1. Complexation and Precipitation Reactions and Titrations
Chapter 17
Complexation and Precipitation Reactions and Titrations

2. 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.

3. Precipitation Titrations
PART 1
Precipitation Titrations

4. 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

5. Example: For the titration of 50. 0 mL of 0. 0500 M Cl– with 0
Example: For the titration of 50.0 mL of M Cl– with 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
0 mL (b) 10.0 mL (c) 25.0 mL (d) 26.0 mL
Solution:

6. (a) 0 mL Ag+ added (At beginning)
[Ag+] = 0, pAg can not be calculated.
[Cl–] = , pCl = 1.30
(b) 10 mL Ag+ added (Before Ve)

7. (c) 25 mL Ag+ added (At Ve)
AgCl(s)  Ag+(aq) + Cl–(aq) Ksp = 1.8×10–10
s = [Ag+]=[Cl–]
Ksp = 1.82×10–10 = s2
[Ag+]=[Cl–]=1.35x10–5
pAg = 4.87
pCl = 4.87

8. (d) 26 mL Ag+ added (After Ve)

9. Construct a titration curve
Example: Titration of 50.0 mL of M Cl– with M Ag+

10. Diluting effect of the titration curves
50.00 mL of M NaCl titrated with M AgNO3,
50.00 mL of M NaCl titrated with M AgNO3.

11. Ksp effect of the titration curves
50.00 mL of a M solution of the anion was titrated with M AgNO3.

12. 2. Titration of a mixture
Titration curves for mL of a solution M in Cl- and M in I- or Br-.
Ksp for
AgCl = 1.82x10-10
AgBr = 5.0x10-13
AgI = 8.0x10-17

13. Example: A mL solution containing Br– and Cl– was titrated with M AgNO3. Ksp(AgBr)=5.0x10–13, Ksp(AgCl)=1.82x10–10.
Which analyte is precipitated first?
The first end point was observed at mL. Find the concentration of the first that precipitated (Br– or Cl–?).
The second end point was observed at 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

14. (b)
Ans
(c)
Ans

15. 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

16. 2). Volhard method: A colored complex (back titration)
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–)

17. Fajans Method: An adsorbed/colored indicator. Titrating Cl– and adding
dichlorofluoroscein for example:
Before Ve
(Cl– excess)
Greenish yellow solution
AgCl(s)
Cl–
1st
layer
Ag+
In–
pink
After Ve
(Ag+ excess)
AgCl(s)
1st
layer

18. 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

19. Complexation Titration
PART 2
Complexation Titration

20. 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
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

21. 1) Chemistry and Properties of EDTA
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 1:1 stoichiometric ratio (metal : ligand = 1 : 1).

22. iv) Six-coordinate structure of metal-EDTA (indeed Y4–complex with metal)

23. v) Equation for Y4– fraction
CT = [H4Y]+[H3Y–]+[H2Y2–]+[HY3–]+[Y4–]
Remark: known pH, then calculate α4,
then calculate [Y4–] = α4 CT

24. vi) Composition of EDTA solutions as a function of pH

25. (Continued)
Figure 17-7 Spreadsheet to calculate 4 for EDTA at selected pH values.

27. 2) EDTA complex i) Formation Constant (KMY) for EDTA complex
Mn+ + Y4– → MY(n–4)

28. 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.

32. EDTA Titration Curves
Example: Calculate the pCa2+ for 50.0 mL of M Ca2+ (buffered at pH 10) with addition of M EDTA of
(a) 0.0 mL), (b) 15.0 mL, (c) 25.0 mL, (d) 26.0 mL
(Kf for CaY2– = 5.0x1010, , αY4– at pH = 0.35 )
Solution:
50.0 x = Veq x 0.01, Veq = 25.0 mL
(a) 0.0 mL EDTA
(At beginning)
pCa2+ = –log[Ca2+] = –log(0.0050) = 2.30

33. (b) 15.0 mL EDTA added
(before equivalence point, 0 < VEDTA < Veq)
pCa2+ = 2.81

34. (c) 25.0 mL EDTA added
(at equivalence point, VEDTA = Veq)
Ca CT → CaY2–
Initial x10-3
Change x x –x
Final x x x10-3 – x
pCa2+ = 6.36

35. (d) 26.0 mL EDTA added
(after equivalence point, VEDTA > Veq)
Ca2+ + CT → CaY2–
[Ca2+] = 1.43 x 10–9 M
pCa2+ = 8.85

36. (Continued)
Figure 17-8 Spreadsheet for the titration of mL of M Ca+2 with M EDTA in a solution buffered at pH 10.0.

37. Effect of the pH
Figure 17-10 Influence of pH on the titration of M Ca+2 with 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.

38. Effect of the Analyte’s Kf
Figure 17-11 Titration curves for 50.0 mL of M solutions of various cations at pH 6.0.
Cations with larger formation constants provide sharp end points even in acidic media.

39. 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 Kf’ value), more complete reaction
- Preventing precipitate with hydroxide, because metal ions formed complex with NH3.

40. 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

41. 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)

42. 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:

43. c) Define Kf’’:
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
EDTA titration in specified pH buffered and specified conc. of auxiliary complexing agent:

44. Example: Calculating the pZn for 50. 0 mL of 0. 0050 M Zn2+ with 0
Example: Calculating the pZn for 50.0 mL of M Zn2+ with 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. (Kf for Zn2+–EDTA is 3.12x1016; for pH 9: 44– 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
(a) 0.0 mL EDTA added
(At beginning)
[Zn2+] = αM x CM = (1.17x10–5)( M) = 5.85x10–8 M
pZn = –log[Zn2+] = –log(5.85 x 10–8) = 7.23

45. (b) 20.0 mL EDTA added
(before equivalence point, 0 < VEDTA < Veq)
[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

46. 25.0 mL EDTA
(at equivalence point, VEDTA = Veq)
CM CT  ZnY2–
Initial – – x10–3
Change +x x –x
Final x x x10–3 – x
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

47. (d) 30.0 mL EDTA (after equivalence point, VEDTA > Veq)
CZn CT → ZnY2– x
6.25x10–4
3.12x10–3
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

48. 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 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 M for one titration and M for the other.

49. 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 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 metallochromic indicators are also acid-base indicators, therefore, the pH control is required for some EDTA titration

50. Common metal-ion indicator

51. Eriochrome Black T (EBT)
MY2– +
H3In
H2In–
(red)
MIn–
HIn2–
(blue)
In3–
(orange)
pKa1=6.3
pKa2=11.6
M: Mg2+, Ca2+, Mn2+,
Zn2+, etc.
At pH 8~10 +
HY3–

52. Indicator’s Transition Range and its Feasibility, EBT as example:
Indicator’s Transition Range and its Feasibility, EBT as example: Example:
Titrating 50 mL of M Ca2+ and Mg2+ with 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)

53. (Continued) SOLUTION: Transition range For Mg2+ titration:
Equation (1) + (3)
Mg2+ + HIn2– + H2O → MgIn– + H3O+

54. (Continued) SOLUTION: (Continuous)
Transition range For Ca2+ titration:
Equation (2) + (3)
Ca2+ + HIn2– + H2O → CaIn– + H3O+

55. (Continued)
Titrating 50 mL of M Ca2+ and M Mg2+ with M EDTA.
10.0
CaIn– + HY3– 
 HIn2– + CaY2–
red
blue
8.0
pCa
6.0 or
pMg
4.0
MgIn– + HY3– 
 HIn2– + MgY2–
red
blue
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
0.0
10.0
20.0
30.0
40.0
Vol of EDTA for Mg2+ , mL

56. How to Minimize Titration Error for Ca2+–EDTA titration
Information:
Mg2+ + EDTA4– → MgEDTA2– Kf = 4.9×108
Ca2+ + EDTA4– → CaEDTA2– Kf = 5.0×1010
Mg2+ + In3– → MgIn– Kf = 1.0×107
Ca2+ + In3– → CaIn– Kf = 2.5×105
Adding few drops of (Na)2Mg–EDTA 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

57. End of Chapter 17