1. Chapter 15 Applications of Aqueous Equilibria Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1

2. 2 Common Ion Effect The shift in equilibrium that occurs because of the addition of an ion already involved in the equilibrium reaction. AgCl(s)  Ag + (aq) + Cl  (aq)

3. Acid Solutions containing Common Ions In section 14.5 we found that the equilibrium concentrations of H + in a 1.0 M HF solution is 2.7 X 10 -2 M, and the percent dissociation of HF is 2.7%. Calculate [H + ] and the percent dissociation of HF in a solution containing 1.0 M HF (K a = 7.2 x 10 -4 ) and 1.0 M NaF. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 3

4. 4 A Buffered Solution... resists change in its pH when either H + or OH  are added. 1.0 L of 0.50 M H 3 CCOOH + 0.50 M H 3 CCOONa pH = 4.74 Adding 0.010 mol of solid NaOH raises the pH of the solution to 4.76, a very minor change.

5. The pH of a Buffered Solution A buffered solution contains 0.50 M acetic acid HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) and 0.50 M of sodium acetate(NaC 2 H 3 O 2 ). Calculate the pH of this solution. What are the major species? HC 2 H 3 O 2, C 2 H 3 O 2 -, Na +, and H 2 O Copyright©2000 by Houghton Mifflin Company. All rights reserved. 5

6. 6 Key Points on Buffered Solutions 1.They are weak acids or bases containing a common ion. 2.After addition of strong acid or base, deal with stoichiometry first, then equilibrium.

7. pH changes in Buffered Solutions Calculate the change in pH that occurs when 0.010 mol of solid NaOH is added to 1.0 L of the buffered solution from the previous example. Compare this pH change with that which occurs when 0.010 mol of solid NaOH is added to 1.0 L of water. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7

8. How does a Buffer Work OH - + HA → A - + H 2 O [H + ] = K a [HA]/[A - ] Copyright©2000 by Houghton Mifflin Company. All rights reserved. 8

9. 9 Henderson-Hasselbalch Equation Useful for calculating pH when the [A  ]/[HA] ratios are known.

10. The pH of a Buffered Solution (weak acid and conj. base) Calculate the pH of a solution containing.75 M lactic acid (K a = 1.4 x 10 -4 ) and.25 M sodium lactate. Lactic acid is a common constituent of biologic systems. For example, it is found in milk and is present in human muscle tissue during exertion. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 10

11. The pH of a buffered Solution (weak base and its conj. acid) A buffered solution contains.25 M NH 3 (K b = 1.8 x 10 -5 ) and.40 M NH 4 Cl. Calculate the pH of this solution Copyright©2000 by Houghton Mifflin Company. All rights reserved. 11

12. Adding Strong Acid to a Buffered Solution Calculate the pH of the solution that results when.10 mol of gaseous HCl is added to 1.0 L of a buffered solution that contains.25 M NH 3 (K b = 1.8 x 10 -5 ) and.40 M NH 4 Cl. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 12

13. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 13 Buffered Solution Characteristics Buffers contain relatively large amounts of weak acid and corresponding base. Added H + reacts to completion with the weak base. Added OH  reacts to completion with the weak acid. The pH is determined by the ratio of the concentrations of the weak acid and weak base.

14. Buffer Capacity the amount of H + or OH - a buffer can absorb without a significant change in the pH a buffer with a large capacity can absorb a large amount of H + or OH - with only a little change in the pH a buffer with a large capacity contains a large concentration of the weak acid and its salt

15. Adding Strong acid to a Buffered Solution #2 Calculate the change in pH that occurs when.010 mol of gaseous HCl is added to 1.0 L of each of the following solutions: Solution A: 5.00 M HC 2 H 3 O 2 and 5.00 M NaC 2 H 3 O 2 Solution B:.050 M HC 2 H 3 O 2 and.050 M NaC 2 H 3 O 2 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 15

16. How to Prepare a Buffer pH of a buffered solution determined by the ratio of [A - ]/[HA] optimal buffering (least change in pH) will occur when [A - ] = [HA] so pH = pK a + log (1) so pH = pK a + 0 so pH = pK a Pick an acid with a K a closest to the desired pH of the buffered solution.

17. Preparing a Buffer A chemist needs a solution buffered at pH 4.30 and can choose from the following acids ( and their sodium salts): a.Chloroacetic acid (K a = 1.35 x 10 -3 ) b.Propanoic acid (K a = 1.3 x 10 -5 ) c.Benzoic acid (K a = 6.4 x 10 -5 ) d.Hypochlorus acid (K a = 3.5 x 10 -8 ) Copyright©2000 by Houghton Mifflin Company. All rights reserved. 17

18. Titration and pH curves Titration used to determine the concentration or amount of an unknown acid or base uses a solution of known concentration (the titrant) uses a buret (for precision) uses an indicator to show the endpoint can be monitored by plotting pH vs. amount of titrant added

19. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 19 Titration (pH) Curve con. Equivalence (stoichiometric) point: Enough titrant has been added to react exactly with the solution being analyzed.

20. Strong Acid-Strong Base Titrations millimoles may be used because of the small quantities used in a titration 1000 millimoles = 1000mmoles = 1 mole Molarity = moles/liter = mmoles/mL mmoles = Volume (in mL) x Molarity

21. Strong Acid-Strong Base Titration Example Titration of 50.0 mL of 0.200 M HNO 3 with 0.100 M NaOH What is the pH at various stages of the titration? Initially...No NaOH has been added pH is determined by H + from the 0.200 M HNO 3 pH = - log (0.200) = 0.699 50.0 mL x 0.200 M H + = 10.0 mmol H + present

22. Con. Still the beginning of the reaction…10.0 mL of 0.100 M NaOH has been added The added OH - will neutralize an equivalent amount of H +, I.e., 10.0 ml x 0.100 M or 1.00 mmole. How much H + is left? 10.0 mmole - 1.00 mmole = 9.0 mmole H + What is the concentration of H + now? 9.0 mmole/ (50.0 mL + 10.0 mL) = 0.15 M pH = -log (0.15) = 0.82…pH is increasing

23. Con. 20.0 mL of NaOH has been added… pH = 0.942 50.0 mL of NaOH as been added… pH = 1.301

24. Con. Strong Acid-Strong Base Titration At the equivalence point…100.0 mL of NaOH has been added 100.0 mL x 0.100 M = 10.0 mmole OH - This is enough OH - to completely react with the H + in solution. At the equivalence point, the pH is 7, the solution is neutral

25. Con. 150.0 mL of NaOH has been added now OH - is in excess, the pH is determined by the excess OH - 150.0 mL x 0.100 M = 15.00 mmoles OH - 15.00 mmoles OH - - 10.0 mmoles H + = 5.0 mmoles OH - in solution 5.0 mmoles/ (50.0 + 150.0 mL) = 0.025 M OH - ….so [H + ] = 4.0 x 10 -13 …pH = 12.40

26. Con. 200.0 mL of NaOH has been added still excess OH - pH = 12.60

27. pH Curve Copyright©2000 by Houghton Mifflin Company. All rights reserved. 27

28. Applications of Aqueous Equilibria Strong Acid-Strong Base Titration pH changes very little initially until close to the equivalent point (lots of H +, added OH - doesn’t change the pH much) Near the equivalence point, there is less H +, so added OH - changes the pH a lot At the equivalence point, the pH = 7.00 Just after the equivalence point, added OH - also changes the pH a lot

29. Weak Acid-Strong Base Titration Calculating the titration curve is like solving a series of buffer problems. Involves a stoichiometry problem where the reaction goes to completion and the concentrations of the weak acid and the conjugate base are calculated Involves an equilibrium problem, calculate pH from this

30. Con. pH curve for this titration is different before the equivalence point from the strong acid- strong base titration after the equivalence point, the titration curves are the same

31. Con. For a weak acid-strong base titration, the pH rises more rapidly in the beginning of the titration, then levels off at the halfway point due to buffering effects. pH at the equivalence point is higher than for a strong acid-strong base titration

32. Weak acid-Strong Base titration The amount of acid determines the equivalence point The pH value at the equivalence point depends on the acid strength the weaker the acid, the higher the pH at the equivalence point

33. Weak Acid-Strong Base Titration Let’s consider the titration of 50.0 ml of 0.10 M acetic acid(HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) with 0.10 M NaOH. We will calculate the pH at various points Representing volumes of added NaOH. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 33

34. pH curve for weak acid-strong base Copyright©2000 by Houghton Mifflin Company. All rights reserved. 34

35. Acid-Base Indicators Two common methods for determining the equivalence point of a titration Use a pH meter and then plot the titration curve. The center of the vertical region of the pH curve indicates the equivalence point. Use an acid-base indicator to mark the end point with a change in color.

36. Acid-Base Indicators con. The endpoint (when the indicator changes color) may not be the same as the equivalence point. An indicator must be chosen based on the acid and base used in the titration so that the endpoint is as close to the equivalence point as possible.

37. Acid-Base Indicators