chapter15 p 17 complexation reactions and titrations chapter

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17Complexation Reactions and Titrations

CHAPTER

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Complexation reaction( 複合反應、錯合反應 )• One of the first uses of these reactions was for

the titrating cations ~ ~ the major topic

• Complexes are colored or absorb ultraviolet radiation ~ ~ the basis for spectrophotometric determine

Complex: [adj.] 錯綜複雜的，合成的 [n] 複合物、錯合物

Complexes: 複數

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§ The Formation of Complexes

• Complex: metal + ligand• Ligand: have at least one pair of unshared electron

s available for bond formation.

(electron donor) Ex: H2O, NH3, Cl-, Br-, I-……~ ~ An ion or a molecule that forms a covalent bond

with a cation or neutral metal atom by donating a pair of electrons, which ate then shared by two. (p450)

449

[Cr(NH3)6]3+

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§ The Formation of Complexes• Coordination number ( 配位數 ): the number of c

ovalent bonds that a cation tends to form with electron donors

• Ex: [Cu(NH3)4]2+, [CuCl4]2-, [Cr(NH3)6]3+

• Complexometric method: titrimetric methods based on complex formation ( 錯離子滴定法 )

• Chelate ( 螯合物 ) : when a metal ion coordinates with two or more donor groups of a single ligand to form a five- or six-member heterocyclic ring.

450

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Cu2+ + C C

H

NH2

O

OHH2

Glycine

CuO

NH

O

NH

CH2H2C

C C

O O

+ 2H+

• Chelate: when a metal ion coordinates with two or more donor groups of a single ligand to form a five- or six-member heterocyclic ring. ( 螯合物 )

• Unidentate: ( 單牙基 ) a ligand that has a single donor group Ex: NH3

• Bidentate: ( 雙牙基 ) a ligand that has two groups available for covalent bonding Ex: glycine

• Tridentate, tetradentate, ……

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§ The Formation of Complexes

• Macrocycles : metal ions and cyclic organic compounds

～～ the organic compounds contain nine or more atoms in the cycle and include at least three heteroatoms, usually O, N, S.

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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 in Chemistry in 1987

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§ Complexation Equilibria

M + L MLML + L ML2ML2 + L ML3

MLn-1 + L MLn

.

.

.

.

.

.

.

.

.

451

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M + L ML

M + 2L ML2

M + 3L ML3

[ML]

[M] [L]= K1

[ML2]

[M] [L]2 = K1 K2

[ML3]

[M] [L]3 = K1K2 K3

β: the overall formation constant

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α: the fraction of the total metal or metal complex concentration existing

M + L MLML + L ML2ML2 + L ML3

MLn-1 + L MLn

.

.

.

.

.

.

.

.

.

M =1 + LLLnLn

1

ML =1 + LLLnLn

L

ML =1 + LLLnLn

L

2

ML =1 + LLLnLn

nLn

n

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Calculation of α for metal complexes

452

M + L MLML + L ML2ML2 + L ML3

MLn-1 + L MLn

.

.

.

.

.

.

.

.

.

M =2 ML =

nCT

[M]ML =

CT

[ML]ML =

CT

[ML2]

CT

[MLn]

CT = CM = [M] + [ML] + [ML2] +…… + [MLn]

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M + L ML

M + 2L ML2

M + 3L ML3

[ML]

[M] [L]= K1

[ML2]

[M] [L]2 = K1 K2

[ML3]

[M] [L]3 = K1K2 K3

[ML] = β1 [M] [L][ML2] = β2 [M] [L]2

[ML3] = β3 [M] [L]3

[MLn] = βn [M] [L]n

CT = CM = [M] + [ML] + [ML2] +…… + [MLn] = [M] +β1[M] [L]+β2[M] [L]2 + ...+βn[M] [L]n

= [M] { 1 + β1[L]+β2[L]2 + …...+βn[L]n }

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M =CM

[M]

=[M] {1 + LLLnLn}

[M]

=1 + LLLnLn

1

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§ The Formation of Insoluble Species

MxAy(s) xMy+ (aq) + yAx- (aq) Ksp = [My+]x [Ax-]y

§ Ligands That Can Protonate

•Side reaction~ involving the metal or the ligand•For ligand ~ if the ligand is weak acid, then ligand can be protonated.

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Complexation with protonating ligands

M + L L : the conjugate base of polyprotic acid Adding acid ~ reduces the concentration of free

L available to complex with M, ~ decrease the effectiveness of L as a complexin

g agent.

Le Châtelier’s principle

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For a diprotic acid : oxalic acid H2Ox

H2Ox, HOx-, Ox2-草酸

[H2Ox] + [HOx-] + [Ox2-]CT =

[H+]2 + Ka1[H+] + Ka1Ka2

0 =CT

[H2Ox]=

[H+]2

[H+]2 + Ka1[H+] + Ka1Ka2

1 =CT

[HOx-]=

Ka1[H+]

[H+]2 + Ka1[H+] + Ka1Ka2

2 =CT

[Ox2-]=

Ka1Ka2

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Conditional Formation Constant

• Conditional Formation Constant

(Effective Formation Constant)

~ ~ the effect of pH on the free ligand concentration in a complexation reaction.

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Fe3+ + Ox2- FeOx+

[FeOx+]

[Fe3+] [Ox2-]

[FeOx+]

[Fe3+] CT = K1 =

At a particular pH value , 2 is constant

[FeOx+]

[Fe3+] CT = K1 '=

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§ Titrations with Inorganic Complexing Agents

AgNO3 + X- AgX(s)

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Figure 17-1Titration curves for complexometric titrations. Titration of 60.0 mL of a solution that is 0.020 M in metal M with (A) a 0.020 M solution of the tetradentate ligand D to give MD as the product; (B) a 0.040 M solution of the bidentate ligand B to give MB2; and (C) a 0.080 M solution of the unidentate ligand A to give MA4. The overall formation constant for each product is 1020.

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§ Aminocarboxylic Acid Titrations • Ethylenediaminetetraacetic acid [EDTA]

H2C

H2CN N

C

C C

C OH

OH

HO

HO

O O

OO

C

C

C

C

458

The EDTA molecule has six potential sites for bonding a metal ion.

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H2C

H2CN N

C

C C

C OH

OH

HO

HO

O O

OO

C

C

C

C

Four carboxyl group~~~ H4Y 代表 EDTA

Acidic Properties of EDTA

H4Y H+ + H3Y-

H3Y- H+ + H2Y2-

H2Y2- H+ + HY3-

HY3- H+ + Y4-

K1 = 1.02 x 10-2

K2 = 2.14 x 10-3

K3 = 6.92 x 10-7

K4 = 5.50 x 10-11

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Figure 17-2Composition of EDTA solutions as a function of pH.

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Figure 17F-1Structure of H4Y and its dissociation products. Note that the fully protonated species H4Y exists as the double zwitterion with the amine nitrogens and two of the carboxylic acid groups protonated. The first two protons dissociate from the carboxyl groups, while the last two come from the amine groups.

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Figure 17-3Structure of a metal/EDTA complex. Note that EDTA behaves here as a hexadentate ligand in that six donor atoms are involved in bonding the divalent metal cation.

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§ Complexes of EDTA and Metal IonsThe reagent combines with metal ions in a 1:1 ra

tio regardless of the charge on the cation.

Ag+ + Y4- AgY3-

Al3+ + Y4- AlY-

Mn+ + Y4- MY(n-4)+

KMY=[MY(n-4)+]

[Mn+] [Y4-]

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Mn+ + Y4- MY(n-4)+

KMY=[MY(n - 4)+]

[Mn+] [Y4-]

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§ Equilibrium Calculations Involving EDTA

A titration curve for the reaction of Mn+ and EDTA

~ ~ ~ a polt of pM versus reagent volume

Mn+ + Y4- MY(n-4)+

KMY=[MY(n - 4)+]

[Mn+] [Y4-]

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From 15H • Tetraacetic acid ~~~

H4Y H3Y- H2Y2- HY3- Y4- CT

[Y4-]

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

[H+]4 + K1[H+]3 + K1K2[H+]2 + K1K2K3[H+] + K1K2K3K4

K1K2K3K4

CT

[Y4-]=

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Conditional Formation Constants

Mn+ + Y4- MY(n-4)+

KMY=[MY(n - 4)+]

[Mn+] [Y4-]

CT

[Y4-]

KMY =[MY(n - 4)+]

[Mn+] 4CT

[MY(n - 4)+]

[Mn+] CTKMY 4=K'MY =

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[H+]4 + K1[H+]3 + K1K2[H+]2 + K1K2K3[H+] + K1K2K3K4

K1K2K3K4

CT

[Y4-]=

K1K2K3K4

D

[H+]4

D

K1[H+]3

D

K1K2[H+]2

D

K1K2K3[H+]

D

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Example 17-1 Calculate the molar Y4- concentration in a 0.0200M

EDTA solution buffered to a pH of 10.00

試求下列溶液的 [Y4-] ，0.0200 M EDTA 溶液， pH 值為 10.0

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Mn+ + Y4- MY(n-4)+

KMY=[MY(n - 4)+]

[Mn+] [Y4-]

Calculation of the cation concentration in EDTA solutions

[MY(n - 4)+]

[Mn+] CTKMY 4=K'MY =

Conditional formation constant

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Example 17-2

Calculate the equilibrium concentration of Ni2+ in a solution with an analytical NiY2- concentration of 0.0150M at pH (a) 3.0 and (b) 8.0

計算在不同 pH 值下， 0.0150M 的 [NiY2-] 溶液中有多少 [Ni2+] 利用 conditional formation constant 解題

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Example 17-3

Calculate the concentration of Ni2+ in a solution that was prepared by mixing 50.0mL of 0.0300M Ni2+ with 50.00mL of 0.05M EDTA. The mixture was buffered to a pH of 3.0

將 50.0mL ， 0.0300M Ni2+ 與 50.00mL ， 0.05M EDTA 溶液混合，並將混合液的 pH 值調整至 3.0 。試計算 Ni2+ 的濃度

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Example 17-4

Use a spreadsheet to construct the titration curve of pCa versus volume of EDTA for 50.00mL of 0.00500M Ca2+ being titrated with 0.0100M EDTA in a solution buffered to a constant pH of 10.0

§ EDTA Titration Curves

467

在 pH 值為 10.0 時，利用 0.0100M EDTA 溶液滴定 50.00mL0.00500M Ca2+ 溶液，並建構其滴定曲線 pCa vs EDTA( 體積 )

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Figure 17-5Spreadsheet for the titration of 50.00 mL of 0.00500 M Ca2+ with 0.0100 M EDTA in a solution buffered at pH 10.0.

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Figure 17-6EDTA titration curves for 50.0 mL of 0.00500 M Ca2+ (K’C

aY=1.75×1010) and Mg2+ (K’MgY=1.72×108) at pH 10.0. Note that because of the larger formation constant, the reaction of calcium ion with EDTA 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.

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Figure 17-7Influence of pH on the titration of 0.0100 M Ca2+ with 0.0100 M EDTA. Note that the end point becomes less sharp as the pH decreases because the complex formation reaction is less complete under these circumstances.

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Figure 17-8Titration curves for 50.0 mL of 0.0100 M solutions of various cations at pH 6.0.

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Figure 17-9Minimum pH needed for satisfactory titration of various cations with EDTA. (From C.N.Reilley and R.W.Schmid, Anal. Chem., 1958,30,947.copyrigh 1958 American Chemical Society. Reprinted with permission of the American Chemical Society.)

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• pH increase ~ ~ [OH-]

§ The Effect of Other Complexing Agents on EDTA Titration Curves

會產生 M(OH)x 的化合物

An auxiliary complexing agent is needed to keep the cation in solution, cause the end points to be less sharp.Auxiliary: 輔助的

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Figure 17-10Influence of ammonia concentration 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.

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Feature 17-5EDTA titration curves when a complexing agent is present

Zn2+ , NH3 EDTA (Y4-)

metal ligand complexes

[Zn(NH3)2+][Zn(NH3)2

2+][Zn(NH3)3

2+][Zn(NH3)4

2+] [ZnY2-]

鋅一般為 4 配位

M =CM

[Zn2+]

CM = [Zn2+] + [Zn(NH3)2+] + [Zn(NH3)22+] + [Zn(NH3)3

2+] + [Zn(NH3)42+]

M 只與 NH3 的濃度， Zn(NH3)x 的形成常數有關

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αM: the fraction of the total metal or metal complex concentration existing

M + L MLML + L ML2ML2 + L ML3

MLn-1 + L MLn

.

.

.

.

.

.

.

.

.

M =1 + LLLnLn

1

M =1 + NH3NH3

NH34NH34

1

K"ZnY = M4KZnY =[ZnY2-]

CTCM

K”ZnY : 在特定 pH 與特定 NH3 濃度下的條件形成常數

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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 NH3 and 0.175M NH4Cl to provide a constant pH of 9.0

在 pH 值為 9.0 的緩衝溶液中 (NH3 +NH4Cl) ，利用 0.0100M EDTA 溶液滴定 50.00mL 0.00500M Zn2+ 溶液，求 pZn值。 EDTA 體積 (a) 20.0 mL (b)25.0 mL (c) 30.0 mL

K”ZnY : 在特定 pH 與特定 NH3 濃度下的條件形成常數

K"ZnY = M4KZnY =[ZnY2-]

CTCM

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Figure 17-11Structure and molecular model of Eriochrome Black T. The compound contains a sulfonic acid group that completely dissociates in water and two phenolic groups that only partially dissociate.

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Figure 17-12Structural formula and molecular model of Calmagite. Note the similarity to Eriochrome Black T (see Figure 17-11).

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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

在 pH 值為 10.0 的溶液中，利用 EBT 當指示劑，滴定 Ca2

+ 與 Mg2+ 時， EBT 的變色範圍。

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Figure 17F-2Typical kit for testing for water hardness in household water.