Complexation and Precipitation Reactions and Titrations

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Chapter 17

• Complexation and Precipitation Reactions and
  Titrations

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What Is a Complexation and Precipitation
Titrations?
A typical precipitation titration, using the
analysis of Ag in an aqueous solution by its
titration with Cl- as an example.
A typical complexation titration, using the
analysis of Ca2 in water by its titration with
EDTA as an example.
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PART 1 Precipitation Titrations
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• 1. Titration curve
• 1) Guidance in precipitation titration
  calculation
• Find Ve (volume of titrant at equivalence point)
• Find y-axis values
• - At beginning
• - Before Ve
• - At Ve
• - After Ve

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• Example For the titration of 50.0 mL of 0.0500 M
  Cl with 0.100 M Ag. The reaction is
• Ag(aq) Cl(aq) ? AgCl(s) K 1/Ksp
  1/(1.821010) 5.6 x 109
• Find pAg and pCl of Ag solution added
• 0 mL (b) 10.0 mL (c) 25.0 mL (d) 26.0 mL
• Solution

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(a) 0 mL Ag added (At beginning) Ag 0, pAg
can not be calculated. Cl 0.0500, pCl 1.30
(b) 10 mL Ag added (Before Ve)
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(c) 25 mL Ag added (At Ve) AgCl(s) ? Ag(aq)
Cl(aq) Ksp 1.81010 s
AgCl Ksp 1.821010
s2 AgCl1.35x105 pAg 4.87 pCl
4.87
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(d) 26 mL Ag added (After Ve)
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• Construct a titration curve
• Example Titration of 50.0 mL of 0.0500 M Cl
  with 0.100 M Ag

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• Diluting effect of the titration curves

1. 50.00 mL of 0.05000 M NaCl titrated with 0.1000 M
   AgNO3,
2. 50.00 mL of 0.00500 M NaCl titrated with 0.01000
   M AgNO3.

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• Ksp effect of the titration curves

50.00 mL of a 0.0500 M solution of the anion was
titrated with 0.1000 M AgNO3.
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2. Titration of a mixture
Titration curves for 50.00 mL of a solution
0.0800 M in Cl- and 0.0500 M in I- or Br-. Ksp
for AgCl 1.82x10-10 AgBr 5.0x10-13 AgI
8.0x10-17
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Example A 25.00 mL solution containing Br and
Cl was titrated with 0.03333 M AgNO3.
Ksp(AgBr)5.0x1013, Ksp(AgCl)1.82x1010.

1. Which analyte is precipitated first?
2. The first end point was observed at 15.55 mL.
   Find the concentration of the first that
   precipitated (Br or Cl?).
3. The second end point was observed at 42.23 mL.
   Find the concentration of the second that
   precipitated (Br or Cl?).

Solution
(a) Ag(aq) Br(aq) ? AgBr(s) K 1/Ksp(AgBr)
2x1012 Ag(aq) Cl(aq) ? AgCl(s) K
1/Ksp(AgCl) 5.6x109 Ans AgBr precipitated first
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(b)
Ans
(c)
Ans
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• 3. Argentometric Titration
• General information
• Define Argentometric Titration A precipitation
  titration in which Ag is the titrant.
• Argentometric Titration classified by types of
  End-point detection
• Volhard method A colored complex (back
  titration)
• Fajans method An adsorbed/colored indicator
• Mohr method A colored precipitate

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2) Volhard method A colored complex (back
titration). Analysing Cl for example
Step 1 Adding excess Ag into sample Ag Cl
? AgCl(s) left Ag Step 2 Removing AgCl(s) by
filtration/washing Step 3 Adding Fe3 into
filtrate (i.e., the left Ag) Step 4 Titrating
the left Ag by SCN Ag SCN ?
AgSCN(s) Step 5 End point determination by red
colored Fe(SCN)2 complex. (when all Ag has been
consumed, SCN reacts with Fe3) SCN Fe3 ?
Fe(SCN)2(aq)
Total mol Ag (mol Ag consumed by Cl)
(mol Ag consumed by SCN)
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• Fajans Method An adsorbed/colored indicator.
  Titrating Cl and adding
• dichlorofluoroscein for example

Cl
Before Ve (Cl excess) Greenish yellow solution
AgCl(s)
1st layer
Ag
In pink
AgCl(s)
After Ve (Ag excess)
1st layer
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4) Mohr Method A colored precipitate formed by
Ag with anion, other than analyte, once the Ve
reached. Analysing Cl and adding CrO42 for
example
Precipitating Cl Ag Cl ? AgCl(s) Ksp 1.8
x 1010 End point determination by red colored
precipitate, Ag2CrO4(s) 2Ag CrO42 ?
Ag2CrO4(s) Ksp 1.2 x 1012
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PART 2 Complexation Titration
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1. Examples of simple Complexation titration
Titrant Analyte Remarks
Hg(NO3)2 Br, Cl, SCN, thiourea Products are neutral Hg(II) complexes, Various indicators used
AgNO3 CN Product is Ag(CN)2 indicator is I titrate to first turbidity of AgI
NiSO4 CN Product is Ni(CN)42 Various indicators used
KCN Cu2, Hg2, Ni2 Product are Cu(CN)22, Hg(CN)2, and Ni(CN)42 various indicators used
Remark Feasibility titration for M L ? ML
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2. EDTA Titration 1) Chemistry and Properties of
EDTA
i) Structure of Ethylenediaminetetraacetic acid
(EDTA)
ii) EDTA is a hexadentate ligand (2 N atoms and 4
O atoms) iii) All metal-EDTA complexes have a
11 stoichiometric ratio (metal ligand 1 1).
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iv) Six-coordinate structure of metal-EDTA
(indeed Y4complex with metal)
Y4
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v) Equation for Y4 fraction
CT H4YH3YH2Y2HY3Y4
Remark known pH, then calculate a4, then
calculate Y4 a4 CT
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vi) Composition of EDTA solutions as a function
of pH
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(Continued)
Figure 17-7 Spreadsheet to calculate ?4 for EDTA
at selected pH values.
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2) EDTA complex
i) Formation Constant (KMY) for EDTA complex
Mn Y4 ? MY(n4)
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ii) pH buffered Conditional Formation Constant
(KMY)
The complexatin reaction can be written as Mn
CT ? MY(n4)
CT the initial (total) concentration of EDTA.
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• EDTA Titration Curves

Example Calculate the pCa for 50.0 mL of 0.0050
M Ca2 (buffered at pH 10) with addition of
0.0100 M EDTA of (a) 0.0 mL), (b) 15.0 mL, (c)
25.0 mL, (d) 26.0 mL (KMY for CaY2 5.0x1010,
a4 at pH 10.00 0.35 )
Solution 50.0 x 0.005 Veq x 0.01, Veq 25.0 mL
(a) 0.0 mL EDTA (At beginning) pCa logCa2
log(0.0050) 2.30
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(b) 1.0 mL EDTA added (before equivalence point,
0 lt VEDTA lt Veq)
pCa2 2.81
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(c) 25.0 mL EDTA added (at equivalence point,
VEDTA Veq)
Ca2 CT ?
CaY2 Initial 0 0
3.33x10-3 Change x x
x Final x x
3.33x10-3 x
pCa2 6.36
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(d) 26.0 mL EDTA added (after equivalence point,
VEDTA gt Veq)
Ca2 CT ? CaY2
Ca2 1.43 x 109 M pCa2 8.85
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(Continued)

• Figure 17-8 Spreadsheet 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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• Effect of the pH

• Figure 17-10 Influence of pH on the titration of
  0.0100 M Ca2 with 0.0100 M EDTA.
• The titration curves for calcium ion in solutions
  buffered to various pH levels. As the conditional
  formation constant becomes less favorable, there
  is a smaller change in pCa in the
  equivalence-point region.

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• Effect of the Analytes Kf

• Figure 17-11 Titration curves for 50.0 mL of
  0.0100 M solutions of various cations at pH 6.0.
• Cations with larger formation constants provide
  sharp end points even in acidic media.

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iii) pH buffered with Auxiliary Complexing Agents
Conditional Formation Constant (KMY)

• Why using Auxiliary Complexing Agents
• Problems of EDTA titration
• - At low pH, the metal-EDTA reaction is
  incomplete.
• - At high pH, metal hydroxide precipitate
  formed.
• Solution for above problem Adding of auxilliary
  complexing agents, e.g., NH3.
• Function of auxilliary complexing agents
• - Increase pH value, increase ratio of Y4
  (higher KMY value), more complete reaction
• - Preventing precipitate with hydroxide, because
  metal ions formed complex with NH3.

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• Fraction of uncomplexed metal
• Zn2 with auxiliary complexing agent NH3 for
  example

a) Cumulative formation constant for Zn2/NH3
system Zn2 NH3 ? ZnNH32 log ?12.21 Zn2
2NH3 ? Zn(NH3)22 log ?24.50 Zn2 3NH3 ?
Zn(NH3)32 log ?36.86 Zn2 4NH3 ?
Zn(NH3)42 log ?48.89
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b) Fraction of uncomplexed Zn2 (aM)
CM Zn2ZnNH32Zn(NH3)22Zn(NH3)32Z
n(NH3)42 Zn2ß1Zn2NH3ß2Zn2NH32ß3
Zn2NH33ß4Zn2NH34 Zn2(1ß1NH3ß2N
H32ß3NH33ß4NH34)
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Example Zn2 and NH3 form the complexes
Zn(NH3)2, Zn(NH3)22, Zn(NH3)32, Zn(NH3)42.
(log ?12.21, log ?24.50, log ?36.86, log
?48.89). If the concentration of unprotonated
NH3 is 0.10 M, find the zinc fraction in the form
of Zn2.
Solution
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c) Define KMY
The complexation reaction can be written as CM
CT ? MY(n4)
CT initial (total) concentration of
EDTA. CM initial (total) concentration of Mn
metal ion
EDTA titration in specified pH buffered and
specified conc. of auxiliary complexing agent
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Example Calculating the pZn for 50.0 mL of
0.0050 M Zn2 with 0.010 M EDTA at a pH 9 and in
the presence of 0.10 M NH3, when adding EDTA (a)
0 mL, (b) 20.0 mL, (c) 25.0 mL, (d) 30.0 mL. (KMY
for Zn2EDTA is 3.12x1016 for pH 9 ?4 is
5.21x10-2 for 0.10 M NH3 ?M is 1.17x105)
Solution (50.0)(0.0050) (0.010)(Veq)
Veq 25.0 mL
(a) 0.0 mL EDTA added (At beginning) Zn2
aM x CM (1.17x105)(0.0050 M) 5.85x108 M
pZn logZn2 log(5.85 x 108) 7.23
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(b) 20.0 mL EDTA added (before equivalence
point, 0 lt VEDTA lt Veq)
Zn2 aM x CM (1.17x105)(7.14 104 M)
8.35 109 M
pZn logZn2 log(8.35 109) 8.08
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• 25.0 mL EDTA
• (at equivalence point, VEDTA Veq)

CM CT ?
ZnY2 Initial
3.33x103 Change x x
x Final x x
3.33x103 x
x CM 4.19 107 M Zn2aM x CM
(1.17x105)(4.19x107 M) 4.90 1012 M
pZn logZn2 11.31
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(d) 30.0 mL EDTA (after equivalence point, VEDTA
gt Veq)
CZn CT ? ZnY2
x
3.12x103
6.25x104
CM 2.63 x 1010 M Zn2 aM x CM
(1.17x105)(2.63 x 1010) 3.08 x 1015 M
pZn logZn2 14.51
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• Effect of the concentration of auxiliary
  complexing agents

• Figure 17-13 Influence of ammonia concentration
  on the end point for the titration of 50.0 mL of
  0.00500 M Zn2.
• Here are two theoretical curves for the titration
  of zinc(II) with EDTA at pH 9.00.
• The equilibrium concentration of ammonia was
  0.100 M for one titration and 0.0100 M for the
  other.

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3. Metal-ion Indicator

• Metal-ion Indicator A chemical that has a
  change in its color or its fluorescence
  properties when it is free in solution or
  complexed to a metal ion.
• MIn EDTA ? MEDTA In
• Color 1
  Color 2
• If the MIn complex strength is too strong, the
  color change occurs after the equivalence point.
• If the MIn complex strength is too weak,
  however, the color change occurs before the
  equivalence point.
• Most metal-ion indicators are also acid-base
  indicators, therefore, the pH control is required
  for some EDTA titration

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• Common metal-ion indicator

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Eriochrome Black T (EBT)
MY2
HY3
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• Indicators Transition Range and its Feasibility,
  EBT as example Example
• Titrating 50 mL of 0.005 M Ca2 and Mg2 with
  0.010 M EDTA at pH 10 using EBT as the indicator.
  Calculate the transition range for Ca2 and Mg2.
• Information
• Mg2 In3 ? MgIn Kf 1.0107 (1)
• Ca2 In3 ? CaIn Kf 2.5105 (2)
• HIn2 H2O ? In3 H3O Ka 2.81012 (3)
• Mg2 EDTA4 ? MgEDTA2 Kf 4.9108 (4)
• Ca2 EDTA4 ? CaEDTA2 Kf 5.01010 (5)

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(Continued)

• SOLUTION
• Transition range For Mg2 titration
• Equation (1) (3)
• Mg2 HIn2 H2O ? MgIn H3O

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(Continued)

• SOLUTION (Continuous)
• Transition range For Ca2 titration
• Equation (2) (3)
• Ca2 HIn2 H2O ? CaIn H3O

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(Continued)
10.0
Titrating 50 mL of 0.005 M Ca2 and 0.005 M Mg2
with 0.010 M EDTA.
8.0
pCa
6.0
or
pMg
4.0
Using EBT as indicator, the MgIn transition
range (not CaIn transition range) cover the pCa
at equivalance point for Ca2-EDTA titration
2.0
10.0
20.0
30.0
40.0
Vol of EDTA for Ca2 , mL
10.0
20.0
30.0
40.0
0.0
Vol of EDTA for Mg2 , mL
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• How to Minimize Titration Error for Ca2EDTA
  titration
• Information
• Mg2 EDTA4 ? MgEDTA2 Kf 4.9108
• Ca2 EDTA4 ? CaEDTA2 Kf 5.01010
• Mg2 In3 ? MgIn Kf 1.0107
• Ca2 In3 ? CaIn Kf 2.5105
• Adding few drops of (Na)2MgEDTA solution into
  analyte solution with EBT indicator, resulted in
• Ca2 MgEDTA2 ? CaEDTA2 Mg2 (1)
• Mg2 In3 ? MgIn (2)
• After all the Ca2 has been titrated, then
• MgIn EDTA4 H ? MgEDTA2 HIn2 (3)
• Net reaction (1)(2)(3)
• Ca2 EDTA4 In3 H ? CaEDTA2 HIn2
• No quantitative effect by the added Mg2 for
  Ca2-EDTA titration

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End of Chapter 17