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Buffer Example and Titration Calculations

pH Change to 1M Acetic Acid/1M Acetate Ion Soln.. Moles H Added. 0. Moles OH- Added. Buffers. Buffer Solutions resist a change in pHBuffers contain relatively large concentrations of eitherAn acid, HA and its conjugate base A-A base, B, and its conjugate acid (BH ) . Buffers. NH4 to react with OH-.

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Buffer Example and Titration Calculations

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    1. Buffer Example and Titration Calculations With your Chemistry 102 Host Dr. Mike Daniel

    2. pH Change to 1M Acetic Acid/1M Acetate Ion Soln.

    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

    5. Buffers

    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 ? H2O + A- OH- + BH+ ? H2O + B

    8. Buffers pH = pKa + log (base/acid) Want pH ? pKa ? 1 pH determined by Ka of acid and ratio of acid/conjugate base or Kb of base and ratio base/conjugate acid

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

    10. Buffer Table

    11. Buffer Choice Choose a pKa near the desired pH pH = pKa + 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 pKb = 4.74 pKa = 14.00 4.74 = 9.26 NH3 / NH4Cl 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. Follow Text page 714 (Brady & Holum) 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. VMHCl = 5 X 10-3 mol -VMNaOH = 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. Acid / Base Titrations Strong A & B

    17. Acid / Base Titrations Strong A & B

    18. Acid / Base Titrations Strong A & B

    19. 25 mL of .2 M Acetic Acid (HAc) 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 VMHAc = 5 X 10-3 mol -VMNaOH = moles of HAc leftover VMNaOH = moles OH- added = moles Ac- made Say 10. mL of .2 M NaOH added to 25 mL of .2 M HAc Acid / Base Titrations Weak Acid with Strong Base

    21. HC2H3O2 ? H+ + C2H3O- Acid / Base Titrations Weak Acid with Strong Base

    22. 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 Kb and C2H3O2- + H2O ? HC2H3O2 + 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 < or > 7 ? pH = 8.88 Acid / Base Titrations Weak Acid with Strong Base

    26. Acid / Base Titrations Weak Acid with Strong Base

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

    28. Acid / Base Titrations Weak Acid with Strong Base

    29. Acid / Base Titrations Weak Base with Strong Acid

    30. Acid / Base Titrations Weak Base with Strong Acid

    31. Acid / Base Titrations Weak Base with Strong Acid

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