1. Chapter 17 Sections 1-3 Common ions, Buffers and Titration © 2012 Pearson Education, Inc.

2. Aqueous Equilibria Sample Exercise 17.1 Calculating the pH When a Common Ion is Involved What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution? Practice Exercise Calculate the pH of a solution containing 0.085 M nitrous acid (HNO 2, K a = 4.5  10  4 ) and 0.10 M potassium nitrite (KNO 2 ).

3. Aqueous Equilibria Sample Exercise 17.2 Calculating Ion Concentrations When a Common Ion is Involved Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Practice Exercise Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid (HCOOH, K a = 1.8  10  4 ) and 0.10 M in HNO 3.

4. Aqueous Equilibria (b) A.NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH – (aq) B.NH 4 + (aq) + Cl - (aq)  NH 3 (aq) + HCl(l) C.H 3 O + (aq) + OH – (aq)  2H 2 O(l) D.NH 4 + (aq) + H 2 O(l)  H 3 O + (aq) + NH 3 (aq) (a) A.There are no spectator ions. B.Cl – C.Both NH 4 + and Cl – D.NH 4 +

5. Aqueous Equilibria Buffers Buffers are solutions of a weak conjugate acid–base pair. They are particularly resistant to pH changes, even when strong acid or base is added. © 2012 Pearson Education, Inc.

6. Aqueous Equilibria A.All listed pairs will not function as buffers. B.HCHO 2 and CHO 2 – will not work as a buffer because HCHO 2 is a weak acid and CHO 2 – is a spectator ion. C.HCO 3 – and CO 3 2 – will not work as a buffer because HCO 3 – is a weak acid and HCO 3 – is a spectator ion. D.HNO 3 and NO 3 – will not work as a buffer because HNO 3 is a strong acid and NO 3 – is a spectator ion.

7. Aqueous Equilibria Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH  to make F  and water. Similarly, if acid is added, the F  reacts with it to form HF and water. © 2012 Pearson Education, Inc.

8. Aqueous Equilibria A.There is no reaction because C 2 H 3 O 2 – and HC 2 H 3 O 2 form a buffer. B.The NaOH reacts with C 2 H 3 O 2 – converting some of it into HC 2 H 3 O 2. C.The NaOH reacts with HC 2 H 3 O 2 converting some of it into C 2 H 3 O 2 –. D.The NaOH is neutralized and all concentrations (HC 2 H 3 O 2 and C 2 H 3 O 2 – ) remain unchanged.

9. Aqueous Equilibria A.There is no reaction because C 2 H 3 O 2 – and HC 2 H 3 O 2 form a buffer. B.The HCl reacts with C 2 H 3 O 2 – converting some of it into HC 2 H 3 O 2. C.The HCl is neutralized and all concentrations (HC 2 H 3 O 2 and C 2 H 3 O 2 – ) remain unchanged. D.The HCl reacts with HC 2 H 3 O 2 converting some of it into C 2 H 3 O 2 –.

10. Aqueous Equilibria Sample Exercise 17.3 Calculating the pH of a Buffer What is the pH of a buffer that is 0.12 M in lactic acid [CH 3 CH(OH)COOH, or HC 3 H 5 O 3 ] and 0.10 M in sodium lactate [CH 3 CH(OH)COONa, or NaC 3 H 5 O 3 ]? For lactic acid, K a = 1.4  10  4.

11. Aqueous Equilibria Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: [H 3 O + ] [A  ] [HA] K a = HA + H 2 OH 3 O + + A  © 2012 Pearson Education, Inc.

12. Aqueous Equilibria Buffer Calculations Rearranging slightly, this becomes [A  ] [HA] K a = [H 3 O + ] Taking the negative log of both side, we get [A  ] [HA]  log K a =  log [H 3 O + ] +  log pKapKa pH acid base © 2012 Pearson Education, Inc.

13. Aqueous Equilibria Buffer Calculations So pK a = pH  log [base] [acid] Rearranging, this becomes pH = pK a + log [base] [acid] This is the Henderson–Hasselbalch equation. © 2012 Pearson Education, Inc.

14. Aqueous Equilibria Continued Practice Exercise Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. (Refer to Appendix D.) Sample Exercise 17.3 Calculating the pH of a Buffer

15. Aqueous Equilibria Sample Exercise 17.4 Preparing a Buffer How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH is 9.00? (Assume that the addition of NH 4 Cl does not change the volume of the solution.) Practice Exercise Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (C 6 H 5 COOH) to produce a pH of 4.00.

16. Aqueous Equilibria pH Range The pH range is the range of pH values over which a buffer system works effectively. It is best to choose an acid with a pK a close to the desired pH. © 2012 Pearson Education, Inc.

17. Aqueous Equilibria A.HClO because it is a stronger weak acid. A salt containing ClO – is also needed. B.HNO 2 because it is a stronger weak acid. A salt containing NO 2 – is also needed. C.HClO because its pK a is closer to pH = 7.0. A salt containing ClO – is also needed. D.HNO 2 because its pK a is closer to pH = 7.0. A salt containing NO 2 – is also needed.

18. Aqueous Equilibria When Strong Acids or Bases Are Added to a Buffer When strong acids or bases are added to a buffer, it is safe to assume that all of the strong acid or base is consumed in the reaction. © 2012 Pearson Education, Inc.

19. Aqueous Equilibria Addition of Strong Acid or Base to a Buffer 1.Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. 2.Use equilibrium to determine the new pH of the solution. © 2012 Pearson Education, Inc.

20. Aqueous Equilibria Sample Exercise 17.5 Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol CH 3 COOH and 0.300 mol CH 3 COONa to enough water to make 1.000 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1). (a) Calculate the pH of this solution after 5.0 mL of 4.0 M NaOH (aq) solution is added. (b) For comparison, calculate the pH of a solution made by adding 5.0 mL of 4.0 M NaOH( aq) solution to 1.000 L of pure water. Practice Exercise Determine (a) the pH of the original buffer described in Sample Exercise 17.5 after the addition of 0.020 mol HCl (b) the pH of the solution that would result from the addition of 0.020 mol HCl to 1.000 L of pure water.

21. Aqueous Equilibria Titration In this technique, a known concentration of base (or acid) is slowly added to a solution of acid (or base). © 2012 Pearson Education, Inc.

22. Aqueous Equilibria A.No change B.Increase C.Decrease D.Need volumes and molarities to answer the question

23. Aqueous Equilibria Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base. © 2012 Pearson Education, Inc.

24. Aqueous Equilibria Titration of a Strong Acid with a Strong Base © 2012 Pearson Education, Inc. From the start of the titration to near the equivalence point, the pH goes up slowly. Just before (and after) the equivalence point, the pH increases rapidly. At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid. As more base is added, the increase in pH again levels off.

25. Aqueous Equilibria A.100.00 mL B.50.00 mL C.25.00 mL D.12.50 mL

26. Aqueous Equilibria Sample Exercise 17.6 Calculating pH for a Strong Acid–Strong Base Titration Calculate the pH when (a) 49.0 mL and (b) 51.0 mL of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. Practice Exercise Calculate the pH when (a) 24.9 mL and (b) 25.1 mL of 0.100 M HNO 3 have been added to 25.0 mL of 0.100 M KOH solution.

27. Aqueous Equilibria A.pH = 4.2 B.pH = 7.0 C.pH = 8.2 D.pH = 9.8

28. Aqueous Equilibria Titration of a Weak Acid with a Strong Base Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. At the equivalence point the pH is >7. Phenolphthalein is commonly used as an indicator in these titrations. © 2012 Pearson Education, Inc.

29. Aqueous Equilibria A.The volume of base does not change but the equivalence point pH increases. B.The volume of base increases but the equivalence point pH does not change. C.The volume of base decreases and the equivalence point pH does not change. D.The volume of base does not change but the equivalence point pH decreases.

30. Aqueous Equilibria Titration of a Weak Acid with a Strong Base At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time. © 2012 Pearson Education, Inc.

31. Aqueous Equilibria Sample Exercise 17.7 Calculating pH for a Weak Acid–Strong Base Titration Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH (K a = 1.8  10  5 ). Practice Exercise (a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (C 6 H 5 COOH, K a = 6.3  10  5. (b) Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH 3.

32. Aqueous Equilibria Sample Exercise 17.8 Calculating the pH at the Equivalence Point Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. Practice Exercise Calculate the pH at the equivalence point when (a) 40.0 mL of 0.025 M benzoic acid (C 6 H 5 COOH, K a = 6.3  10  5 ) is titrated with 0.050 M NaOH; (b) 40.0 mL of 0.100 M NH 3 is titrated with 0.100 M HCl.

33. Aqueous Equilibria Titration of a Weak Acid with a Strong Base With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle. © 2012 Pearson Education, Inc.

34. Aqueous Equilibria A.The volume of base does not change but the equivalence point pH increases. B.The volume of base increases but the equivalence point pH does not change. C.The volume of base decrease sand the equivalence point does not change. D.The volume of base does not change but the equivalence point pH increases.

35. Aqueous Equilibria Titrations of Polyprotic Acids When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation. © 2012 Pearson Education, Inc.

36. Aqueous Equilibria Titration of a Weak Base with a Strong Acid The pH at the equivalence point in these titrations is <7, so using phenolphthalein would not be a good idea. Methyl red is the indicator of choice. © 2012 Pearson Education, Inc.

37. Aqueous Equilibria A.No, methyl red does not change color at equivalence point. B.No methyl red’s color change is not sharp and gives only a rough estimate of the equivalence point pH. C.Yes, methyl red changes color in the pH region between 4 and 6 in a steeply sloping region. D.Yes, methyl red changes color in the pH region between 8–10 in a steeply sloping region.

38. Aqueous Equilibria A.The nearly vertical equivalence point portion of the titration curve is large for a weak acid-strong base titration, and fewer indicators undergo their color change so quickly because the change is difficult to monitor. B.The nearly vertical equivalence point portion of the titration curve is smaller for a weak acid-strong base titration, and fewer indicators undergo their color change within this narrow range. C.Many indicators do not change colors at the equivalence points of weak acid-strong base titrations. D.Equivalence points at pH’s other than 7.00 are difficult to determine.