Transcript Slide 1

Chapter 17
Complexation and Precipitation
Reactions and Titrations
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.
PART 1
Precipitation Titrations
1.
1)


Titration curve
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
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/(1.82×10–10) = 5.6 x 109
Find pAg and pCl of Ag+ solution added
(a) 0 mL (b) 10.0 mL (c) 25.0 mL (d) 26.0 mL
Solution:
(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)
mmol Cl  Left  original mmol Cl   precipitated mmol Cl 

0.0500 mmol Cl   
0.100 mmol Ag  1 mmol Cl  


  50.0 mL Cl 

  10.0 mL Ag 


 
1
mL
Cl
1
mL
Ag
1
mmol
Ag

 

 1.50 mmol Cl 
Vtotal  50.0 mL  10.0 mL  60.0 mL
1.5 mmol Cl 
[Cl ] 
 2.50  102 M
60.0 mL
K sp 1.82  1010

9
[ Ag ] 


7.28

10
M

2
[Cl ] 2.50  10
pCl  1.60

pAg  8.14
(c) 25 mL Ag+ added (At Ve)
AgCl(s)  Ag+(aq) + Cl–(aq)
s = [Ag+]=[Cl–]
Ksp = 1.82×10–10 = s2
[Ag+]=[Cl–]=1.35x10–5
pAg = 4.87
pCl = 4.87
Ksp = 1.8×10–10
(d) 26 mL Ag+ added (After Ve)
mmol Ag  Left  original mmol Ag   precipitated mmol Ag  withVe of Ag 

0.100 mmol Ag   
0.100 mmol Ag  


  26.0 mL Ag 
   25.0 mL Ag 



1
mL
Ag
1
mL
Ag

 

 0.10 mmol Ag 
Vtotal  50.0 mL  26.0 mL  76.0 mL
0.10 mmol Ag 
[ Ag ] 
 1.32  103 M
76.0 mL
pAg  2.88

 Construct a titration curve
Example: Titration of 50.0 mL of 0.0500 M Cl–
with 0.100 M Ag+
 Diluting effect of the titration curves
(A) 50.00 mL of 0.05000 M NaCl
titrated with 0.1000 M AgNO3,
(B) 50.00 mL of 0.00500 M NaCl
titrated with 0.01000 M
AgNO3.
 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.
2. Titration of a mixture
Titration curves for 50.00 mL
of a solution 0.0800 M in Cland 0.0500 M in I- or Br-.
Ksp for
AgCl = 1.82x10-10
AgBr = 5.0x10-13
AgI = 8.0x10-17
Example: A 25.00 mL solution containing Br– and Cl– was
titrated with 0.03333 M AgNO3. Ksp(AgBr)=5.0x10–13,
Ksp(AgCl)=1.82x10–10.
(a) Which analyte is precipitated first?
(b) The first end point was observed at 15.55 mL. Find the
concentration of the first that precipitated (Br– or Cl–?).
(c) 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



15.55
mL
Ag
1
L
Ag
0.3333
mol
Ag
(b)





1
1000mLAg
1 L Ag
1 mol Br 
1
1000mL




0.02073
M
Br
Ans
1 mol Ag 25 mL
1L
(c) (42.23- 15.55)mL Ag
1

1 L Ag
0.3333mol Ag





1000mLAg
1 L Ag
1 molCl 
1
1000mL




0.03557
M
Cl
Ans
1 mol Ag 25 mL
1L
3. Argentometric Titration
1) General information:
 Define Argentometric Titration: A precipitation
titration in which Ag+ is the titrant.
 Argentometric Titration classified by types of Endpoint detection:
– Volhard method: A colored complex (back titration)
– Fajans method: An adsorbed/colored indicator
– Mohr method: A colored precipitate
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–)
3) Fajans Method: An adsorbed/colored indicator.
Titrating Cl– and adding
dichlorofluoroscein for example:
Before Ve
(Cl– excess)
Greenish
yellow
solution
AgCl(s)
After Ve
(Ag+ excess)
AgCl(s)
Cl–
1st
layer
Ag+
1st
layer
In–
pink
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 10–10
End point determination by red colored precipitate,
Ag2CrO4(s):
2Ag+ + CrO42– → Ag2CrO4(s)
Ksp = 1.2 x 10–12
PART 2
Complexation Titration
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
[ML]
Kf 
 108
[M][L]
2. EDTA Titration
1) Chemistry and Properties of EDTA
i)
Structure of Ethylenediaminetetraacetic acid (EDTA):
HOOC CH2
HOOC CH2
:N CH2 CH2 N :
CH2 COOH
CH2 COOH
ii) EDTA is a hexadentate ligand (2 N atoms and 4 O
atoms)
iii) All metal-EDTA complexes have a 1:1 stoichiometric
ratio (metal : ligand = 1 : 1).
iv) Six-coordinate structure of metal-EDTA (indeed Y4–complex
with metal)
Y4– :
v) Equation for Y4– fraction
[Y4 ]
4 =
CT

K1K 2 K3K 4
[H3O ]4 +K1[H3O ]3  K1K 2 [H3O ]2  K1K 2K 3[H3O ]  K1K 2K 3K 4
CT = [H4Y]+[H3Y–]+[H2Y2–]+[HY3–]+[Y4–]
Remark: known pH, then calculate α4,
then calculate [Y4–] = α4 CT
vi) Composition of EDTA solutions as a function of pH
(Continued)
Figure 17-7 Spreadsheet to calculate 4 for EDTA at selected
pH values.
2) EDTA complex
i)
Formation Constant (KMY) for EDTA complex
Mn+ + Y4– → MY(n–4)
K MY
[MY (n 4) ]
[MY (n 4) ]

 n+
n+
4
[M ][Y ] [M ] 4 C T
ii) pH buffered Conditional Formation Constant (K’MY)
The complexatin reaction can be written as:
Mn+ + CT → MY(n–4)
CT : the initial (total) concentration of EDTA.
K'MY   4 K MY
[MY(n4) ]

[Mn+ ]CT

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, α4 at pH 10.00 = 0.35 )
Solution:
50.0 x 0.005 = Veq x 0.01, Veq = 25.0 mL
K'MY  K M  4   0.35   5.0 x1010   1.75 x1010
(a) 0.0 mL EDTA
(At beginning)
pCa = –log[Ca2+] = –log(0.0050) = 2.30
(b) 1.0 mL EDTA added
(before equivalence point, 0 < VEDTA < Veq)
[Ca 2 ] 
(0.0050 M)(50.0 mL)  (0.010 M)(15.0 mL)
 1.54 x10 3 M
50.0 mL + 15.0 mL
pCa2+ = 2.81
(c) 25.0 mL EDTA added
(at equivalence point, VEDTA = Veq)
(0.010 M)(25.0 mL)
[CaY ] 
 3.33 x10 3 M
50.0 mL +25.0 mL
2
Ca2+ +
Initial 0
Change x
Final x
CT
0
x
x
→
CaY2–
3.33x10-3
–x
3.33x10-3 – x
[CaY 2 ] 3.33 x10 3  x
'
10


K

1.75
x
10
MY
[Ca 2 ]C T
x2
x  [Ca 2  ]  4.36 x 10 7 M
pCa2+ = 6.36
(d) 26.0 mL EDTA added
(after equivalence point, VEDTA > Veq)
Ca2+ + CT → CaY2–
CT 
(0.010 M)(26.0  25.0 mL)
 1.32 x 10 4 M
50.0 mL +26.0 mL
(0.010 M)(25.0 mL)
[CaY ] 
 3.29 x10 3 M
50.0 mL +26.0 mL
2
[CaY 2 ]
3.29 x10 3
'
10


K

1.75
x
10
MY
[Ca 2 ]C T [Ca 2 ](1.32x10 -4 )
[Ca2+] = 1.43 x 10–9 M
pCa2+ = 8.85
(Continued)
Figure 17-8 Spreadsheet for the titration of 50.00 mL of 0.00500 M Ca+2 with
0.0100 M EDTA in a solution buffered at pH 10.0.
 Effect of the pH
Figure 17-10 Influence of pH on the titration of 0.0100 M Ca+2 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.
 Effect of the Analyte’s 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.
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.
 Fraction of uncomplexed metal
Zn2+ with auxiliary complexing agent NH3 for
example:
a)
Cumulative formation constant for Zn2+/NH3 system:
Zn2+ + NH3  ZnNH32+
log 1=2.21
Zn2+ + 2NH3  Zn(NH3)22+
log 2=4.50
Zn2+ + 3NH3  Zn(NH3)32+
log 3=6.86
Zn2+ + 4NH3  Zn(NH3)42+
log 4=8.89
b) Fraction of uncomplexed Zn2+ (αM)
CM = [Zn2+]+[ZnNH32+]+[Zn(NH3)22+]+[Zn(NH3)32+]+[Zn(NH3)42+]
=[Zn2+]+β1[Zn2+][NH3]+β2[Zn2+][NH3]2+β3[Zn2+][NH3]3+β4[Zn2+][NH3]4
=[Zn2+](1+β1[NH3]+β2[NH3]2+β3[NH3]3+β4[NH3]4)
[ Zn 2 ]
M 
CM
1

1  1[ NH 3 ]   2 [ NH 3 ]2   3 [ NH 3 ]3   4 [ NH 3 ]4
Example: Zn2+ and NH3 form the complexes Zn(NH3)2+,
Zn(NH3)22+, Zn(NH3)32+, Zn(NH3)42+. (log 1=2.21, log 2=4.50,
log 3=6.86, log 4=8.89). If the concentration of unprotonated
NH3 is 0.10 M, find the zinc fraction in the form of Zn2+.
Solution:
1
αM 
1  β1[NH 3 ]  β 2 [NH 3 ]2  β 3 [NH 3 ]3  β 4 [NH 3 ]4
1

1  (10 2.21 )(0.10)  (10 4.50 )(0.10) 2  (10 6.86 )(0.10) 3  (108.89 )(0.10) 4
 1.17x10 5
c) Define KMY’’:
The complexation reaction can be written as:
CM + CT → MY(n–4)
CT : initial (total) concentration of EDTA.
CM: initial (total) concentration of Mn+ metal ion
K MY (n  4 )
[MY (n 4) ]
[MY (n 4) ]


n+
4
[M ][Y ]  M C M  4 C T
EDTA titration in specified pH buffered and
specified conc. of auxiliary complexing agent:
K
"
MY
  4 M K MY
[MY (n 4) ]

CM CT
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 Zn2+–EDTA is 3.12x1016; for pH 9: 4 is
5.21x10-2; for 0.10 M NH3: M is 1.17x10–5)
Solution:
(50.0)(0.0050) = (0.010)(Veq)
Veq = 25.0 mL
K"MY  4 M K f   5.21x102 1.17 x105  (3.12 x1016 )  1.9 x1010
(a) 0.0 mL EDTA added
(At beginning)
[Zn2+] = αM x CM = (1.17x10–5)(0.0050 M) = 5.85x10–8 M
pZn = –log[Zn2+] = –log(5.85 x 10–8) = 7.23
(b) 20.0 mL EDTA added
(before equivalence point, 0 < VEDTA < Veq)
(0.0050 M)(50.0 mL)  (0.010 M)(20.0 mL)
CM 
50.0 mL + 20.0 mL
 7.14  10 4 M
[Zn2+] = αM x CM = (1.17x10–5)(7.14 × 10–4 M) = 8.35 × 10–9 M
pZn = –log[Zn2+] = –log(8.35 × 10–9) = 8.08
(c) 25.0 mL EDTA
(at equivalence point, VEDTA = Veq)
(0.0050 M)(50.00 mL)
[ZnY ] 
 3.33  103 M
50.0 mL+25.0 mL
2
CM + CT  ZnY2–
Initial –
–
3.33x10–3
Change +x
+x
–x
Final x
x
3.33x10–3 – x
[ZnY 2 ] 3.33  103  x
"
10

=K

1.9

10
MY
C M CT
x2
x = CM = 4.19 × 10–7 M
[Zn2+]=αM x CM = (1.17x10–5)(4.19x10–7 M) = 4.90 × 10–12 M
pZn = –log[Zn2+] = 11.31
(d) 30.0 mL EDTA (after equivalence point, VEDTA > Veq)
(0.010 M)(30.0 mL)  (0.005 M)(50.0 mL)
CT 
 6.25x10 4 M
30.0 mL+50.0 mL
(0.010 M)(25.0 mL)
[ZnY ] 
 3.12  10 3 M
30.0 mL+50.0 mL
2
CZn + CT
→
ZnY2–
x
6.25x10–4
3.12x10–3
[ZnY 2 ]
3.12  10 3
"
10


K

1.9

10
f
C M CT
C M  6.25  10 4
CM = 2.63 x 10–10 M
[Zn2+] = αM x CM = (1.17x10–5)(2.63 x 10–10) = 3.08 x 10–15 M
pZn = –log[Zn2+] = 14.51

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 Zn+2.
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.
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.
M–In + EDTA  M–EDTA + In
Color 1
Color 2
– If the M–In complex strength is too strong, the color change
occurs after the equivalence point.
– If the M–In 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
 Common metal-ion indicator

Eriochrome Black T (EBT)
H3In
MY2–
pKa1=6.3 +
pKa2=11.6
In3–
H2In–
HIn2–
(orange)
(red)
(blue)
At pH 8~10
MIn–
(red)
+
HY3–
M: Mg2+, Ca2+, Mn2+,
Zn2+, etc.
51/57
 Indicator’s 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 = 1.0×107
(1)
Ca2+ + In3– → CaIn–
Kf = 2.5×105
(2)
HIn2– + H2O → In3– + H3O+
Ka = 2.8×10–12
(3)
Mg2+ + EDTA4– → MgEDTA2–
Kf = 4.9×108
(4)
Ca2+ + EDTA4– → CaEDTA2–
Kf = 5.0×1010
(5)
52/57
(Continued)
• SOLUTION:
– Transition range For Mg2+ titration:
Equation (1) + (3)
Mg2+ + HIn2– + H2O → MgIn– + H3O+
[MgIn- ][H3O+]
5
Ka  K f 

2.8

10
[HIn2- ][Mg2+]
[MgIn- ] 1.0  1010
[Mg ]=

2[HIn ] 2.8  105
2+
[MgIn- ]
Set ratio 10 and 0.1 for
(Red to Blue)
2[HIn ]
Resulted in [Mg2+] 3.6x105 and 3.6x10 7
Transition range : pMg  5.4  1.0
53/57
(Continued)
• SOLUTION: (Continuous)
– Transition range For Ca2+ titration:
Equation (2) + (3)
Ca2+ + HIn2– + H2O → CaIn– + H3O+
[CaIn- ][H3O+]
7
Ka  K f 

7

10
[HIn2- ][Ca2+ ]
[CaIn- ] 1.0  1010
[Ca ]=

2[HIn ]
7  107
2+
[CaIn- ]
Set ratio 10 and 0.1 for
(Red to Blue)
2[HIn ]
Resulted in [Ca2+] 1.4x103 and 1.4x10 5
Transition range : pCa  3.8  1.0
54/57
(Continued)
Titrating 50 mL of
0.005 M Ca2+ and
0.005 M Mg2+ with
0.010 M EDTA.
10.0
CaIn– + HY3– 
red
HIn2– + CaY2–
8.0
blue
pCa 6.0
or
pMg4.0
Using EBT as indicator,
2.0
the MgIn– transition
range (not CaIn–
transition range) cover
the pCa at equivalance
point for Ca2+-EDTA
titration
MgIn– + HY3– 
red
HIn2– + MgY2–
blue
10.0
20.0
Vol of EDTA for
0.0
10.0
30.0
Ca2+ ,
40.0
mL
20.0
30.0
40.0
Vol of EDTA for Mg2+ , mL
55/57
 How to Minimize Titration Error for Ca2+–EDTA
titration
Information:
Mg2+ + EDTA4– → MgEDTA2–
Ca2+ + EDTA4– → CaEDTA2–
Mg2+ + In3– → MgIn–
Ca2+ + In3– → CaIn–
Kf = 4.9×108
Kf = 5.0×1010
Kf = 1.0×107
Kf = 2.5×105
– Adding few drops of (Na)2Mg–EDTA solution into analyte
solution with EBT indicator, resulted in:
Ca2+ + MgEDTA2– → CaEDTA2– + Mg2+
Mg2+ + In3– → MgIn–
– After all the Ca2+ has been titrated, then
MgIn– + EDTA4– + H+ → MgEDTA2– + HIn2–
– Net reaction (1)+(2)+(3):
Ca2+ + EDTA4– + In3– + H+ → CaEDTA2– + HIn2–
* No quantitative effect by the added Mg2+ for Ca2+-EDTA
titration
(1)
(2)
(3)
56/57
End of Chapter 17