03 Metal Complex Equilibria

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METAL COMPLEX EQUILIBRIA Many metal ions, especially transition metals, form coordinate covalent bonds with molecules or anions having a lone pair of e-s. – This type of bond formation is essentially a Lewis acid-base reaction – For example, the silver ion, Ag+, can react with ammonia to form the Ag(NH3)2+ ion.

TDCVillar Chem 32

METAL COMPLEX EQUILIBRIA M

+

Ag+

+



L 2 :NH3 ⇌

ML Ag(NH3)+2

A metal (Lewis acid) - is an electron deficient species, thus accept electrons from a ligand Ligand – (Lewis base) is a complexing agent an e- rich, and thus, e- donating species e.g. H2O, :NH3 Coordination Compound - compounds formed from combination of metal ions with a complexing agents Coordination number – The actual number of point of attachment (pair of e’s) accepted by the metal.

Metal when dissolved in water react with water to form hydrated compound. CuSO4 + 6 H2O

⇌

Cu(H2O)62+ + SO42-

Cu(H2O)62+ + 4 NH3 Fe(H2O)63+ + SCN- ⇌

⇌

Ag+

2 :NH3 ⇌

+

Ag(NH3)+2

METAL COMPLEX EQUILIBRIA M

+

L



ML

Coordination number – maximum number of coordinating ligands. N = 4 tetrahedral or square planar N = 5 trigonal bipyramidal N = 6 octahedral

It is common practice to omit the water and write the reaction in simplified form. Cu2+ + 4 NH3 ⇌ Cu(NH3)42+ + 4 H2O

Cu(NH3)42+ + 4 H2O

FeSCN2+ + 6H2O

Fe3+ + SCN- ⇌

FeSCN2+ + 6H2O

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Ligand

Ligand

monodentate - if it donates a single pair of electrons

monodentate - if it donates a single pair of electrons

e.g. X-, CN- , OH-, pyridine, H2O, :NH3

Ag+

+

2 :NH3 ⇌

Ag(NH3)+2

Ligand

6 :NH3 ⇌

Co2+ +

Ligand

bidentate ligand - donates two pairs of electrons.

bidentate ligand - donates two pairs of electrons.

e.g. ethylenediamine, :NH2CH2CH2H2N: [Co(en)3]3+ [Pt(en)2]2+ Co2+

+3

Ni2+ + 2

Dimethylglyoxime (DMG)

Ligand Polydentate ligand/multidentate ligand – also called a chelating agent

e.g. EDTA, trien (triethylenetetraamine)

The aqueous silver ion forms a complex ion with ammonia in steps. Ag+

+

:NH3 ⇌

Ag(NH3)+ + :NH3 ⇌

Ag(NH3)+ Ag(NH3)+2

When you add these equations, you get the overall equation for the formation of Ag(NH3)2+.

Cu2+ + trien

Ag+

+

2 :NH3 ⇌

Ag(NH3)+2

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The formation constant for Ag(NH3)2+ is:

M



X  MX

K f1 



Kf 

[ Ag ( NH 3 ) 2 ] [ Ag  ][ NH 3 ]2

MX

The formation constant, Kf , is the equilibrium constant for the formation of a complex ion from the aqueous metal ion and the ligands.

Ag+

2 :NH3 ⇌

+

Ag(NH3)+2



X  MX 2

MX  M X 

K f2 

MX  M X 

Net reaction: M

 2 X  MX

Kf Kf 

2

1

2

MX 2  MX X 2

The value of Kf for Ag(NH3)2+ is 1.7 x 107.

The overall, or cumulative formation constants are denoted βn. For the reaction: M

 2 X  MX

Kf Kf 

2

1

2

MX 2  MX X 2

Cumulative formation constant

2  K f K f 1

[𝑀𝑋𝑛 ] 𝑀𝑋𝑛−1 ][𝑋]𝑛

 n  K f K f K n 1

Reciprocal, or inverse value of Kf. Also called instability constant (Kinst) The equation for the dissociation of Ag(NH3)2+ is 

𝛽𝑛 =

X  MX n

2

2

Dissociation constant, Kd

Ag ( NH 3 ) 2 (aq )

MX n 1 

Sample Problem: A divalent metal ion reacts with a ligand to form a 1:1 complex. Find the concentration of the metal ion in a solution prepared by mixing 1L 0.20 M M 2+ and 1L of 0.20 M ligand (L). kf = 1.0x108

Ag  (aq )  2 NH 3 (aq)

The equilibrium constant equation or Kd is Kd 

1 [ Ag  ][ NH 3 ]2  K f [ Ag ( NH 3 ) 2  ]

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Distribution of Metal among Several Complexes Silver ion forms a stable 1:1 complex with trien. Calculate the [Ag+] when 25.00 mL of 0.0100M AgNO3 is added to 50.00 mL of 0.015 M trien Kf = 5.0 x 107

Calculation of Alpha Values for Metal Complexes Consider stepwise formation of Ag(NH3)2+

Ag+

1.

:NH3 ⇌

+

K f1 

Ag(NH3)+

Ag ( NH )   2.1 x 10 Ag NH  

3



Ag(NH3)+ + :NH3 ⇌

2.

3

3

Ag(NH3)+2

Ag( NH )   8.2 x 10 Ag( NH NH  

Kf 

3 2

3

Distribution of Metal among Several Complexes Calculation of Alpha Values for Metal Complexes

Overall reaction: Ag+

2 :NH3 ⇌

+

Ag(NH3)+2

Overall formation constant:

 2  K f K f  1.7 x 107 1

Calculation of Alpha Values for Metal Complexes

Kf



1



K f1 5.)

3

Ag NH  3



1.)

 0   Ag 

Ag  

1   AgNH 

Ag ( NH )  

Ag ( NH )    Ag ( NH NH  Ag(NH )  3

6.)



2.)



3 )2





3 2

3 2



C Ag



3

C Ag

3

 2   Ag ( NH





3



Ag( NH )   

3 2

C Ag

Then,

0 

Ag  Ag   Ag ( NH )   Ag ( NH )  







3

from K f 2





C Ag  

3



K f2 

Ag   Ag ( NH )   Ag ( NH ) 

MBE:

2

Ag ( NH )  AgNH    from

3

Distribution of Metal among Several Complexes

3.

4.)

3



2

3 2

0 

3 2

3

substitute 4.) to 5.)

Ag(NH )   

3 2

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General equation for the alpha metal and alpha complexes

ASSIGN: Derive 1.) α1 [or 𝛼𝐴𝑔(𝑁𝐻3 )+ ] and 2.) α2 [or 𝛼𝐴𝑔(𝑁𝐻3 )2 + ]

Complexes of EDTA and metal ions Ethylenediaminetetraacetic acid (EDTA)

𝛼𝑀 =

1 1 + 𝛽1 [𝐿] + 𝛽2 [𝐿]2 + 𝛽3 [𝐿]3 + ⋯ + 𝛽𝑛 [𝐿]𝑛

𝛼𝑀𝐿 =

𝛽1 [𝐿] 1 + 𝛽1 [𝐿] + 𝛽2 [𝐿]2 + 𝛽3 [𝐿]3 + ⋯ + 𝛽𝑛 [𝐿]𝑛

𝛼𝑀𝐿 =

𝛽𝑛 [𝐿]𝑛 1 + 𝛽1 [𝐿] + 𝛽2 [𝐿]2 + 𝛽3 [𝐿]3 + ⋯ + 𝛽𝑛 [𝐿]𝑛

Complexes of EDTA and metal ions Has six potential sites for bonding with metal -

O

O -

O

O

N N

O

O -O

Oethylenediaminetetracetate (EDTA)

EDTA metal complex

EDTA metal complex Kf 

Ag+ General reaction of EDTA

Mn+ + Y4-

MY

(n-4)+

MY  M Y 

+

y4-

Al3+

+

y4-

Zn2+

+

y4-

n4

n

4

AgY3-

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Acid/Base Properties of EDTA

Fractional compositional diagram of EDTA

EDTA is a hexaprotic system (H6Y2+) with 4 carboxylic acids and 2 ammoniums:

We usually express the equilibrium for the formation of complex ion in terms of the Y4- form (all six protons dissociated). You should not take this to mean that only the Y4- form reacts

Fractional compositional diagram of EDTA Concentration of EDTA and all its forms will depend on pH

Effect of pH on the composition of EDTA As in acid base equilibria, we can define α for each species as the fraction of EDTA e.g.  Y 4

Y  4

At acidic pH H4Y will predominate and at basic pH Y4- will predominate.

Y  4

Y  4

C EDTA

Y   H Y  H Y  H Y  HY  Y  4-

H Y  H Y 2

6

5



-

4

3

2-

3-

4-

2

At moderately acidic pH H2Y2(pH 3-6) predominates

Effect of pH on the composition of EDTA Following our previous derivation in acid base equilibria

Y  4

Ka1 Ka2 Ka3 Ka4 D

Conditional Formation Constants Also called effective formation constants pH dependent equilibrium constant that apply at a single pH only

D= [H+]4 + ka1 [H+]3 + ka1ka2[H+]2 + ka1ka2ka3[H+] + ka1ka2ka3ka4

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Conditional Formation Constants Recall A divalent metal ion reacts with a ligand to form a 1:1 complex. Find the concentration of the metal ion in a solution prepared by mixing 1L 0.20 M M 2+ and 1L of 0.20 M ligand (L). kf = 1.0x108

Ca2+ + Y4Kf =

CaY2-

[CaY2-] [Ca2+][Y4-]

Since there are some other forms of EDTA

Y 

4

Y   

𝐾𝑓 =

CT

4

Y 4

Substitution in the formation constant expression

[𝐶𝑎𝑌 2− ] 𝐶𝑎2+ 𝐶𝐸𝐷𝑇𝐴 𝛼𝑌 4−

Conditional Formation Constants

Kf 

K'f 

MY  M Y  n4

n

[𝐶𝑎𝑌 2− ] 𝐶𝐸𝐷𝑇𝐴 𝛼𝑌 4−

𝐶𝑎2+

CEDTA

[𝐶𝑎𝑌 2− ] 𝐾𝑓 = 𝐶𝑎2+ [𝑌 4− ]

𝐾𝑓 =

Conditional Formation Constants

4

Y 

Then,

[Ca2+] ≠ [Y4-]

[Ca2+] = CEDTA

Then,

Conditional Formation Constants

kf = 5.0x1010

4

MY    K M C

Rearranging

𝐾𝑓 𝛼𝑌 4− =

[𝐶𝑎𝑌 2− ] 𝐶𝑎2+ 𝐶𝐸𝐷𝑇𝐴

= 𝐾′𝑓

Conditional Formation Constants

Sample problem: Calculate the fraction of EDTA present as Y4- in a solution at pH 8.00 and pH 11.0 and its mole percent. Ka1=1.02 x 10-2 Ka2=2.14 x 10-3 Ka3=6.92 x 10-7 Ka4=5.50 x 10-11

n4

n

4

f

T

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Ca2+

+

y4-

CaY2-

EDTA Titration Derive a curve (pM as a function of volume of EDTA) for the titration of 50.0 mL of 0.0050 M Ca2+ (buffered at pH=10) with 0.0100 M EDTA. Ve=25.0 mL

Effect of Auxilliary complexing agents on metal ion concentrations Metals reacts most completely with EDTA at basic solutions

Zn2+

⇌

y4-

+

ZnY2-

But the problem is that many of them also form insoluble hydroxides or basic oxides which makes the reaction slow making the titration impossible. Zn2+ +

⇌

OH-

Zn(OH)2

To avoid this problem we often have to use an auxiliary complexing agent e.g., NH3, tartrate, citrate, or triethanolamine to react with the metal ion and prevent its precipitation when the solution is made basic

Assign: Factors affecting the titration curve Discuss the effect of the ff: 1. The value of the formation constant. 2. The concentrations of EDTA and metal ion. 3. The pH of the solution

Effect of Auxilliary complexing agents on metal ion concentrations Complexing agent  This is a ligand that binds strongly enough to the metal to prevent hydroxide precipitation, but weak enough to be displaced by EDTA

 NH3 is especially useful for this purpose because it forms soluble complexes with many transition metals and when mixed with its conjugate acid ammonium ion, it form a basic pH buffer.

Effect of Auxilliary complexing agents on metal ion concentrations

Effect of Auxilliary complexing agents on metal ion concentrations

Complexing agent

Complexing agent

Zn2+ + Kf  1

⇌ Zn(NH3)2+

NH3

Zn( NH )  Zn NH  +

= 1.62 x

NH3

Zn( NH ) 

102

⇌

Zn( NH ) NH  3 2 2

3

Zn( NH )  Zn( NH ) NH  3 3 2

3 2

Zn(NH3)22+

Zn(NH3)32+

= 2.29 x 102

3

+ NH3 ⇌

Zn( NH )  Zn( NH ) NH 

Zn(NH3)42+

2

2

K f2 

NH3 ⇌ Zn(NH3)32+

+

2

Kf3 

3

2

3

Zn(NH3)2+

Zn(NH3)22+

2

3

= 1.95 x 102

K f4 

3 4 2

3 3

= 1.07 x 102

3

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Effect of Auxilliary complexing agents on metal ion concentrations

Effect of Auxilliary complexing agents on metal ion concentrations

Complexing agent

Sample problem: Calculate the concentration of Zn2+ in a 0.0100 M zinc nitrate solution buffered at pH 9.15 with NH 3/NH4Cl. The free ammonia concentration in the buffer is 0.0800 M.



CZn  Zn 2 

Zn(NH )  2

2

 Zn  2

 + Zn(NH )  Zn(NH )  2

3 2

+

2

+

3 3

 Zn 

2

+ Zn(NH 3 )

3 4

Zn   Zn  2

Zn'

2

CZn

1 2 3 4 1  K f NH 3   K f K f NH 3   K f K f K f NH 3   K f K f K f K f NH 3 1

1

2

1

2

3

1

2

3

4

EDTA Titration in the presence of complexing agent Sample problem: Calculate the pZn2+ for solutions prepared by adding 20.00, 25.00 and 30.00 mL of 0.0100 M EDTA to 50.00 mL of 0.00500 M Zn2+. Assume that both the Zn2+ and EDTA are 0.0100 M in NH3 to provide a constant pH of 9.0

𝐾𝑍𝑛𝑦2− = 3.0 x 1016

𝛼𝑦 4− = 5.0 𝑥 10−2 𝛼𝑍𝑛2+ = 1.17 𝑥 10−5

Titrations involving Unidentate Ligand 1. Liebig method for cyanide Analyte: CNTitrant: AgNO3 Indicator: self-indicating Titration reaction: Ag+ +

2CN- →

Ag(CN)2-

Indicator reaction: Ag+ + Ag(CN)2- → Ag(CN)2-

Sample problem: How many grams of NaCN are present in a solution that is titrated just to a permanent turbidity with 26.05 ml of AgNO 3 solution containing 8.125 g of AgNO3 per liter. AgNO3 = 169.9 g/mole NaCN = 49.01

Titrations involving Unidentate Ligand 2. Nickel determination by titration with CNAnalyte: Ni2+ Titrant: KCN Indicator: suspended AgI(s) Condition: Analysis is done in ammoniacal solution Titration reaction: Ni(NH3)42+ +4CN- →

Ni(CN)4 2- +

Indicator reaction: AgI + 2CN- →

Ag(CN)2-

+

Ag+

+

2CN-

→

Ag(CN)2-

4NH3 I-

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Titrations involving Polydentate Ligand (Chelometry)

Titrations involving Polydentate Ligand (Chelometry) EDTA-most popular chelon Preparation of EDTA Most commonly available are the free acid H4Y and the disodium salt, Na2H2Y Standardization of EDTA Unit of concentration: M or Titer 1o standard: CaCO3

Sample Problem: A CaCO3 solution that will be used to standardize EDTA was prepared by dissolving 2.8000 g of solid CaCO3 in 100 mL dilute HCl. A 20.00 mL aliquot was taken for titration with EDTA consuming 29.00 mL of the titrant to reach the endpoint. Express the concentration of EDTA in molarity and in CaCO3 titer. Na2H2Y2•2H2O = 372.24 g/mole CaCO3 = 100.09 g/mole

Indicators for EDTA titrations The indicator is usually a weaker chelate forming ligand. MgIn-

(Color 1)

+ Y4- →

MgY2- + In3-

(Color 2)

EBT(Eriochrome Black T)

Common metal ion indicator 1. Direct titration – EDTA is used to titrate the metal directly - Mg2+ and other divalent ions can be determined

Sample Problem: A 100.0 mL aliquot of a city drinking water was treated with a small amount of an NH3-NH4Cl buffer to bring the pH to 10. After the addition of Calmagite indicator, the solution required 21.46 mL of 5.140 x 10-3 M EDTA. Calculate the water hardness in terms of ppm CaCO3.

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Sample Problem:

2. Back-titrations – used for analysis of metallic ions that form very stable complexes with EDTA but for which a satisfactory indicator is not available.

M2+ + unr Y4-

Y4-

⇌

+ Mg2+

MY2- + unr Y4⇌

MgY2-

3. Replacement (displacement) reactions – replacing the analyte with an acceptable substitute Hg2+ does not have a satisfactory indicator, but a displacement titration is feasible. Hg2+ is treated with excess Mg(EDTA)2- to displace Mg2+, which is titrated with standard EDTA. Hg2+ +

MgY2- ⇌

Mg2+ +

A 25.00 mL aliquot of a solution containing Hg2+ in dilute HNO3 was treated with 10.00 mL of 0.04882 M EDTA and the pH was adjusted to 10.00 mL with an NH3 buffer. Two drops of EBT were added and the unreacted EDTA was backtitrated with 0.01137 M Mg2+, requiring 24.66 mL to reach the endpoint. What is the molarity of Hg2+ in the sample?

HgY2-

Widely used technique for controlling interferences and titrating mixtures of ions Ex CN- F-, thiourea, (NH2)2CS Masking agent – an auxiliary ligand that preferentially forms highly stable complexes with the potential interference - used to prevent one element from interfering the analysis of another element CN- masks Cd2+, Zn2+, Hg2+, Co2+, Cu+, Ag+, Ni2+, Pd2+, Pt2+, Fe2+, and Fe3+, but not Mg2+, Ca2+, Mn2+, or Pb2+. e.g. when CN- is added to a solution containing Cd2+ and Pb2+, only Pb2+ reacts with EDTA.

Widely used technique for controlling interferences and titrating mixtures of ions Ex CN- F-, thiourea (NH2)2CS

Demasking agent – causes the release of a metal from masking complex CN- complexes can be demasked with formaldehyde:

Masking agent – an auxiliary ligand that preferentially forms highly stable complexes with the potential interference - used to prevent one element from interfering the analysis of another element e.g. Al3+ in a mixture of Mg2+ - Al3+ can be measured by first masking the Al3+ with F-, thereby leaving only the Mg2+ to react with EDTA.

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Sample Problem: A 25.00 mL sample of unknown containing Fe3+ and Cu2+ required 16.06 mL of 0.05083 M EDTA for complete titration. A 50.00 sample of the unknown was treated with NH4F to protect the Fe3+. The Cu3+ was reduced and masked by addition of thiourea. Upon addition of 25.00 mL of 0.05083 M EDTA, the Fe3+ was liberated from its fluoride complex and formed an EDTA complex. The excess EDTA required 19.77 mL of 0.01883 M Pb2+ to reach an endpoint using xylenol orange. Find the concentration of Cu2+ and Fe3+ in M.

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