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Pn Syazni Zainul Kamal PPK Bioproses

Chapter 4 : Classical Methods in Techniques of Analytical Chemistry : Titrimetric Methods of Analysis. Pn Syazni Zainul Kamal PPK Bioproses. CO4: ABILITY TO DIFFERENTIATE AND CALCULATE CONCENTRATION OF ANALYTES OF VARIOUS TITRIMETRIC METHODS (ACID-BASE, COMPLEXATION, REDOX, PRECIPITATION).

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Pn Syazni Zainul Kamal PPK Bioproses

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  1. Chapter 4 : Classical Methods in Techniques of Analytical Chemistry : Titrimetric Methods of Analysis Pn Syazni Zainul Kamal PPK Bioproses

  2. CO4: ABILITY TO DIFFERENTIATE AND CALCULATE CONCENTRATION OF ANALYTES OF VARIOUS TITRIMETRIC METHODS (ACID-BASE, COMPLEXATION, REDOX, PRECIPITATION)

  3. What is Titrimetry?? • Any method in which volume is the signal. Also known as volumetric method.

  4. Types of Titrimetric Methods • Classified into four groups based on type of reaction involve; • Acid-base titrations • Complexometric titrations • Redox titrations • Precipitation titrations

  5. Acid base titration • Neutralization - acid is reacted with equivalent amount of base (reaction in acid base titration) • Titration curve – the independent variable (X) is the volume of the titrant, while the dependent variable (Y) is the pH of the solution (which changes depending on the composition of the two solutions). • Titrant: A reagent of a known concentration • Equivalence point: the point at which all of the starting solution, eg an acid, has been neutralized by the titrant, eg a base.

  6. titrant

  7. End point/ equivalence point – signals the completion of the reaction (the point at which all of the starting solution, usually an acid, has been neutralized by the titrant, usually a base).

  8. Strong acid vs strong base (easy titration)

  9. In the case of strong acid and strong base titration, both the titrant & analyte are completely ionized • Example 1 : titration of 0.100 M HCl with 0.100M NaOH : H+ + Cl- + Na+ + OH- H2O + NaCl

  10. Before titration started – only have HCl. [H+]=[HX] • Titration proceed – part of H+ is removed from solution as H2O. [H+] decrease gradually. pH increase [H+]=[remaining HX] • At equivalence point - we have solution of NaCl. Neutralization complete. HCl had been neutralize by NaOH. [H+]= √Kw • Excess NaOH added – [OH-] increase. pH is determine by concentration of OH- [OH-]=[excess titrant]

  11. Equation governing a strong acid (HX) titration Fraction F titrated Present Equation F=0 HX [H+]=[HX] 0<F<1 HX/X- [H+]=[remainingHX] F=1 X- [H+]= √Kw F>1 OH-/X- [OH-]=[excess titrant]

  12. A 50.0ml aliquot of 0.100 M HCl is titrated with 0.200 M NaOH. Calculate the pH of the solution after addition of 0.0, 5.0, 10.0, 25.0, 35.0, 45.0, 55.0 of NaOH Molarity = total moles of solute (mol)/vol. of solution (ml)

  13. Solution : a) Addition 0f 0.0ml NaOH [H+]=[HX] pH = - log [H+] = - log 0.100 = 1.00

  14. b) Addition of 5.0ml NaOH [H+]=[remaining HX] Initial mmol H+ = 0.100 M x 50.0 ml = 5.00 mmol mmol OH- added = 0.200 M x 5.0 ml = 1.00 mmol mmol H+ left = 4.00 mmol in 55.0ml [H+] = 4.00 mmol = 0.0727 M (5.0 + 50.0)ml pH = - log [H+] = - log 0.0727 = 1.14

  15. c) Addition of 25.0 ml NaOH Initial mmol H+ = 0.100 M x 50.0 ml = 5.00 mmol mmol OH- added = 0.200 M x 25.0 ml = 5.00 mmol The equivalence point is reached. All H+ had reacted with OH-. The H+ now is from H2O.NaCl does not contribute to any pH value. [H+]= √Kw = √1.00x10-14 = 1.00x10-7 pH = - log 1.00x10-7 = 7.00

  16. c) Addition of 35 ml NaOH Initial mmol H+ = 0.100 M x 50.0 ml = 5.00 mmol mmol OH- added = 0.200 M x 35.0 ml = 7.00 mmol OH- excess 2.0 mmol [OH-]=[excess titrant] [OH+] = 2.0 mmol = 0.0235 M (35.0 + 50.0)ml pOH = - log [OH+] = - log 0.0235 = 1.63 pH = 14.00 – 1.63 = 12.37

  17. STRONG BASE VERSUS STRONG ACID

  18. Equation governing a strong base (BOH) titration Fraction F titrated Present Equation F=0 BOH [OH-]=[BOH] 0<F<1 BOH/B- [OH-]=[remainingBOH] F=1 B+ [H+]= √Kw F>1 H+/B+ [H+]=[excess titrant]

  19. Assignment 1 Question 1 • A 50.0ml aliquot of 0.100 M NaOH is titrated with 0.200 M HCl. Calculate the pH of the solution after addition of 0.0, 5.0, 10.0, 25.0, 35.0, 45.0, 55.0 of HCl. Plot the titration curve in a graph paper.

  20. WEAK ACID VERSUS STRONG BASE

  21. Weak acid Vs Strong base • Titration of 100 ml 0.1 M acetic acid with 0.1 M sodium hydroxide • Neutralization reaction : HOAc + Na+ + OH- H2O + NaOAc • Acetic acid is neutralized to water and equivalent amount of sodium acetate.

  22. Before titration started – only have HOAc. [H+]= √Ka . CHA • Titration started - some of HOAc convert to NaOAc & buffer system set up. • Titration proceed – pH increase slowly as the ratio [OAc-]/[HOAc] change. pH=pKa + log CA- CHA • Midpoint of titration – [OAc-] = [HOAc] (pH=pKa)

  23. At equivalence point – we have solution of NaOAc. NaOAc is a bronsted base (it hydrolyze in water to form OH- and undissociate HOAc) [OH-]=√Kw . CA- Ka • Excess NaOH added – pH is determine by concentration of OH-

  24. Titration of HOAc with NaOH

  25. Equation governing a weak-acid (HA) Fraction F titrated Present Equation F=0 HA [H+]= √Ka . CHA 0<F<1 HA/A- pH=pKa + log CA- CHA F=1 A- [OH-]=√Kw . CA- Ka F>1 OH-/A- [OH-]= [excess titrant]

  26. Exercise 1 A 50.0 ml aliquot of 0.100 M acetic acid is titrated with 0.100 M NaOH. Calculate the pH of the solution after the addition of 0.0, 5.0, 25.0, 45.0, 50.0, 55.0 and 75.0 ml of NaOH. The dissociation constant for acetic acid is 1.75 x 10-5

  27. Solution • Addition of 0.0 ml NaOH At 0 ml, we have a solution of only 0.100 M HOAc: HOAc H+ + OAc- Initial (M) 0.100 0 0 Equilibrium (M) 0.100 – x x x [H+][OAc-] = (x)(x) = 1.75 x 10-5 [HOAc] 0.100 x = [H+] = 1.32 x10-3 M pH = - log 1.32 x10-3 M = 2.88

  28. b) Addition of 5.0 ml NaOH Initial mmol HOAc = 50.0ml x0.100 M =5.00mmol mmol NaOH added = 5.0ml x 0.100 M =0.50mmol =mmol OAc- form mmol HOAc left =4.50mmol pH=pKa + log CA- CHA = - log 1.75 x10-5 + log 0.50 = 3.81 4.50

  29. c) Addition of 50.0ml NaOH Initial mmol HOAc = 50.0ml x0.100 M =5.00mmol mmol NaOH added = 50.0ml x 0.100 M =5.00mmol Equivalence point is reached. The solution contain salt of weak acid (conjugate base), NaOAc, hence the hydrolysis of salt will take place. All HOAc has been converted to OAc- (5.00 mmol in 100ml or 0.0500 M) NaOAc Na+ + OAc- OAc- + H2O HOAc + OH- [OH-]=√Kw . CA- Ka =√1.00 x10-14 x 0.0500 = 5.35x10-6 1.75x10-5 pOH = -log 5.35x10-6 = 5.27 pH = 14.00 – 5.27 = 8.72

  30. d) Addition of 55.0 ml NaOH Initial mmol HOAc = 50.0ml x0.100 M =5.00mmol mmol NaOH added = 55.0ml x 0.100 M =5.50mmol Excess OH- =0.50 mmol in 105ml At 55.0 ml, we have solution of NaOAc and excess added NaOH. So pH is calculated by concentration of excess OH-. [OH-] = 0.50 mmol = 4.76x10-3 (50+55)ml pOH = - log [OH-] = - log 4.76x10-3 = 2.32 pH = 14.00 – 2.32 = 11.68

  31. WEAK BASE VERSUS STRONG ACID

  32. Weak base vs strong acid • Titration curve reverse of those for a weak acid versus strong base • Titration of 100ml 0.1M ammonia with 0.1M HCl NH3 + H+ + Cl- NH4+ + Cl-

  33. Before titration started – only have 0.1 M NH3. [OH+]= √Kb . Cb • Titration started - some of NH3 convert to NH4+ & buffer system set up. • Titration proceed – pH decrease slowly as the ratio [NH3]/[NH4+] change. pH=(pKw – pKb) + log CB CBH+ • Midpoint of titration – [NH4+] = [NH3] (pH=14 - pKb)

  34. At equivalence point – we have solution of NH4Cl. Neutralization complete. NH3 had been neutralize by HCl. [H+]=√Kw . CBH+ Kb • Excess NaOH added – pH is determine by concentration of H+

  35. Assignment 1 Question 2 • Sketch a titration curve based on the pH values calculated from the addition of 0.0, 10.0, 25.0, 40.0, 50.0, 55.0 and 60.0 ml of 0.100M HCl into 50.0ml 0.100M NH3. Kb is 1.75x10-5. • Submit on : 16 march 2011 (Wednesday class)

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