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Complexation Reactions
 widely used in analytical chemistry
Gravimetry

Ni/DMG

Titrimetric Methods Metal-EDTA
Spectrophotometry

Metal-dithiocarbamate

Complex Formation
 complexes are formed from the reaction of
metal ions with electron-pair donors

 complexes are also called coordination
compounds
 ligands (donor species) must have at least one
pair of unshared electrons for bond formation
- anion, cation, or neutral molecule with the
ability to donate e- pair

Ligands
Some common ligand groups

water

ammonia

halides

carbonyls

hydroxides

mercaptans

 in aqueous systems, ligands are typically anionic
or polar neutral species

Ligands
Classification of ligands according to dentate
number

Unidentate
single donor group - ammonia
Bidentate
two donor groups - glycine
Multidentate
variable number based on need - EDTA

Coordination Number
 number of covalent bonds that the metal tends
to form with the ligand
 typical values are 2,4, and 6

 complexes formed due to coordination may be
(+) charged, (-) charged, or neutral

Coordination Number
Copper(II) with a coordination number of 4

Cu(NH3)42+
Cu(NH2CH2COO)2
CuCl42-

Monodentate Ligands
 possess only one accessible donor group

Example
Water, H2O – most metal ions exist as aquo
complexes in water
Ag(H2O)2+

Cu(H2O)42+

Fe(H2O)63+

 charge and coordination number are NOT related

Monodentate Ligands
Common monodentate ligands

Neutral
H2O
NH3
RNH2

Anionic
OHCNS2RCOOSCN-

Bidentate Ligands
 form two bonds with metal ion

Example
Ethylene diamine (en) – NH2CH2CH2NH2

Bidentate Ligands
Common bidentate ligands
 8-hydroxyquinoline

Bidentate Ligands
Common bidentate ligands
 Dimethylgyloxime - DMG

Bidentate Ligands
Common bidentate ligands
 1,10-phenanthroline

Chelate
 complex formed when metal ion coordinates
with two or more donor groups of a single
ligand to form a 5- or 6-membered ring

O

C

O

O

C

NH

CH
2

Cu
H2 C

NH

O

Chelate
 chelating agent – refers to the multidentate
ligand

 chelon – refers to the multidentate ligand that
forms stoichiometric, stable, 1:1
soluble complex

Chelon
Ethylenediamine tetraacetic acid (EDTA)
 most widely used complexometric titrant
 forms 1:1 complexes with most metals
 forms stable, water soluble complexes

 high formation constant

EDTA
 the disodium salt is often used since it is more
soluble

 has 6 donor groups

Soluble Complexes
Complexation Reaction
M + L  ML

Kf 

ML
[ M][L]

 Kf - formation constant

Soluble Complexes
Stepwise Complexation Reaction
M + L  ML

ML + L  ML2
ML(n-1) + L  MLn

K f1 
Kf2 

K fn 

ML
[ M][L]

ML2 
[ML][L]

MLn 
[ ML(n 1) ][L]

Soluble Complexes
Stepwise Complexation Reaction
Example
Formation of Ag(NH3)2+ complex
Ag+ + NH3  AgNH3+
AgNH3+ + NH3  Ag(NH3)2+

[ AgNH3  ]
K f1 
[ Ag ][ NH3 ]

Kf2

[ Ag(NH3)2  ]


[ AgNH3 ][ NH3 ]

Soluble Complexes
Overall Formation
Example
Formation of Ag(NH3)2+ complex

Ag+ + 2NH3  Ag(NH3)2+
K f  K f 1K f 2

[ Ag(NH3)2  ]

[ Ag ][ NH3 ]2

Distribution of Metal Among
Several Complexes
Consider the Ag(NH3)2+ complex:

MBE:

F = [Ag+] + [AgNH3+] + [Ag(NH3)2+]

0 = fraction of Ag+

1 = fraction of AgNH3+
2 = fraction of Ag(NH3)2+

Distribution of Metal Among
Several Complexes
[Ag  ]
[Ag  ]
β0 



F
[Ag  ]  [AgNH3 ]  [Ag(NH3 )2 ]




[AgNH3 ]
[AgNH3 ]
β1 




F
[Ag ]  [AgNH3 ]  [Ag(NH3 )2 ]




[Ag(NH3 )2 ]
[Ag(NH3 )2 ]
β2 




F
[Ag ]  [AgNH3 ]  [Ag(NH3 )2 ]

Distribution of Metal Among
Several Complexes
 from Kf1

[AgNH3+] = Kf1[Ag+][NH3]
 from Kf
[Ag(NH3)2+] = Kf1Kf2[Ag+][NH3]2

Distribution of Metal Among
Several Complexes
 substituting and simplifying equations
[Ag  ]
β0 
[Ag  ]  K f1 [Ag  ][NH3 ]  K f1K f2 [Ag  ][NH3 ]2

1
β0 
1  K f1 [NH3 ]  K f1K f2 [NH3 ]2

Distribution of Metal Among
Several Complexes
 deriving the other ’s
K f1 [NH3 ]
β1 
1  K f1 [NH3 ]  K f1K f2 [NH3 ]2

K f1K f2 [NH3 ]2
β2 
1  K f1 [NH3 ]  K f1K f2 [NH3 ]2

Distribution of Metal Among
Several Complexes
Beta Expressions
 the denominator for each beta expression is
identical
 the denominator is an ascending power series in
[L], starting with [L]0 to [L]n

1  K f 1[L]  K f 1K f 2[L]2  K f 1K f 2K f 3[L]3
...
9  1016
[Cd 2  ][Y 4  ]

Conditional or Effective
Formation Constants
 EDTA is a weak acid, H6Y2+
cEDTA  [H6Y 2 ]  [H5Y  ]  [H4Y ]  [H3Y  ]  [H2Y 2 ]  [HY 3 ]  [Y 4  ]

Y 4 

[Y 4  ]

cEDTA

[Y 4  ]  cEDTAY 4

[CdY 2  ]
Kf 
[Cd 2  ]Y 4  cEDTA

[CdY 2  ]
Kf ' 
[Cd 2  ]cEDTA

Competition with Other
Ligands
Auxiliary Complexing Agent

 a second ligand in a complexation titration that
initially binds with the analyte but is displaced
by the titrant
Example
EDTA (Y4-) and cadmium ions (Cd2+) in the
presence of NH3-buffer

Competition with Other
Ligands
Auxiliary Complexing Agent

+

Cd2+ + Y4-  CdY2-



NH3
Cd(NH3)2+
[CdY 2  ]
Kf 
 2
...
0200
M EDTA solution buffered at pH 10
...

EDTA (Ka1 = 1
...
1 x 10-3,
Ka3 = 7
...
8 x 10-11)

Ans: 8
...
0 (b) = 8
...
Assume that
 2  4
...

Cd

EDTA (Ka1 = 1
...
1 x 10-3,
Ka3 = 7
...
8 x 10-11)

Conditional or Effective
Formation Constants
Cd2+ + Y4-  CdY2[CdY 2  ]
Kf 
 2
...
0,

Y 4  4
...
9  1016)(4
...
5  10 4 )
K f ' '  6
...
0,

Y 4  6
...
9  1016)(6
...
5  10 4 )
K f ' '  8
...
0150 M at pH (a) 3
...
0?
Ni2+ + Y4-  NiY2K NiY 2

[NiY 2  ]

 4
...
0
[Ni2  ] 

0
...
2  10 5 M
(4
...
5  10 11)

Conditional or Effective
Formation Constants
 at pH 8
...
0150
[Ni ] 
 8
...
2  1018)(5
...
50 x 10-3 M at pH 9
...
Assume Cu 2  5
...

Cu2+ + Y4-  CuY2KCuY 2

[CuY 2  ]

 6
...
3  1018
Cu 2 CT ,Cu 2 Y 4  CT ,Y 4 

K f ' '  KCuY 2  Cu 2  Y 4 

[CuY 2  ]

CT ,Cu 2 CT ,Y 4 
[CuY 2  ]

Kf ' '

CT ,Y 4  CT ,Cu 2

CT 2

[CuY 2  ]
Kf ' ' 
CT 2

[CuY 2  ]
CT 
Kf ' '

Conditional or Effective
Formation Constants
[CuY 2  ]
CT 
Kf ' '
7
...
897  10  8 M
(6
...
20  10 5)(0
...
20  10 5)(1
...
86  10 13 M

Conditional or Effective
Formation Constants
Example
Calculate the concentration of Cu2+ in a 0
...
The free NH3 concentration in the
Buffer is 0
...


Cu-NH3 complex
log Kf1 = 4
...
43, log Kf3 = 2
...
48

Conditional or Effective
Formation Constants
Cu(NO3)2  Cu2+ + 2NO3CT  0
...
33  10 8

[Cu 2 ]  Cu2 CT  (1
...
5000)  6
...
volume EDTA
for 50
...
00500 M Ca2+ being titrated with
0
...
0
...
0  1010
[Ca 2  ][Y 4  ]

Titrations with EDTA
Calculating the Conditional Constant
KCaY 2

[CaY 2  ]

[Ca 2  ]Y 4  CT

K f '  Y 4   KCaY 2

[CaY 2  ]

[Ca 2  ]CT

K f '  (0
...
0  1010)  2
...
00500 M
pCa   log[Ca 2 ]  2
...
00 mL Titrant
Ca2+

+

Y4-



CaY2-

Initial: (50
...
00500M)=0
...
250 mmol 0
...
150 mmol

[Ca 2  ] 

(10
...
0100M)=0
...
150 mmol
 CT 
 2
...
00  10
...
100 mmol

pCa  2
...
00 mL Titrant
Ca2+

+

Y4-



CaY2-

Initial: (50
...
00500M)=0
...
250 mmol 0
...
050 mmol

[Ca 2  ] 

(20
...
0100M)=0
...
050 mmol
 CT 
 7
...
00  20
...
200 mmol

pCa  3
...
00 mL Titrant, Equivalence Point
Ca2+

+

Y4-



CaY2-

Initial: (50
...
00500M)=0
...
250 mmol 0
...
00mL)(0
...
250mmol

0

0
...
250 mmol

 3
...
00  25
...
00 mL Titrant, Equivalence Point
Ca2+

+
KCaY 2

Y4-



CaY2-

[CaY 2  ]

 5
...
03  1010
[Ca 2  ]CT

Titrations with EDTA
Addition of 25
...
03  1010
[Ca 2  ]2
[CaY 2  ]
[Ca 2  ] 

Kf '

3
...
05  10  7 M
2
...
05  10 7 M)  6
...
00 mL Titrant, Postequivalence Region
Ca2+
Initial:

Y4-



CaY2-

0
...
300mmol-0
...
13  10 3 M

0
...
250 mmol

[Y 4  ]  6
...
00 mL Titrant, Postequivalence Point
[CaY 2  ]  3
...
13  10 3 M
[Y 4  ]  6
...
03  1010
[Ca 2  ]CEDTA

[CaY 2  ]
[Ca 2  ] 
K f ' CEDTA

Titrations with EDTA
Addition of 30
...
13  10 3
[Ca 2  ] 
 2
...
03  1010)(6
...
47  10 10 M)  9
...
00
5
...
00
20
...
00
25
...
00
30
...
00
50
...
00

pCa
2
...
44
2
...
15
3
...
39
8
...
61
10
...
31
10
...
00

10
...
00

6
...
00

2
...
00
0
...
00

20
...
00

40
...
0100 M EDTA

50
...
00

70
...
Direct titration
2
...
Spectrophotometric Methods
4
...
Displacement Methods

EDTA Titration Techniques
Direct Titration
 standard EDTA solution is added to the sample
until an appropriate end point signal is observed
Example
Mg2+
Mg2+ + H2Y2-  MgY2- + 2H+

MgIn- + H2Y2-  MgY2- + HIn2-

EDTA Titration Techniques
Potentiometric Methods
 potential measurements are used for end point
detection

 applicable for metal ions for which specific ion
selective electrodes are available

EDTA Titration Techniques
Spectrophotometric Methods
 measurement of UV/visible absorption is used for
end point detection

 use an instrument (spectrophotometer) rather
than relying on visual determination of end point

EDTA Titration Techniques
Back-Titration Methods
 for metal ions that react too slowly with EDTA

Example
Al3+
Al3+ + H2Y2-  AlY- + H2Y2-

Complexation reaction

Fe3+ + H2Y2-  FeY- + H2Y2-

Titration reaction

Fe3+ + salicylate  Fe-salicylate

Indicator reaction

EDTA Titration Techniques
Displacement Methods
 addition of measured excess of Mg-EDTA or
Zn-EDTA complex into the analyte solution
 since Kf for Mg-EDTA is usually less than those of
other metals, displacement reaction occurs

Example
Hg2+
Hg2+ + MgY2-  HgY2- + Mg2+
Mg2+ + H2Y2-  MgY2- + 2H+

Solubility Equilibrium/Precipitation
Reactions
Consider BaSO4
BaSO4(s)



Ba2+(aq)

+

SO42-(aq)

 BaSO4 is only slightly soluble in water



K sp  Ba2   SO4 2 



 Ksp – solubility product constant

Solubility Equilibrium/Precipitation
Reactions
Prediction of Precipitation
If
Qsp < Ksp

Type of Sol’n
Unsaturated

Qsp = Ksp

Saturated

Qsp > Ksp

Supersaturated

No precipitation
occurs
Solid and solution
in equilibrium
Precipitation occurs

Solubility Equilibrium/Precipitation
Reactions
Prediction of Precipitation
Example
Should precipitation occur when 50
...
0 x 10-4 M Ca(NO3)2 is mixed with 50
...
0 x 10-4 M NaF?
Ksp CaF2 = 3
...
0 x 10-4 M)(50
...
025 mmol

NaF



(2
...
00 mL)=
0
...
025 mmol

Na+
0
...
050 mmol

+

F0
...
025  0
...
00  100
...
5  10 12

 Qsp < Ksp, no precipitation occurs

2

Solubility Equilibrium/Precipitation
Reactions
Selective Precipitation

 use of reagent whose anion forms a ppt with only
one of the metal ions in the mixture

 used to separate mixtures of metal ions in
aqueous solutions

Solubility Equilibrium/Precipitation
Reactions
Selective Precipitation
Example
A solution contains 1
...
0 x 10-3
M Pb2+
...
9 x 10-9) or CuI
(Ksp = 5
...
9  10 9
 1
...
0  10 3

 CuI will precipitate first

5
...
30  10  8 M
1
...
One liter of saturated Ag2CrO4 solution contains
0
...
Calculate the
solubility product constant of Ag2CrO4
...
Calculate the molar solubilities, concentrations of
constituent ions, and solubilities in grams per
liter for (a) AgCl (Ksp = 1
...
5 x 10-17)
...
What [Ba2+] is necessary to start the precipitation of
BaSO4 in a solution that is 0
...
Ksp BaSO4 = 1
...


Solubility Equilibrium/Precipitation
Reactions
Factors Affecting Solubility
 Common Ion Effect
 Complex Ion Formation

 pH

Factors Affecting Solubility
Common Ion Effect
Example
Calculate the molar solubility of Pb(IO3)2 in (a) water
(b) 0
...
Ksp of Pb(IO3)2 = 2
...

Pb(IO3)2(s)



Pb2+(aq)

+

2IO3-(aq)

Initial:

solid

0

0

:

solid

+x

+2x

Equil:

solid

+x

+2x

Factors Affecting Solubility
A
...
97  10 5 M

K sp  (x)(2x)2  4 x 3

[IO3 ]  2x  7
...
5  10 13)
x3
 3
...
10

0

:

solid

+x

+2x

Equil:

solid

0
...
10)(4 x 2)  0
...
10  x)(2x)

2

x

K sp
0
...
4

2
...
91  10  7 M
0
...
10  x  0
...
6  10 6 M

Factors Affecting Solubility
Assignment
1
...
015 M KBr
solution
...
0 x 10-13
2
...
15 M K2SO4
solution
...
5 x 10-5

Factors Affecting Solubility
Effect of pH
 solubility of ppt depends on the pH

Example
Calculate the molar solubility of CaC2O4 in a solution
that has been buffered so that its pH is constant and
equal to 4
...
Ksp CaC2O4 = 1
...


Factors Affecting Solubility
Effect of pH
CaC2O4(s)



H2C2O4 
0
K sp  [Ca 2  ][C2O4 2  ]
2

K sp  [Ca ][2CT ]

Ca2+(aq)

+

HC2O4- 
1

C2O42-(aq)
C2O422

[Ca 2 ]  [C2O42 ]  [HC2O4  ]  [H2C2O4 ]  CT

K sp
 [Ca 2  ]2
2

2

[Ca ] 

K sp
2

Factors Affecting Solubility
Effect of pH
2

[Ca ] 

K sp
2

1
...
96  10 5 M
0
...
351)(6
...
44  10 5 M

Factors Affecting Solubility
Effect of pH
1
...
0 x 10-1 M and
(b) 1
...
Ksp CuS = 8
...

2
...
00
...
4 x 10-14
...
010 M NH3
(free uncomplexed NH3)
...
8  10 10

Ag+

+

NH3



Ag(NH3)+

K f 1  2
...
1  10 3

Ag(NH3)2+

Factors Affecting Solubility
Complex Ion Formation
K sp  [0CT ][Cl  ]

K sp
 [CT ][Cl  ]
0

K sp
[Cl ] 
0
 2

[Cl ] 

K sp
0

[Cl  ] 

1
...
172 M
9
6
...
09  10 9)(0
...
05  10 9 M

Factors Affecting Solubility
Separating Ions by pH Control: Sulfide Separations
Relevant Equilibria
MS(s)



M2+(aq)

+

S2-(aq)

K sp  [M2  ][S2  ]

H2S + H2O  H3O+ + HS-

K a1  9
...
3  10 14

H2S + 2H2O  2H3O+ + S-

K a1K a 2  1
...
1 M
0
...
1 M

H2S + 2H2O  2H3O+ + S-

K a1K a 2

[H3O  ]2[S2  ]

[H2S]

Factors Affecting Solubility
Separating Ions by pH Control: Sulfide Separations
K a1K a 2

[H3O  ]2[S2  ]

[H2S]

[H3O  ]2[S2  ]
1
...
1

1
...
2  10 22 
 [M2  ]
 [H O  ]2 

3



Factors Affecting Solubility
Separating Ions by pH Control: Sulfide Separations
Steps:
1
...

2
...
e
...
00 x 10-4 M
...
Calculate the [S2-] required to ppt the more soluble cation
...
Determine the [S2-] range that can separate the 1st and 2nd
cations
...
Calculate the pH conditions based on the calculated [S2-]
...
Determine the pH range
...
Find the conditions under which Cd2+ and Tl+
can, in theory, be separated quantitatively with H2S
from a solution that is 0
...

Ksp CdS = 1
...
0 x 10-22

Precipitation Titration
Precipitation Titration
 a titration in which the reaction between the
analyte and the titrant forms an insoluble
precipitate
 limited applications due to the slow rate of
formation of most precipitates

Precipitation Titration
Requirements of a Precipitation Titration
 rate of precipitation and attainment of solubility
must be rapid
 precipitation reaction should be quantitative
and stoichiometric
 a suitable means must be available for locating
or identifying the equivalence point

Precipitation Titration
Titration Curve
 the titration curve follows the change in either
the analyte’s or the titrant’s concentration as a
function of the volume of titrant added
Example
Analysis of I- using Ag+ as titrant
Ag+(aq)

+

I-(aq)



AgI(s)

 plot pAg or pI against the volume of titrant

Precipitation Titration
Titration Curve
Example
Construct a titration curve for the titration of 50
...
0500 M Cl- with 0
...
Ksp AgCl =
1
...
Plot pAg or pCl against the volume of
titrant
...
6  10 9
K sp 1
...
0500 M
pCl   log[Cl  ]   log 0
...
30

[ Ag ]  0
pAg  indeterminate

Precipitation Titration
Addition of 10
...
00mL)(0
...
500mmol

: (10
...
100M)=
1
...
500-1
...
500 mmol
 0
...
00 mL

1
...
025 M  1
...
00 mL Titrant
K sp  [ Ag ][Cl  ]

K sp
1
...
2  10  9 M
[Cl  ]
0
...
2  10

9

 8
...
00 mL Titrant, Equivalence Point
Ag+(aq)
Initial:

Cl-(aq)



AgCl(s)

(50
...
0500M)=
2
...
00mL)(0
...
500mmol
Final:

+

x

K sp  [ Ag ][Cl  ]
K sp  x 2

2
...
500

x

x  1
...
87

Precipitation Titration
Addition of 40
...
00mL)(0
...
500mmol

4
...
500mmol

1
...
500 mmol
 0
...
00 mL

0

pAg  1
...
00 mL Titrant
K sp  [ Ag ][Cl  ]

K sp
1
...
08  10  8 M
[ Ag ]
0
...
08  10

8

 7
...
00

1
...
00

1
...
31

10
...
60

8
...
00

1
...
93

20
...
15

7
...
00

4
...
89

30
...
54

2
...
00

7
...
93

40
...
97

1
...
00

8
...
68

50
...
14

1
...
100 M Ag+

50

60

Precipitation Titration
Effect of Titrant Concentration
 increased sharpness of
the break for the more
concentrated solution

A = 0
...
0100 M AgNO3

Precipitation Titration
Effect of Reaction Completeness
 completeness, K=1/Ksp
 smaller Ksp gives much
sharper breaks at the
end point

Precipitation Titration
Evaluating the End Point for Argentometric Titrations
 Formation of a colored precipitate: The Mohr
Method
 Formation of a colored complex: The Volhard
Method
 Adsorption Indicators: The Fajans Method

Precipitation Titration
The Mohr Method
 Na2CrO4 or K2CrO4 serves as an indicator
Ag+(aq) + X-(aq)  AgX(s)

Titration Reaction

2Ag+(aq) + CrO42-(aq)  Ag2CrO4(s)

Indicator Reaction

reddish brown

 CrO42- imparts a yellow color; low concentrations
are used

Precipitation Titration
The Mohr Method
 results to a positive error (more titrant used)
 to compensate, a reagent blank is analyzed
 solution is maintained at slightly alkaline pH

 if pH > 10, silver precipitates as silver hydroxide

Precipitation Titration
The Volhard Method
 Fe3+ serves as an indicator
 silver ions are titrated with a standard solution
thiocyanate ion
SCN-(aq) + Ag+(aq)  AgSCN(s)

Titration Reaction

SCN-(aq) + Fe3+(aq)  FeSCN2+(aq)

Indicator Reaction

red

Precipitation Titration
The Volhard Method
 titration is carried out in acidic solution to
prevent precipitation of Fe3+ as the hydrated
oxide

Precipitation Titration
The Fajans Method
 utilizes adsorption indicators
 organic compound that tends to be adsorbed onto
the surface of the precipitate
 color when adsorbed to the precipitate is
different from that when it is in solution

Precipitation Titration
The Fajans Method
Example
Anionic dye dichlorofluorescein
 before the end point it is solution and imparts
a greenish yellow color
 at the end point, adsorbs on the surface of
precipitate and imparts a pink color

Precipitation Titration
Exercise
The %w/w I- in a 0
...
After adding 50
...
05619 M
AgNO3 and allowing the precipitate to form, the
remaining silver was back titrated with 0
...
14 mL to reach the end point
...


Ans: 17
...
2000 g sample
requires 39
...
4103 g of KSCN
per 100 mL) for the precipitation of silver?

Ans: 90
...
5000
g
...
00 mL of 0
...
50 mL of 0
...
What is the %SrCl2 in the sample?

Ans: 53
...
300 mg of Fe2O3
...
0500 M AgNO3 would be required to
titrate 50
...
3 mL

Problem Set in Complexometry
1
...
500 M
ammonium ion and 0
...

log Kf Ni(II)-EDTA = 18
...
36, log Kf2 = 1
...
55, log Kf4 = 1
...
85, log Kf6 = 0
...
26
2
...
0
containing 0
...

log Kf Hg(II)-EDTA = 21
...
0, log Kf2 = 16
...
83, log Kf4 = 2
...
00, pKa2 = 2
...
16, pKa4 =10
...
A standard solution of calcium was prepared by dissolution
of 200
...
Then the
solution was boiled to remove CO2 and was diluted to
250
...
When 25
...
62 mL of the EDTA solution was
required
...

4
...
00-mL aliquot of a nickel solution was diluted with
distilled water and ammonia-ammonium chloride buffer,
then treated with 15
...
0100 F EDTA solution
...
01500
F magnesium chloride solution, of which 4
...
Calculate the molar concentration of the
original nickel(II) solution
...
A standard solution of EDTA is prepared, and by titration
each mL is found to complex with Mg in 10
...
300 g of MgCl2 per liter
...
60 mL of the
standard EDTA
...
A sample consisting of 0
...
3000 g
MgCl2 was dissolved in HCl, a buffer added, and the soln
...
000 L
...

What was the hardness of the water expressed in terms
of ppm CaCO3?

Problem Set in Complexometry
7
...
To a
25
...
00 mL of
0
...
The pH is adjusted and the excess EDTA
is backtitrated with 1
...
70
mL
...

8
...
00 mL of 0
...
02000 M EDTA in a solution buffered to pH 11
...

Calculate pSr values after the addition of 0
...
00,
24
...
90, 25
...
10, 26
...
00 mL of titrant
...
63

Answers to Problems
1
...

3
...

5
...

7
...


Kf’ = 5
...
49 x 10-12
8
...
45 x 10-2 M Ni
271 ppm
407
...
01412 M
Vol, mL
pSr
0
...
00
10
...
30
24
...
57
24
...
57
25
...
37
25
...
16
26
...
16
30
...
86



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