1. Titrations

2. Strong Acid with Strong Base Starting pH pH = -log[F Acid ] Just before the Equivalence Point [H + ] = (V acid ·F acid -V base ·F base )/(V sol +V base ) Very Sharp Equivalence Point pH = 7.00 Excess base pH = 14 – (-log[OH - ]) [OH - ] = F base ·{(V base -V*)/(V base +V sol )]}

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 + ] = F acid ·{(V acid -V*)/(V acid +V sol )]}

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 K a = [H + ] 2 /(F Acid – [H + ]) pH = -log[H + ] HA + OH - → A- + H 2 O Before the Equivalence Point use H-H eq pH = pKa + log [(F NaOH ·V NaOH )/(F HA ·V HA - F NaOH ·V NaOH )] ½ way to Equivalence point, pH = pKa

11. At Equivalence point: use weak base dissociation K b = [OH - ] 2 /(F base – [OH - ]), where F base = F HA ·V HA /(V sol +V eq ) Excess base [OH - ] = F base ·{(V base -V*)/(V base +V sol )]} 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 K b = [OH - ] 2 /(F base – [OH - ]) pH = 14.00 – (-log[OH - ]) A - + H + → HA Before the Equivalence Point use H-H eq pH = pKa + log [(F A- ·V A- - F HCl ·V HCl )/ (F HCl ·V HCl )] ½ way to Equivalence point, pH = pKa

16. At Equivalence point: use weak acid dissociation K b = [H + ] 2 /(F acid – [H + ]), where F ACID = F A- ·V A- /(V sol +V eq ) Excess HCl [H + ] = F acid ·{(V acid -V*)/(V acid +V sol )]} 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 H 2 A with strong base Starting pH – weak acid equilibrium ½ way to first eq.: pH = pKa 1 At first equivalence pt: pH = (pKa 1 + pKa 2 )/2 ½ way to 2 nd eq.: pH = pKa 2 At 2 nd 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 = (pKa 1 + pKa 2 )/2 = 6.00 ½ way to 2 nd eq.: pH = 9.00 At 2 nd 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 (GH 2 +, GH, G - ) pKa 1 = 2.350, pKa 2 = 9.778

24. Alpha Fraction Plot for Glycine

26. Triprotic system Example: H 3 PO 4

27. Titration of a triprotic acid with strong base Starting pH – weak acid equilibrium ½ way to first eq.: pH = pKa 1 At first equivalence pt: pH = (pKa 1 + pKa 2 )/2 ½ way to 2 nd eq.: pH = pKa 2

28. At 2 nd equiv. pt.: pH = (pKa 2 + pKa 3 )/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. H 3 PO 4 (pKa’s 2.2, 7.2, 12.2)