1. Indicators for Acid-Base Titrations (Sec. 9-6)

2. transition range needs to match the endpoint pH as closely as possible in order to minimize titration error

3. Acid-Base indicators are themselves weak acids….. e.g. phenolthalein H 2 In = HIn - = In 2-

5. Ch 10: Acid-Base Titrations phenolthalein 8.0-9.6 Automated titrators determine the endpoint electronically by numerically calculating the 2 nd derivative

6. Acid-Base Titrations Curves - pH (or pOH) as a function of mL of titrant added mL base  pH  analyte = strong acid titrant = strong base mL acid  pH  analyte = strong base titrant = strong acid 1 2 3 4

7. I. Strong Acid-Strong Base Titration Curves (Sec. 10-1) equivalence pt. volume: 50 mL of 0.100 M HCl is titrated with 0.100 M NaOH. Calculate the titration curve for the analysis. 1 Initial pH

8. 2 pH before the equivalence pt. 3 pH at the equivalence pt.

9. 4 pH after the equivalence pt.

10. mL base  pH  [H + ] = C HA so pH = -log C HA Strong Acid - Strong Base Titration (both monoprotic) (analyte) (titrant) Eq. Pt. pH = 7 [H + ] = M a V a - M b V b V total [OH-] = M b (V b beyond eq.pt.) V total

11. methyl red 4.2-6.2 phenolthalein 8.0-9.6

12. Titration Error phenolthalein 8.0-9.6 0.02 mL/50 mL =0.04% error!

13. II. Weak Acid-Strong Base Titration Curve (Sec. 10-2) HA = H + + A - 50 mL of a 0.100 M soln of the weak acid HA, K a = 1.0 x 10 -5, is titrated with 0.100 M NaOH. Calculate the titration curve for the analysis.

14. equivalence pt. volume: 1 Initial pH

15. 2 pH before the equivalence pt.

16. 4 pH after the equivalence pt. = same as SA-SB titration 3 pH at the equivalence pt.

17. mL base  pH  Weak Acid - Strong Base Titration (both monoprotic) (analyte) (titrant) Eq. Pt. Hydrolysis of the conjugate base [OH-] = M b (V b beyond eq.pt.) V total Buffer region 1/2 eq. pt. pH = pK a

19. Ch 11: Titrations in Diprotic Systems Biological Applications - Amino Acids (Sec. 11-1) low pH high pH R = (CH 3 ) 2 CHCH 2 -

20. Finding the pH in Diprotic Systems (Sec. 11-2) The strength of H 2 L + as an acid is much, much greater than HL - K a1 = 10 -2.328 = 4.7 x 10 -3 K a2 = 10 -9.744 = 1.8 x 10 -10 So assume the pH depends only on H 2 L + and ignore the contribution of H + from HL. 1. The acidic form H 2 L +

21. Calculate the pH of 0.050M H 2 L +

22. 2. The basic form L - K a1 = 10 -2.328 = 4.7 x 10 -3 K a2 = 10 -9.744 = 1.8 x 10 -10 Strengths of conjugate bases: for L - K b1 = K w /K a2 = 1.01 x 10 -14 /1.8 x 10 -10 = 5.61 x 10 -5 for HLK b2 = K w /K a1 = 1.01 x 10 -14 /4.7 x 10 -3 = 2.1 x 10 -12 Since the second conj. base HL is so weak, we'll assume all the OH- comes from the L - form.

23. Example: Calculate the pH of a 0.050M solution of sodium leucinate

24. The Intermediate Form The pH of a Zwitterion Solution - Leucine (HL form)

27. [H + ] 2 = K a1 K a2 -log [H + ] 2 = - log K a1 - log K a2 2 pH = pK a1 + pK a2 assume: K w K a1 << K a1 K a2 C HL K a1 << C HL pH of a solution of a diprotic zwitterion

28. Example: pH of the Intermediate Form of a Diprotic Acid Potassium hydrogen phthalate, KHP, is a salt of the intermediate form of phthalic acid. Calculate the pH pf 0.10M KHP and 0.010M KHP.

29. Titration Curve for the Amino Acid Leucine

30. equivalence pt. volumes (V e1 & V e2 ) = pts B and D: 1 st and 2 nd half eq. pt's = pt A: init. pH (H 2 L + treat as monoprotic weak acid) =

31. pt C: 1 st eq. pt (HL) = pt E: 2nd eq. pt (L - ) =

33. Example p. 233: Titration of Sodium Carbonate (soda ash) Calculate the titration curve for the titration of 50.0 mL of 0.020 M Na 2 CO 3 with 0.100 M HCl. equivalence pt. volumes (V e1 & V e2 ) =

35. pt C: 1 st eq. pt (HCO 3 - ) = pt E: 2nd eq. pt (H 2 CO 3 treat as monoprotic weak acid) = pts B and D: 1 st and 2 nd half eq. pt's = pt A: init. pH (CO 3 2- treat as monoprotic weak base) =

36. pt E: 2nd eq. pt (H 2 CO 3 treat as monoprotic weak acid) =

38. Buffers of Polyprotic Acids and Bases H3PO4HPO42- PO43-H2PO4-