无机分析化学上-第五章配位反应

EDTA乙二胺四乙酸(H4Y)乙二胺四乙酸二钠盐(Na2H2Y)H2

H2HOOCH2C-OOCH2CCH2COOHCH2COO-：：：：·

NH+

CC

·NH+EDTAH6Y2+

＝H+

+

H5Y+H5Y+

＝H+

+

H4YH4Y

＝H+

+

H3Y-H3Y-

＝H+

+H2Y2-2H

Y2-

＝H+

+

HY3-HY3-

＝H+

+

Y4-a1K

==

10-0.90[H+][H5Y][H6Y]Ka2==

10-1.60[H+][H4Y]5[H

Y]Ka3==

10-2.00[H+][H3Y][H

Y]a6K

==

10-10.26[HY][H

Y]2[H+][Y]Ka5==

10-6.164[H+][H2Y][H3Y][H+][HY]a4K

==

10-2.67EDTA:

-pH图246810

12

14pHH6Y2+5H

Y+H3Y-H4YH2Y2-HY3-Y4-1.0

0.80.

60.

40.

20.

00M-EDTA螯合物的构型OCaOOONCOH2CH2CCH2NCH2CCH2H2CCOCOOEDTA

通常与金属离子形成1:1的螯合物多个五元环某些金属离子与EDTA的形成常数lgKlgKlgKlgKNa+1.7Mg2+

8.7Ca2+

10.7Fe2+

14.3Al3+

16.1Hg2+

21.8Th4+

23.2Zn2+

16.5Cd2+

16.5Fe3+

25.1Bi3+

27.9Pb2+

18.0Cu2+

18.8一、配合物的稳定常数(K,

)M

+

L=

ML5.5

配合物的离解平衡[M][L]KML=

[ML]

●●●●●●●●●K

表示相邻络合物之间的关系M

+L

=MLML

+

L

=

ML2MLn-1

+

L

=

MLn累积稳定常数

表示络合物与配体之间的关系逐级稳定常数Ki[ML][M][L][ML2]1K

=[ML][L]K2=[MLn][MLn-1][L]nK

=[ML][M][L][ML2]1=K1=[M][L]22=K1K2=[MLn][M][L]nnn=K1K2

···

K

=[ML]

=

1

[M]

[L][ML2]

=

2

[M]

[L]2●●●[MLn

]=

n

[M]

[L]n●●●M

+L

=MLML

+

L

=

ML2MLn-1

+

L

=

MLncM=[M]+[ML]+[ML2]+…+[MLn]=[M](1+

1

[L]+

2

[L]2+…+

n

[L]n)溶液中各级配合物的分布分布分数δM=[M]/cM

=

1/(1+1[L]+2[L]2+…+n[L]n)δML=[ML]/cM

=

1[L]/(1+1[L]+2[L]2+…+n[L]n)=δM1[L]δMLn=[MLn]/cM

=

n[L]n/(1+1[L]+2[L]2+…+n[L]n)=δMn[L]n●●●4-+Y +H

＝

HY3-3-+HY +

H

＝

H2Y2-H2Y2-

+

H+

＝

H3Y-H3Y-

+

H+

＝

H4YH4Y +

H+

＝

H5Y+H5Y+

+

H

+

＝

H6Y2+K1==1010.261=K1=

1010.262=K1K2=

1016.423=K1K2K3=

1019.091a6K1K2==

106.16a5K1K3==

102.67a4K1K4==

102.004=K1K2K3K4=

1021.095=K1K2..K5=

1022.69Ka3K5==

101.601Ka26K

==

10

0.906=K1K2..K6

=

1023.591Ka1EDTA的有关常数离解常数Ka1Ka2Ka3Ka4Ka5Ka610-0.9010-1.6010-2.0010-2.6710-6.1610-10.26逐级常数K1K2K3K4K5K61010.26106.16102.67102.00101.60100.90累积常数β1β2β3β4β5β61010.261016.421019.091021.091022.691023.59二、副反应系数和条件稳定常数M

+

Y

=

MY主反应副反应OH-LH+OH-MOHMHY

MOHYNNYH+HY●●●●●●ML●●●●●●●●●●●●M(OH)n

MLn

H6YMYMY1.副反应系数：未参加主反应组分的浓度[X]与游离的平衡浓度[X]的比值，用表示。

[M]

[M]M=Y

=

[Y] [Y]

[MY]MY=

[MY]M+Y=MY主反应副反应OH-LH+OH-MOHNNYH+HYMHY

MOHY●●●●●●ML●●●●●●●●●●●●M(OH)n

MLn

H6Y[M]

[Y][MY](1)配位剂的副反应系数Y:Y(H)：酸效应系数Y(N)：共存离子效应系数

[Y]

[Y]Y=NNYH+HY●●●H6YYM

+Y=MY=

1

Y酸效应系数Y(H)：[Y]Y(H)=(1+1[H+]+2[H+]2+…+6[H+]6)

[Y]

[Y]+[HY]+[H2Y]+…+[H6Y]=

=

[Y][Y]+[Y][H+]1+[Y][H+]22+…+[Y][H+]66[Y]=Y(H)

≥1[Y]

=

[Y] Y(H)例1

计算pH5.00时EDTA的Y(H)解Y(H)

=(1+1[H+]+2[H+]2+…+6[H+]6)=1+10-5.00＋10.26+10-10.00+16.42+10-15.00+19.09+10-20.00＋21.09+10-25.00+22.69+10-30.00+23.59=1+105.26+106.42+104.09

+101.09+10-2.31+10-6.41=106.45lg

Y(H)～pH图EDTA的酸效应系数曲线lgY(H)共存离子效应系数Y(N)Y(N)===

1+

KNY[N]

[Y]

[Y][Y]+[NY][Y]Y(N)=

1+KN1Y[N1]+KN2Y[N2]+…+KNnY[Nn]=Y(N1)+Y(N2)+…+Y(Nn)-(n-1)

[Y]

[Y]+[N1Y]+[N2Y]+…+[NnY]=

=[Y]

[Y]多种共存离子Y的总副反应系数Y=

Y(H)

+

Y(N)

-1

=

[Y]

=Y[Y]+[HY]+[H2Y]+

·+[NY][Y][Y](2)

金属离子的副反应系数

MM

+ Y

=

MYOH-LMOH●●●ML●●●M(OH)n

MLnMM(OH)=1

+1[OH-]+

2[OH-]2+

…+

n[OH-]nM(L)

=1+1[L]

+2[L]2+…+n[L]n多种配位剂共存

M

=

M(L1)+

M(L2)

+…+

M(Ln)-(n-1)M(L)

=1+1[L]

+2[L]2+…+n[L]n64200

2

4

6

8

10

12

14pHAlFeIIIBiZnPbCdCuFeIIlgM(OH)~pH例用EDTA滴定Zn2+至化学计量点附近,pH=11.00，[NH3]=0.10mol·L

-1,计算lg

Zn查附录表,pH=11.00时,lg

Zn(OH)=5.4Zn(NH3)=1+[NH3]1+[NH3]2

2+[NH3]3

3+[NH3]4

4=1+10-1.00+2.27+10-2.00+4.61+10-3.00+7.01+10-4.00+9.06=1+101.27+102.61+104.01+105.06=105.10lg

Zn=

lg(

Zn(NH3)+

Zn(OH)-1)=lg(105.1+105.4-1)=

5.6例

用EDTA滴定Zn2+至化学计量点附近,pH

=

9.0，cNH3=0.10mol·L

-1,计算lg

Zn(NH3)c(NH3)=[NH3]+[NH4+]+

[Zn(NH3)i

]忽略Zn(NH3)=1+[NH3]1+[NH3]2

2+[NH3]3

3+[NH3]4

4=103.23c

NH

==

0.03

molL-110-9.4

0.110-9.0

+10-9.4NH3=δ0(3)

配合物的副反应系数

MYM

+ Y

=MYH+OH-MHY

MOHYMY酸性较强MY(H)=1+KMHY×[H+]碱性较强MY(OH)=1+KM(OH)Y×[OH-]计算：pH=3.0、5.0时的lg

ZnY(H)(KZnHY

=

103.0

)pH=3.0,

αZnY(H)=1+10-3.0+3.0=2

,

lgαZnY(H)=

0.3pH=5.0,

αZnY(H)=1+10-5.0+3.0=1, lgαZnY(H)=

0lg

KMY

=

lgKMY

-

lgM

-

lgY

+

lg

MY≈

lgKMY

-

lgM

-

lgY=

lgKMY

-lg(

M(A1)+

M(A2)

+…+

M(An)-(n-1))-

lg

(Y(H)

+

Y(N)

-1)KMY==KMY[MY'][M'][Y']MYMY2.条件稳定常数（酸度、试剂浓度一定）例计算pH2.0和5.0时的lgK

ZnYlg

K

=lg

K

-

lg

-lg

+

lg

ZnY

ZnY

Zn

YZnYlgKZnY

=16.5,

lgKZnHY

=3.0pH=2.0:lgY(H)=13.79, lg

Zn(OH)=0

ZnY(H)=1+[H+]KZnHY

=1+10-2.0+3.0=101.0,

lgZnY(H)=1.00lgK

=lgK

-lg

-lg

+lg

=

3.71ZnY

ZnY

Zn

Y

ZnYpH=5.0:lgY(H)=6.45, lg

Zn(OH)=0

ZnY(H)=1+[H+]KZnHY

=1+10-5.0+3.0=1,

lgZnY(H)=0.00