1. Chapter 17 Complexation and Precipitation Reactions and Titrations 1/57

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 Ca 2+ in water by its titration with EDTA as an example. 2/57

3. PART 1 Precipitation Titrations 3/57

4. 1.Titration curve 1)Guidance in precipitation titration calculation Find V e (volume of titrant at equivalence point) Find y-axis values: - At beginning - Before V e - At V e - After V e 4/57

5. 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/K sp = 1/(1.82×10 –10 ) = 5.6 x 10 9 Find pAg and pCl of Ag + solution added (a) 0 mL (b) 10.0 mL (c) 25.0 mL (d) 26.0 mL Solution: 5/57

6. (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 V e ) 6/57

7. (c)25 mL Ag + added (At V e ) AgCl (s)  Ag + (aq) + Cl – (aq) K sp = 1.8×10 –10 s = [Ag + ]=[Cl – ] K sp = 1.82×10 –10 = s 2 [Ag + ]=[Cl – ]=1.35x10 –5 pAg = 4.87 pCl = 4.87 7/57

8. (d)26 mL Ag + added (After V e ) 8/57

9. Construct a titration curve Example: Titration of 50.0 mL of 0.0500 M Cl – with 0.100 M Ag + 9/57

10. Diluting effect of the titration curves (A)50.00 mL of 0.05000 M NaCl titrated with 0.1000 M AgNO 3, (B)50.00 mL of 0.00500 M NaCl titrated with 0.01000 M AgNO 3. 10/57

11. K sp effect of the titration curves 50.00 mL of a 0.0500 M solution of the anion was titrated with 0.1000 M AgNO 3. 11/57

12. 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 -. K sp for AgCl = 1.82x10 -10 AgBr = 5.0x10 -13 AgI = 8.0x10 -17 12/57

13. Example: A 25.00 mL solution containing Br – and Cl – was titrated with 0.03333 M AgNO 3. K sp (AgBr)=5.0x10 –13, K sp (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/K sp (AgBr) = 2x10 12 Ag + (aq) + Cl – (aq)  AgCl (s) K = 1/K sp (AgCl) = 5.6x10 9 Ans: AgBr precipitated first 13/57

14. (b) (c) Ans 14/57

15. 3.Argentometric Titration 1)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 15/57

16. 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 Fe 3+ 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 Fe 3+ ) SCN – + Fe 3+ → Fe(SCN) 2+ (aq) Total mol Ag + = (mol Ag + consumed by Cl – ) + (mol Ag + consumed by SCN – ) 16/57

17. Before V e (Cl – excess) Greenish yellow solution AgCl (s) Cl – 1st layer After Ve (Ag + excess) AgCl (s) Ag + In – pink 1st layer 3)Fajans Method: An adsorbed/colored indicator. Titrating Cl – and adding dichlorofluoroscein for example: 17/57

18. 4)Mohr Method: A colored precipitate formed by Ag + with anion, other than analyte, once the V e reached. Analysing Cl – and adding CrO 4 2– for example: Precipitating Cl – : Ag + + Cl – → AgCl (s) K sp = 1.8 x 10 –10 End point determination by red colored precipitate, Ag 2 CrO 4(s) : 2Ag + + CrO 4 2– → Ag 2 CrO 4(s) K sp = 1.2 x 10 –12 18/57

19. PART 2 Complexation Titration 19/57

20. TitrantAnalyteRemarks Hg(NO 3 ) 2 Br –, Cl –, SCN –, thiourea Products are neutral Hg(II) complexes, Various indicators used AgNO 3 CN – Product is Ag(CN) 2 – ; indicator is I – ; titrate to first turbidity of AgI NiSO 4 CN – Product is Ni(CN) 4 2– ; Various indicators used KCNCu 2+, Hg 2+, Ni 2+ Product are Cu(CN) 2 2–, Hg(CN) 2, and Ni(CN) 4 2– ; various indicators used 1.Examples of simple Complexation titration Remark: Feasibility titration for M + L  ML 20/57

21. 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). 2.EDTA Titration 1)Chemistry and Properties of EDTA 21/57

22. iv)Six-coordinate structure of metal-EDTA (indeed Y 4– complex with metal) Y 4– : 22/57

23. C T = [H 4 Y]+[H 3 Y – ]+[H 2 Y 2– ]+[HY 3– ]+[Y 4– ] Remark: known pH, then calculate α 4, then calculate [Y 4– ] = α 4 C T v)Equation for Y 4– fraction 23/57

24. vi)Composition of EDTA solutions as a function of pH 24/57

25. Figure 17-7 Spreadsheet to calculate  4 for EDTA at selected pH values. (Continued) 25/57

26. 26/57

27. M n+ + Y 4– → MY (n–4) 2)EDTA complex i)Formation Constant (K MY ) for EDTA complex 27/57

28. The complexatin reaction can be written as: M n+ + C T → MY (n–4) C T : the initial (total) concentration of EDTA. ii)pH buffered Conditional Formation Constant (K’ MY ) 28/57

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32. Example: Calculate the pCa for 50.0 mL of 0.0050 M Ca 2+ (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 (K MY for CaY 2– = 5.0x10 10, α 4 at pH 10.00 = 0.35 ) Solution: 50.0 x 0.005 = V eq x 0.01, V eq = 25.0 mL (a)0.0 mL EDTA (At beginning) pCa = –log[Ca 2+ ] = –log(0.0050) = 2.30 EDTA Titration Curves 32/57

33. (b)1.0 mL EDTA added (before equivalence point, 0 < V EDTA < V eq ) pCa 2+ = 2.81 33/57

34. (c)25.0 mL EDTA added (at equivalence point, V EDTA = V eq ) pCa 2+ = 6.36 Ca 2+ + C T → CaY 2– Initial 0 0 3.33x10 -3 Change x x –x Final x x 3.33x10 -3 – x 34/57

35. (d)26.0 mL EDTA added (after equivalence point, V EDTA > V eq ) Ca 2+ + C T → CaY 2– [Ca 2+ ] = 1.43 x 10 –9 M pCa 2+ = 8.85 35/57

36. 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. (Continued) 36/57

37. 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 pH 37/57

38. 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. Effect of the Analyte’s K f 38/57

39. iii)pH buffered with Auxiliary Complexing Agents Conditional Formation Constant (K MY ”)  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., NH 3. Function of auxilliary complexing agents: - Increase pH value, increase ratio of Y 4– (higher K MY ’ value), more complete reaction - Preventing precipitate with hydroxide, because metal ions formed complex with NH 3. 39/57

40. Fraction of uncomplexed metal Zn 2+ with auxiliary complexing agent NH 3 for example: a)Cumulative formation constant for Zn 2+ /NH 3 system: Zn 2+ + NH 3  ZnNH 3 2+ log  1 =2.21 Zn 2+ + 2NH 3  Zn(NH 3 ) 2 2+ log  2 =4.50 Zn 2+ + 3NH 3  Zn(NH 3 ) 3 2+ log  3 =6.86 Zn 2+ + 4NH 3  Zn(NH 3 ) 4 2+ log  4 =8.89 40/57

41. C M = [Zn 2+ ]+[ZnNH 3 2+ ]+[Zn(NH 3 ) 2 2+ ]+[Zn(NH 3 ) 3 2+ ]+[Zn(NH 3 ) 4 2+ ] =[Zn 2+ ]+β 1 [Zn 2+ ][NH 3 ]+β 2 [Zn 2+ ][NH 3 ] 2 +β 3 [Zn 2+ ][NH 3 ] 3 +β 4 [Zn 2+ ][NH 3 ] 4 =[Zn 2+ ](1+β 1 [NH 3 ]+β 2 [NH 3 ] 2 +β 3 [NH 3 ] 3 +β 4 [NH 3 ] 4 ) b)Fraction of uncomplexed Zn 2+ (α M ) 41/57

42. Example: Zn 2+ and NH 3 form the complexes Zn(NH 3 ) 2+, Zn(NH 3 ) 2 2+, Zn(NH 3 ) 3 2+, Zn(NH 3 ) 4 2+. (log  1 =2.21, log  2 =4.50, log  3 =6.86, log  4 =8.89). If the concentration of unprotonated NH 3 is 0.10 M, find the zinc fraction in the form of Zn 2+. Solution: 42/57

43. The complexation reaction can be written as: C M + C T → MY (n–4) C T :initial (total) concentration of EDTA. C M :initial (total) concentration of M n+ metal ion c)Define K MY ’’: EDTA titration in specified pH buffered and specified conc. of auxiliary complexing agent: 43/57

44. Example: Calculating the pZn for 50.0 mL of 0.0050 M Zn 2+ with 0.010 M EDTA at a pH 9 and in the presence of 0.10 M NH 3, when adding EDTA (a) 0 mL, (b) 20.0 mL, (c) 25.0 mL, (d) 30.0 mL. (K MY for Zn 2+ –EDTA is 3.12x10 16 ; for pH 9:  4 is 5.21x10 -2 ; for 0.10 M NH 3 :  M is 1.17x10 –5 ) (a)0.0 mL EDTA added (At beginning) [Zn 2+ ] = α M x C M = (1.17x10 –5 )(0.0050 M) = 5.85x10 –8 M pZn = –log[Zn 2+ ] = –log(5.85 x 10 –8 ) = 7.23 Solution: (50.0)(0.0050) = (0.010)(V eq ) V eq = 25.0 mL 44/57

45. (b)20.0 mL EDTA added (before equivalence point, 0 < V EDTA < V eq ) [Zn 2+ ] = α M x C M = (1.17x10 –5 )(7.14 × 10 –4 M) = 8.35 × 10 –9 M pZn = –log[Zn 2+ ] = –log(8.35 × 10 –9 ) = 8.08 45/57

46. (c)25.0 mL EDTA (at equivalence point, V EDTA = V eq ) C M + C T  ZnY 2– Initial – – 3.33x10 –3 Change +x +x –x Final x x 3.33x10 –3 – x x = C M = 4.19 × 10 –7 M [Zn 2+ ]=α M x C M = (1.17x10 –5 )(4.19x10 –7 M) = 4.90 × 10 –12 M pZn = –log[Zn 2+ ] = 11.31 46/57

47. (d)30.0 mL EDTA (after equivalence point, V EDTA > V eq ) C Zn + C T → ZnY 2– 3.12x10 –3 x 6.25x10 –4 C M = 2.63 x 10 –10 M [Zn 2+ ] = α M x C M = (1.17x10 –5 )(2.63 x 10 –10 ) = 3.08 x 10 –15 M pZn = –log[Zn 2+ ] = 14.51 47/57

48. 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. Effect of the concentration of auxiliary complexing agents 48/57

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 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 49/57

50. Common metal-ion indicator 50/57

51. Eriochrome Black T (EBT) H 3 InH 2 In – (red) MIn – (red) HIn 2– (blue) In 3– (orange) pKa 1 =6.3pKa 2 =11.6 M: Mg 2+, Ca 2+, Mn 2+, Zn 2+, etc. At pH 8~10 MY 2– + HY 3– 51/57

52. Indicator ’ s Transition Range and its Feasibility, EBT as example: Example: Titrating 50 mL of 0.005 M Ca 2+ and Mg 2+ with 0.010 M EDTA at pH 10 using EBT as the indicator. Calculate the transition range for Ca 2+ and Mg 2+. Information: Mg 2+ + In 3– → MgIn – K f = 1.0×10 7 (1) Ca 2+ + In 3– → CaIn – K f = 2.5×10 5 (2) HIn 2– + H 2 O → In 3– + H 3 O + K a = 2.8×10 –12 (3) Mg 2+ + EDTA 4– → MgEDTA 2– K f = 4.9×10 8 (4) Ca 2+ + EDTA 4– → CaEDTA 2– K f = 5.0×10 10 (5) 52/57

53. SOLUTION: – Transition range For Mg 2+ titration: Equation (1) + (3) Mg 2+ + HIn 2 – + H 2 O → MgIn – + H 3 O + (Continued) 53/57

54. SOLUTION: (Continuous) – Transition range For Ca 2+ titration: Equation (2) + (3) Ca 2+ + HIn 2 – + H 2 O → CaIn – + H 3 O + (Continued) 54/57

55. 2.0 4.0 6.0 8.0 10.0 20.030.0 40.0 10.020.030.0 40.0 0.0 Vol of EDTA for Ca 2+, mL Vol of EDTA for Mg 2+, mL Titrating 50 mL of 0.005 M Ca 2+ and 0.005 M Mg 2+ with 0.010 M EDTA. CaIn – + HY 3–  HIn 2– + CaY 2– red blue MgIn – + HY 3–  HIn 2– + MgY 2– red blue Using EBT as indicator, the MgIn – transition range (not CaIn – transition range) cover the pCa at equivalance point for Ca 2+ -EDTA titration pCa pMg or (Continued) 55/57

56. How to Minimize Titration Error for Ca 2+ – EDTA titration Information: Mg 2+ + EDTA 4– → MgEDTA 2– K f = 4.9×10 8 Ca 2+ + EDTA 4– → CaEDTA 2– K f = 5.0×10 10 Mg 2+ + In 3– → MgIn – K f = 1.0×10 7 Ca 2+ + In 3– → CaIn – K f = 2.5×10 5 – Adding few drops of (Na) 2 Mg – EDTA solution into analyte solution with EBT indicator, resulted in: Ca 2+ + MgEDTA 2 – → CaEDTA 2 – + Mg 2+ (1) Mg 2+ + In 3 – → MgIn – (2) – After all the Ca 2+ has been titrated, then MgIn – + EDTA 4– + H + → MgEDTA 2– + HIn 2– (3) – Net reaction (1)+(2)+(3): Ca 2+ + EDTA 4– + In 3– + H + → CaEDTA 2– + HIn 2– * No quantitative effect by the added Mg 2+ for Ca 2+ -EDTA titration 56/57

57. End of Chapter 17 57/57