; the mole percent of [Y”] is 85%

at pH = 12, a,= 0.98; the mole percent of [Y*] is 98%

1.1 < 1015

1.3 x 1016 1.3 xX 1075 7.9 x 1035

1.6 X 10%

log Kary |

18.80 -

16.50 | 16,46 21.80 18.04

16.13 25.1 25.9

23.2

10.0 20.0 30.0 40.0 50.0 60.0 Volume of 0.0100 M EDTA, mL

Example 12-3: Calculate the equilibrium concentration of Cu”” in a solution with an analytical CuY~ concentration of 0.0150 M at (a) pH 4.0 and (b) pH 10

Cu** + Y* <=> CuY” Keuy2- = 6.3x10"°

[CuY“] [CuY“] [CuY“ ]

2- T— = ›Œ{; =——

K’ is called as a conditional constant, which depends on pH

C_,, =[Cu*]4+[Cuy7 ]

CuY2 -

Cu?

Because K is very large (6.3x10"°), [Cu2*] << [CuY7]

Coy = [Cu**]+[CuY~ ]=[CuY~ ]=0.0150M

Example 12-4: Calculate the pCu”” in a solution prepared by mixing 50.00 mL of 0.0300 M Cu* with 50.0 mL of 0.0500 M EDTA at (a) pH 4.0 and (b) pH 10

[Y*]= a, C,,

C = excess of EDTA + dissociation from CuY” = 0.0100 M + [Cu**]

[Cut2] << excess of EDTA

Any Questions?

IV The Derivation of EDTA Titration Curves

Example 12-5: Calculate the pCa for an assay of calcium

ion using EDTA as the titrant The titration medium is

buffered to oH = 10 50.00 mL of 0.00500 M Ca* will be

titrated with 0.0100 M EDTA

A) 0.00 mL of EDTA titrant added: no EDTA has been added,

pCa = - log [Ca”] = - log [0.00500] = 2.30

B) Vepta = 10.00 mL < V,,: Pre-equivalence point of pCa:

The titration medium contains excess [Ca**] and calcium ion which results from dissociation of the CaY”

Since, due to LeChatelier's Principle, the excess Ca** will suppress

the dissociation of the complex and the conditional formation

constant is > 10'°, we can assume that the [Ca] due to dissociatior

of the complex is negligible

Hence [Ca] = excess [Ca]

0.05000Lx0.00500 M - 0.01000Lx0.0100M

(0.05000 + 0.01000) L pCa = - log (0.00250) = 2.60

C) Vepta = 25.00 mL: At the equivalence point:

At this point, neither EDTA nor Ca” are in excess, and due to the large value for

Kuy', one can assume that the dissociation of the CaY~ complex is negligible:

[Ca**] << [CaY7]

Hence: Cc„¿: =[Ca ]+[CaY~ ]=[CaY~ ]=3.33x10°M

[Ca**] = Cr = [Y“] + [HY”] + [H;Y] + [H;Y] + [H„Y]

D) Vepta = 35.00 mL > V,,: After the equivalence point pCa:

Beyond the equivalence point, [CaY*] and [Y”] are obtained directly, and the [Ca*’] is obtained from the equilibrium constant expression:

C,;= excess of EDTA + dissociation from CaY~

= excess of EDTA + [Ca‘"] ([Ca*?] << excess of EDTA)

Note: How to determine the pre-, at, or post-equivalence point:

Calculate moles of [M™] = CyVwy (in this case, it is Cca2+V caz+) and moles of EDTA = CepraV epta Determine which one is titrant (in this case, it is EDTA)

1) Ifmoles of Ca** > moles of EDTA, it is pre-equivalence point (excess of Ca”’):

Such as at

(A) Coa2+ = 0.00500 M; V ca2+ = 50.00 mL

CEpTA = 0.0100 M; V epTa =0.00 mL

moles of Ca’* =2.50x10* moles > moles of EDTA = 0

Hence: it is pre-equivalence point

(B) Cca2+= 0.00500 M; V ca2+ = 50.00 mL

CEpTA = 0.0100 M; V epTa =10.00 mL

moles of Ca’* =2.50x10* moles > moles of EDTA = 1.00x10* moles

1) Ifmoles of Ca** = moles of EDTA, it is at equivalence point

(no excess of either Ca** or EDTA): Such as at

(C) Cca2+ = 0.00500 M; V ca2+ = 50.00 mL

Cepta = 0.0100 M; V epta =25.00 mL

moles of Ca” =2.50x10 moles = moles of EDTA = 2.50x10* moles

Hence: at equivalence point

2) lfmoles of Ca” < moles of EDTA, it is post-equivalence point (excess of

EDTA):

Such as at (D) Ccaz = 0.00500M; Vca¿.= 50.00 mL

CEDTA = 0.0100 M; VEDTA =35.00 mL

moles of Ca” =2.50x10 moles < moles of EDTA = 3.50x10' moles

Hence: itis postequivalence point

E) Please note the effect of increasing pH on the relative change of pCa before and after the equivalence point

Note that the endpoint becomes sharper as pH increases

G) Note which shows the dependence of K,, on pH and the

minimum pH values at which different metals tons can be

successfully titrated when no other complexing agents are

V The Effect of Other Complexing Agents on EDTA

Titration Curves

> Sometimes it is necessary to add a second complexing agent to the titration

medium to maintain the analyte metal ion in solution (many metal ions from

insoluble hydroxides or oxides at slightly acidic, neutral or basic pH values)

>» Another reason for adding a second complexing agent is to "mask" or remove interfering ions also present in the sample matrix

>» The second complexing agent usually has a higher affinity for the interfering ion (greater Kyy) than the EDTA and prevents it from reacting with the EDTA In this case, the second complexing agent is called a "masking agent”

> Finally, most buffers will complex metal ions because they also contain

functional groups which can form coordinate covalent bonds (-OH, -COOH, -

NH, etc.) and their effect on the free metal ion concentration must be

considered when calculating the pM

In order to account for the presence of secondary complexing agents, we must know that they are in solution!

If we define the cy to be the fraction of free metal ion in solution, then:

Formation Equilibrium Constants (Both Stepwise and Overall) for Complex lon Formation

Consider the formation of a complex ton

- K,, K,, K,, K, are the stepwise formation

constants for reactions 1 - 4, respectively

Example 12-6: consider the titration of Zn'? in an NH,ˆ/NH; buffer

To determine [Zn**] we must consider the Zn-ammonia complexes that form in the titration medium

and substituting the above into the expression for Cy in [Eq 1]:

Cy= [Zn*7]{1 +Ky[NH3] +K;K[NH3]° + K,K2K3[NH3]° + KiKoK3K,[NH3]°}

KE„ =Kux0/0u _!

CC,

Please note that the effect of a secondary complexing agent is to increase the pM (decrease

[M"*]) before and at the equivalence point

After the equivalence point, there is excess EDTA and the effect of the secondary

complexing agent is inconsequential For titrations beyond the equivalence point use K,,,' =

OnKyy to solve for [M*"] If you choose to use K,,,", then multiply c,, by o,

VI Indicators for EDTA Titrations

A metal ion indicator is a compound whose color changes when it binds to a metal ion For a metal ion indicator to be useful, it must bind the metal ion less strongly than EDTA does

A Eriochrome Black T is a typical metal ion indicator

For the titration of Mg** with EDTA, we can write the reaction:

MgIn’ (red) + Y* <=> MgY~ + Hin (blue)

In

|HIn“ |

What influences the chemical speciation of Eriochrome Black T?

1 pH, because Eriochrome Black T is a weak acid:

H;O + HạIn' (red) <=> HIn* (blue) + HạO” K,=5x 10"

HO + HIn* (blue) <=> In* (orange) + H3;0* Kạ = 2.8 x 10”?

It is necessary to adjust the pH > 7 so that the blue form pre-dominates in the absence of metal ion

2 pM of metal, the K; for Min must be known

Mgˆ” +ln” <=> Mgln K; = 1.0x10/

Example 12-7: Determine the transition ranges for Eriochrome Black T

in titration of Mg** and Ca** at pH 10.0 given

(a) H,O + Hin* (blue) <=> In* (orange) +H,0* K,=2.8x1074 [1]

se (b) Mg2* + In3- <=> MgIn- (red) K, = 1.0 x 107 [2]

(c) Ca2* + In* <=> Caln- (red) K, = 2.5 x 10° [3]

The detectable color change occurs when [MglIn‘]/[HIn2"] changes from 10 to 0.1

lf K, is multiplied by K,in Eq [2] we have: H,O + Hin? (blue) + Mg** <=> Mgin: (red) + H,O*

[Mg**]=

Indicator: Eriocrome Black T

For [Ca]:

If Kz is multiplied by K;in Eq [2] we have: H,O + Hin* (blue) + Ca”” <=> Caln' + HạO”

[HO *]lIn"] [Caln] [H;O'J[CaIn]

a [HIn” ] [Ca7llIln'] [HIn?7][Ca7]

[H,O'” J[Cahn ] 7.0x10 ˆ[HIn” ]

[Ca**]=

[Caln | _

substituting [HạO”] = 1.0 x 10°”, and [HIn”] I0 or 0.1

The range of [Ca?*] over which the color changes from red to blue occurs:

[Ca**] = 1.58 x 10° to 1.58 x 10°

pCa = 3.8 + 1.0

Hence: Mg* has greater affinity for In*

VI — Titraton Methods Employing EDTA

A Direct Titration

1 Methods based on indicators for the analyte ion

2 Methods based on indicators for an added ion (addition of Mg* to EDTA in titration of Ca*? with Eriochrome Black T)

At the endpoint: MgIn’ (red) + HY* > MgY* + Hin” (blue)

3 Potentiometric Methods: measurement of [M"’] using ion-selective electrodes

B Back-Titration: select an EDTA-Metal ion complex that is less

stable than that of the analyte-EDTA complex so that

MgY~ + M™ > MY"ế + Mg**

The liberated Mg** can then be titrated with a standard EDTA solution

Summary

Y Basic concepts & terms

#4 Coordination number

=" Ligand and chelate

= Unidentate, bidentate, tridentate

= Formation constant

=" Conditional formation constants: K’yy, K’ wy

Equilibrium Calculations for Complex-formation titration

Y Derive EDTA titration curves

Y Calculation of related concentrations

vV Effect of other complexing agents

v Theory and Calculation of Indicators for EDTA titration

Titration Methods employing EDTA

= Direct titration

"= Back-titration

Homework

> 12-C, D, 1, 2, 3, 7, 8, 24, 34

Before working on Homework,

Practice with all examples that we discussed in the class and examples in the textbook!!