Titrations

1. Titrations

2. Strong Acid with Strong Base • Starting pH pH = -log[FAcid] • Just before the Equivalence Point [H+] = (Vacid·Facid-Vbase·Fbase)/(Vsol+Vbase) • Very Sharp Equivalence Point pH = 7.00 • Excess base pH = 14 – (-log[OH-]) [OH-] = Fbase·{(Vbase-V*)/(Vbase+Vsol)]}

3. 25.00 mL of 0.10 M HCl titrated with 0.10 M NaOH

4. 25.00 mL of ? M HCl titrated with 0.10 M NaOH

5. V* = 12.50 mL of NaOH [HCl] = (12.50 mL)(0.1000 M)/(25.00 mL) = 0.0500 M

6. Strong Base with Strong Acid • Starting pH pH = 14 – {-log[Base] • Very Sharp Equivalence Point pH = 7.00 • Excess acid pH = -log[H+]) [H+] = Facid·{(Vacid-V*)/(Vacid+Vsol)]}

7. 25.00 mL of 0.10 M NaOH titrated with 0.10 M HCl

8. 25.00 mL of ? NaOH titrated with 0.20 M HCl

9. V* = 12.50 mL of HCl [NaOH] = (12.50 mL)(0.2000 M)/(25.00 mL) = 0.1000 M

10. Titrating Monoprotic Weak Acids with Strong Base • Starting pH: Use weak acid dissociation Ka = [H+]2/(FAcid – [H+]) pH = -log[H+] HA + OH- → A- + H2O • Before the Equivalence Point use H-H eq pH = pKa + log [(FNaOH·VNaOH)/(FHA·VHA- FNaOH·VNaOH)] ½ way to Equivalence point, pH = pKa

11. At Equivalence point: use weak base dissociation Kb = [OH-]2/(Fbase – [OH-]), where Fbase = FHA·VHA/(Vsol+Veq) • Excess base [OH-] = Fbase·{(Vbase-V*)/(Vbase+Vsol)]} pH = 14 – (-log[OH-])

12. Ka 1.0·10-5

14. Example for Ka = 10-5 • Starting pH [H+] = ((1∙10-5∙0.1)1/2 = 0.0010 M pH = 3.00 • ½ way – pH = pKa = 5.00 • At Equiv. pt [OH-] = ((1∙10-9∙0.05)1/2 = 7.1·10-6 M pH = 14.00 – [-log(7.1·10-6)] = 8.85

15. Titrating Monoprotic Weak Bases with Strong Acid • Starting pH: Use weak base dissociation Kb = [OH-]2/(Fbase – [OH-]) pH = 14.00 – (-log[OH-]) A- + H+ → HA • Before the Equivalence Point use H-H eq pH = pKa + log [(FA-·VA-- FHCl·VHCl)/ (FHCl·VHCl)] ½ way to Equivalence point, pH = pKa

16. At Equivalence point: use weak acid dissociation Kb = [H+]2/(Facid – [H+]), where FACID = FA-·VA-/(Vsol+Veq) • Excess HCl [H+] = Facid·{(Vacid-V*)/(Vacid+Vsol)]} pH = log[H+])

17. Titrate with 0.10 M HCl

18. Example for Ka = 10-7 • Starting pH [OH-] = ((1∙10-7∙0.1)1/2 = 0.00010 M pH = 10.00 • ½ way – pH = pKa = 7.00 • At Equiv. pt [H-] = ((1∙10-7∙0.05)1/2 = 7.1·10-5 M pH = 4.15

19. Titration of H2A with strong base • Starting pH – weak acid equilibrium • ½ way to first eq.: pH = pKa1 • At first equivalence pt: pH = (pKa1 + pKa2)/2 • ½ way to 2nd eq.: pH = pKa2 • At 2nd equiv. pt.: weak base equilibrium

22. Example • Starting pH – [H+] = (10-3*0.1)1/2 = 0.0100 M • ½ way to first eq.: pH = 3.00 • At first equivalence pt: pH = (pKa1 + pKa2)/2 = 6.00 • ½ way to 2nd eq.: pH = 9.00 • At 2nd equiv. pt.: [OH-] = (10-5*0.05)1/2 = 7.07∙10-4 M, pH = 10.85

23. Amino acids • Amphoteric • Can act as an acid and a base • Glycine (GH2+, GH, G-) pKa1 = 2.350, pKa2 = 9.778

24. Alpha Fraction Plot for Glycine

26. Triprotic system • Example: H3PO4

27. Titration of a triprotic acid with strong base • Starting pH – weak acid equilibrium • ½ way to first eq.: pH = pKa1 • At first equivalence pt: pH = (pKa1 + pKa2)/2 • ½ way to 2nd eq.: pH = pKa2

28. At 2nd equiv. pt.: pH = (pKa2 + pKa3)/2 • ½ way to third equiv. pt.: pH = pKa3 • At third equiv pt: weak base equilibrium

29. pKa’s are 4, 7, and 10

30. H3PO4 (pKa’s 2.2, 7.2, 12.2)