1. Chapter 15 Buffers Titration pH Curves
acid-base equilibria
Chapter 15
Buffers
Titration
pH Curves

2. HA(aq) + H2O(l) H3O+(aq) + A–(aq)
Review!
HA(aq) + H2O(l) H3O+(aq) + A–(aq)
If HA is a weak acid, identify which side dominates at equilibrium. Identify and the bases in solution. Which base is stronger? Which base is weaker?
If HA is a strong acid, identify which side dominates at equilibrium. Which base is stronger? Which base is weaker?
Make comparisons between the relative strengths of the conjugate base and water acting as a base.
For weak acid, the conjugate base will always be a stronger base than water since the weak acid has reactants as its major species.
For strong acids, the conjugate base will always be a weaker base than water since products are the major species.

3. How good is Cl–(aq) as a base? Is A–(aq) a good base?
Review!
How good is Cl–(aq) as a base?
Is A–(aq) a good base?
The bases from weakest to strongest are:
Cl–, H2O, A–

4. Buffers Solutions that resist changes in pH
Made of weak acids or weak bases containing a common ion.
Common ion is the conjugate base/conjugate acid

5. Common Ion Effect
Shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction.
An application of Le Châtelier’s principle.

6. Common Ion Effect HCN(aq) + H2O(l) H3O+(aq) + CN-(aq)
Addition of NaCN will shift the equilibrium to the left because of the addition of CN-, which is already involved in the equilibrium reaction.
A solution of HCN and NaCN is less acidic than a solution of HCN alone.
Ensures that concentrations of both HA and A- are high
Allows for “absorption” of both H+ and OH-
Resists change in pH

7. What happens when you add acid/base to a buffered solution?
Strong base added to buffered solution
Strong acid added to buffered solution

8. Solving Problems with Buffered Solutions

9. Example 15.1
Calculate [H+], the pH, and the percent dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2 x 10-4).
H+ = 0.26
pH = 1.57
percent dissociation = H/HF x 100 = 2.6%
Notice that the 5% rule is the same calculation as % dissociation.

10. Example 15.1 b
Calculate [H+], the pH, and the percent dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF.
H+ = 7.2E-4
pH = 3.14
percent dissociation = H/HF x 100 = 0.072%.

11. Example 15.2
A buffered solution contains 0.50M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and 0.50M sodium acetate (NaC2H3O2). Calculate the pH.
pH = 4.74

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13. Example 15.3
Calculate the change in pH that occurs when mol solid NaOH is added to 1.0L of the buffered solution described in the previous example. Compare this pH change with that which occurs when 0.10 mol solid NaOH is added to 1.0L water.

14. Example 15.3
After reaction with NaOH, main controller of pH is the dissociation of acetic acid.

15. Characteristics of Buffers
Contain large amounts of weak acid and corresponding conjugate base.
Added H+ reacts to completion with the conjugate base.
Added OH- reacts to completion with the weak acid.

16. Characteristics of Buffers
The pH in the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. As long as this ratio remains virtually constant, the pH will remain virtually constant.
[A–] / [HA]
[B] / [BH+]
This will be the case as long as the concentrations of the buffering materials (HA and A– or B and BH+) are large compared with amounts of H+ or OH– added.
A consequence of this is…

17. Henderson-Hasselbalch
For a particular buffering system (conjugate acid–base pair), all solutions that have the same ratio [A–] / [HA] will have the same pH.
Can find pH directly from A- and HA concentrations
Notice what happens when these are equal!
pKa of the weak acid to be used in the buffer should be as close as possible to the desired pH.

18. Practice practice practice!

19. AP Exam, 1986
In water, hydrazoic acid, HN3, is a weak acid that has an equilibrium constant, Ka, equal to 2.810–5 at 25ºC. A litre sample of a molar solution of the acid is prepared.
(a) Write the expression for the equilibrium constant, Ka, for hydrazoic acid.
(b) Calculate the pH of this solution at 25ºC.
(c) To litre of this solution, 0.80 gram of sodium azide, NaN3, is added. The salt dissolved completely. Calculate the pH of the resulting solution at 25ºC if the volume of the solution remains unchanged.
(d) To the remaining litre of the original solution, litre of molar NaOH solution is added. Calculate the [OH–] for the resulting solution at 25ºC.

20. AP Exam, 1992
The equations and constants for the dissociation of three different acids are given below.
HCO3– ↔ H+ + CO32– Ka = 4.2×10–7
H2PO4– ↔ H+ + HPO42– Ka = 6.2×10–8
HSO4– ↔ H+ + SO42– Ka = 1.3×10–2
(a) From the systems above, identify the conjugate pair that is best for preparing a buffer with a pH of 7.2. Explain your choice.
(b) Explain briefly how you would prepare the buffer solution described in (a) with the conjugate pair you have chosen.
(c) If the concentrations of both the acid and the conjugate base you have chosen were doubled, how would the pH be affected? Explain how the capacity of the buffer is affected by this change in concentrations of acid and base.
(d) Explain briefly how you could prepare the buffer solution in (a) if you had available the solid salt of the only one member of the conjugate pair and solution of a strong acid and a strong base.

21. Acid-base titrations

22. Titrations Determine amount of an acid or base in solution.
Graphically understand chemistry
Graphically determine Ka and compare acid strength
(Equivalence point)

23. Titration curves
Plotting the pH of the solution being analyzed as a function of the amount of titrant added.
Equivalence (Stoichiometric) Point – point in the titration when enough titrant has been added to react exactly with the substance in solution being titrated.
(can determine how much acid or base was in your original solution).

24. Titration of: 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH
pH = e.p. for strong acid–strong base titrations
Enough OH- added to react with all H+
Relatively little pH change in beginning: early on there is a large amount of H+

25. Titration of: 100.0 mL of 0.50M NaOH with 1.0 M HCl

26. Weak acid-Strong base titrations
E.P is NOT 7.
Always greater than 7
Why?
Amount of acid, not strength, determines equivalence point.
Strength of acid, not amount, determines pH at the equivalence point.

27. Weak acid-Strong base titrations
Titration of:
50.0 mL of M HC2H3O2 with M NaOH
Weak acid-Strong base titrations
pH increases rapidly in beginning (b/c weak acid).
∆pH levels off because of buffering
Occurs where [HA] = [A-]

28. Titration of: 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH
For weak acid or weak base titrations:
At half the equivalence point…
[H+] = Ka
pH = pKa

29. Ka trends in weak acid titrations

30. Acid-Base Indicators
Marks the end point of a titration by changing color (endpoint).
The equivalence point is not necessarily the same as the end point (but they are ideally as close as possible).
Acid and Base forms of phenolphthalein

31. Choosing the best indicators

32. Different indicators change color and a specific pH
0.10 M HCl with 0.10 M NaOH
0.1 M H3COOH with 0.1 M NaOH

33. Determining the effective range of an indicator
The color transition (endpoint) occurs over a range of pH determined by:
When titrating an acid: pH = (pKa(indicator) -1)
When titrating a base: pH = (pKa(indicator) +1)
It is best to choose an indicator whose endpoint is as close as possible to the equivalence point.

34. END