1. Buffer Example and Titration Calculations

2. pH Change to 1M Acetic Acid/1M Acetate Ion Soln. Moles H + Added 0 Moles OH - Added

3. Buffers Buffer Solutions resist a change in pH Buffers contain relatively large concentrations of either –An acid, HA and its conjugate base A - –A base, B, and its conjugate acid (BH + )

4. Buffers NH 4 + to react with OH -

5. Buffers NH 3 to react with H +

6. Buffers When H + is added, it reacts essentially to completion with the weak base present H + + A -  HA or H + + B  BH +

7. Buffers When OH - is added, it reacts essentially to completion with the weak acid present OH - + HA  H 2 O + A - OH - + BH +  H 2 O + B

8. Buffers pH = pK a + log (base/acid) Want pH  pK a  1 pH determined by K a of acid and ratio of acid/conjugate base or K b of base and ratio base/conjugate acid

9. Buffer Choice Want pH  pK a  1 How do I make a pH 4.0 buffer? Choose a pK a near the desired pH

10. Buffer Table Formic AcidK a 1.8 X 10 -4 pK a 3.74 Barbituric Acid9.8 X 10 -5 4.01 Butanoic Acid1.52 X 10 -5 4.82

11. Buffer Choice Choose a pK a near the desired pH pH = pK a + log (base/acid) 4.0 = 3.74 + log (base/acid) 0.26 = log (base/acid) 10.26 = 1.8 = (Na formate / formic acid)

12. Basic Buffer Choice Ammonia pK b = 4.74 pK a = 14.00 – 4.74 = 9.26 NH 3 / NH 4 Cl used to buffer around pH 9.26

13. Buffer Capacity As long as ratio remains virtually constant, the pH will be virtually constant This is true as long as concentrations of buffering materials (HA/A - ) or (B/BH + ) are large compared with H + or OH - added.

14. 25 mL of.2 M HCl titrated with.2 M NaOH Equivalence Point – Where Stoichiometric amounts of acid and base have been added End Point – Where indicator color change occurs Acid / Base Titrations – Strong A & B

15. VM HCl = 5 X 10 -3 mol -VM NaOH = moles of H + leftover till equivalent point reached At equivalence point, 5 X 10 -3 mol NaCl/.050 L solution pH = 7 Acid / Base Titrations – Strong A & B

16. Equivalence Point

17. Acid / Base Titrations – Strong A & B HCl/NaCl buffer

18. Acid / Base Titrations – Strong A & B Phenolphthalein Bromcresol Green

19. 25 mL of.2 M Acetic Acid (HA) titrated with.2 M NaOH Initial pH calculated as before Acid / Base Titrations – Weak Acid with Strong Base

20. During titration up to equivalence point VM HA = 5 X 10 -3 mol -VM NaOH = moles of HA leftover VM NaOH = moles OH - added = moles Ac - made Say 10. mL of.2 M NaOH added to 25 mL of.2 M HA Acid / Base Titrations – Weak Acid with Strong Base

21. HC 2 H 3 O 2  H + + C 2 H 3 O - Acid / Base Titrations – Weak Acid with Strong Base.005 mol-.002 mol 0.002 mol I.003mol/.035 L 0.002 mol /.035L C-x+x E.086 M-x .086 x.057 M +x .057

22. HAc/Ac - buffer Acid / Base Titrations – Weak Acid with Strong Base

23. At Equivalence Point – all acetic acid converted to Acetate ion At Equivalence Point you have a Sodium Acetate Solution Acid / Base Titrations – Weak Acid with Strong Base

24. At Equivalence Point you have a Sodium Acetate Solution To determine pH Use K b and C 2 H 3 O 2 - + H 2 O  HC 2 H 3 O 2 + OH - To determine [OH - ] and [H + ] and pH Acid / Base Titrations – Weak Acid with Strong Base

25. At Equivalence Point you have a Sodium Acetate Solution pH 7 ? pH = 8.88 Acid / Base Titrations – Weak Acid with Strong Base

26. Equivalence Pt pH = 8.88

27. After Equivalence Point VM NaOH – VM HAc(initial) = moles OH - in total volume. From [OH - ] determine [H + ] and pH Acid / Base Titrations – Weak Acid with Strong Base

28. Phenolphthalein Bromocresol Green

29. Acid / Base Titrations – Weak Base with Strong Acid

30. NH3/NH4+ buffer

31. Acid / Base Titrations – Weak Base with Strong Acid Phenolphthalein Methyl Red