 Download Download Presentation Chapter 17 Additional Aspects of Aqueous Equilibria

# Chapter 17 Additional Aspects of Aqueous Equilibria

Download Presentation ## Chapter 17 Additional Aspects of Aqueous Equilibria

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1. Lecture Presentation Chapter 17Additional Aspects of Aqueous Equilibria Subhash Goel South GA State College Douglas, GA © 2012 Pearson Education, Inc.

2. CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO(aq) The Common-Ion Effect • Consider a solution of acetic acid: • If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left. © 2012 Pearson Education, Inc.

3. The Common-Ion Effect “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.” © 2012 Pearson Education, Inc.

4. Sample Exercise 17.1Calculating 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? Solution First, because CH3COOH is a weak electrolyte and CH3COONa is a strong electrolyte, the major species in the solution are CH3COOH (a weak acid), Na+ (which is neither acidic nor basic and is therefore a spectator in the acid–base chemistry), and CH3COO– (which is the conjugate base of CH3COOH). Second, [H+] and, therefore, the pH are controlled by the dissociation equilibrium of CH3COOH:

5. Sample Exercise 17.1Calculating the pH When a Common Ion is Involved Continued Third, we tabulate the initial and equilibrium concentrations : The equilibrium concentration of CH3COO (the common ion) is the initial concentration that is due to CH3COONa (0.30 M) plus the change in concentration (x) that is due to the ionization of CH3COOH. Now we can use the equilibrium-constant expression: The dissociation constant for CH3COOH at 25 C is from Appendix D; addition of CH3COONa does not change the value of this constant. Substituting the equilibrium constant concentrations from our table into the equilibrium expression gives: Because Ka is small, we assume that x is small compared to the original concentrations of CH3COOH and CH3COO– (0.30 M each). Thus, we can ignore the very small x relative to 0.30 M, giving:

6. Sample Exercise 17.1Calculating the pH When a Common Ion is Involved Continued The resulting value of x is indeed small relative to 0.30, justifying the approximation made in simplifying the problem. Finally, we calculate the pH from the equilibrium concentration of H+(aq): x = 1.8  105 M = [H+] pH = –log(1.8  105) = 4.74 Comment In the absence of sodium acetate a 0.30 M solution of CH3COOH would have a pH of 2.64

7. [H3O+] [F] [HF] Ka = = 6.8  104 The Common-Ion Effect Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Ka for HF is 6.8  104. © 2012 Pearson Education, Inc.

8. HF(aq) + H2O(l) H3O+(aq) + F(aq) The Common-Ion Effect Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M. © 2012 Pearson Education, Inc.

9. HF(aq) + H2O(l) H3O+(aq) + F(aq) The Common-Ion Effect Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M. © 2012 Pearson Education, Inc.

10. HF(aq) + H2O(l) H3O+(aq) + F(aq) The Common-Ion Effect Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M. © 2012 Pearson Education, Inc.

11. (0.10) (x) (0.20) 6.8  104 = (0.20) (6.8  104) (0.10) The Common-Ion Effect = x 1.4  103 = x © 2012 Pearson Education, Inc.

12. The Common-Ion Effect • Therefore, [F] = x = 1.4  103 [H3O+] = 0.10 + x = 0.10 + 1.4  103 = 0.10 M • So, pH = log (0.10) pH = 1.00 © 2012 Pearson Education, Inc.

13. Buffers • A buffer resists changes in pH because it contains both acid to neutralize added OH- ions and a base to neutralize added H+ ions. • Solutions of a weak conjugate acid–base pair fulfill this requirement and acts as buffers. • They are particularly resistant to pH changes, even when strong acid or base is added. • Examples: CH3COOH and CH3COO- ion solution or NH4+ ion and NH3 solution or HF and F- ions solution. © 2012 Pearson Education, Inc.

14. 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. © 2012 Pearson Education, Inc.

15. Buffers Similarly, if acid is added, the F reacts with it to form HF and water. © 2012 Pearson Education, Inc.

16. HA + H2O H3O+ + A [H3O+] [A] [HA] Ka = Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: © 2012 Pearson Education, Inc.

17. [A] [HA] [H3O+] Ka = base [A] [HA] log [H3O+] + log log Ka = pKa acid pH Buffer Calculations Rearranging slightly, this becomes Taking the negative log of both side, we get © 2012 Pearson Education, Inc.

18. [base] [acid] pKa = pH  log [base] [acid] pH = pKa + log Buffer Calculations • So • Rearranging, this becomes • This is the Henderson–Hasselbalch equation. © 2012 Pearson Education, Inc.

19. Henderson–Hasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4  104. © 2012 Pearson Education, Inc.

20. [base] [acid] pH = pKa + log (0.10) (0.12) pH = log (1.4  104) + log Henderson–Hasselbalch Equation pH = 3.85 + (0.08) pH = 3.77 © 2012 Pearson Education, Inc.

21. 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 pKa close to the desired pH. © 2012 Pearson Education, Inc.

22. Sample Exercise 17.4Preparing a Buffer How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00? (Assume that the addition of NH4Cl does not change the volume of the solution.) Solution Plan The major species in the solution will be NH4+, Cl–, and NH3. Of these, the Cl– ion is a spectator (it is the conjugate base of a strong acid). Thus, the NH4+–NH3 conjugate acid–base pair will determine the pH of the buffer. The equilibrium relationship between NH4+ and NH3 is given by the base-dissociation reaction for NH3: The key to this exercise is to use this Kb expression to calculate [NH4+]. pOH = 14.00 – pH = 14.00 – 9.00 = 5.00 [OH–] = 1.0  105M [NH3] = 0.10 M Solve We obtain [OH] from the given pH: and so Because Kb is small and the common ion [NH4+] is present, the equilibrium concentration of NH3 essentially equals its initial concentration:

23. Sample Exercise 17.4Preparing a Buffer Continued We now use the expression for Kb to calculate [NH4+]: Thus, for the solution to have pH = 9.00, [NH4+] must equal 0.18 M. The number of moles of NH4Cl needed to produce this concentration is given by the product of the volume of the solution and its molarity: (2.0 L)(0.18 mol NH4CI/L) = 0.36 mol NH4CI

24. 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.

25. Addition of Strong Acid or Base to a Buffer • Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. • Use the Henderson–Hasselbalch equation to determine the new pH of the solution. © 2012 Pearson Education, Inc.

26. Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added. © 2012 Pearson Education, Inc.

27. Calculating pH Changes in Buffers Before the reaction, since mol HC2H3O2 = mol C2H3O2 pH = 4.74 (given) © 2012 Pearson Education, Inc.

28. Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC2H3O2(aq) + OH(aq)  C2H3O2(aq) + H2O(l) © 2012 Pearson Education, Inc.

29. Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC2H3O2(aq) + OH(aq)  C2H3O2(aq) + H2O(l) © 2012 Pearson Education, Inc.

30. (0.320) (0.280) pH = 4.74 + log Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + 0.06 pH pH = 4.80 © 2012 Pearson Education, Inc.

31. 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.

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

33. Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly. © 2012 Pearson Education, Inc.

34. Titration of a Strong Acid with a Strong Base Just before (and after) the equivalence point, the pH increases rapidly. © 2012 Pearson Education, Inc.

35. Titration of a Strong Acid with a Strong Base 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. © 2012 Pearson Education, Inc.

36. Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off. © 2012 Pearson Education, Inc.

37. Sample Exercise 17.6Calculating 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. Solve The number of moles of H+ in the original HCl solution is given by the product of the volume of the solution and its molarity: Likewise, the number of moles of OH– in49.0 mL of 0.100 M NaOH is Because we have not reached the equivalence point, there are more moles of H+ present than OH–. Each mole of reacts OH–with 1 mol of H+. The volume of the reaction mixture increases as the NaOH solution is added to the HCl solution. Thus, at this point in the titration, the volume in the titration flask is 50.0 mL + 49.0 mL = 99.0 mL = 0.0990 L

38. Sample Exercise 17.6Calculating pH for a Strong Acid–Strong Base Titration Continued Thus, the concentration of H+(aq) in the flask is The corresponding pH is –log(1.0  103) = 3.00 Solve In this case the volume in the titration flask is Hence, the concentration of OH–(aq) in the flask is and we have 50.0 mL + 51.0 mL = 101.0 mL = 0.1010 L pOH = –log(1.0 103) = 3.00 pH = 14.00 – pOH = 14.00 – 3.00 = 11.00

39. 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. Titration of a Weak Acid with a Strong Base © 2012 Pearson Education, Inc.

40. 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.

41. 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.

42. Sample Exercise 17.7Calculating 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 CH3COOH (Ka = 1.8 105). Solution Analyze We are asked to calculate the pH before the equivalence point of the titration of a weak acid with a strong base. Plan We first must determine the number of moles of CH3COOH and CH3COO– present after the neutralization reaction. We then calculate pH using Ka, [CH3COOH], and [CH3COO–]. Solve Stoichiometry Calculation: The product of the volume and concentration of each solution gives the number of moles of each reactant present before the neutralization: The 4.50 103 of NaOH consumes 4.50 103 of CH3COOH: The total volume of the solution is 45.0 mL + 50.0 mL = 95.0 mL = 0.0950 L

43. Sample Exercise 17.7Calculating pH for a Weak Acid–Strong Base Titration Continued The resulting molarities of CH3COOH and CH3COO– after the reaction are therefore Equilibrium Calculation: The equilibrium between CH3COOH and CH3COO– must obey the equilibrium-constant expression for CH3COOH: Solving for [H+] gives Comment We could have solved for pH equally well using the Henderson–Hasselbalch equation.

44. Sample Exercise 17.8Calculating the pH at the Equivalence Point Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH. Analyze We are asked to determine the pH at the equivalence point of the titration of a weak acid with a strong base. Because the neutralization of a weak acid produces its anion, which is a weak base, we expect the pH at the equivalence point to be greater than 7. Plan The initial number of moles of acetic acid equals the number of moles of acetate ion at the equivalence point. We use the volume of the solution at the equivalence point to calculate the concentration of acetate ion. Because the acetate ion is a weak base, we can calculate the pH using Kb and [CH3COO–]. Solve The number of moles of acetic acid in the initial solution is obtained from the volume and molarity of the solution: Moles = M  L = (0.100 mol)/L(0.0500 L) = 5.00 103mol CH3COOH Hence, 5.00 103 mol of [CH3COO–] is formed. It will take 50.0 mL of NaOH to reach the equivalence point. The volume of this salt solution at the equivalence point is the sum of the volumes of the acid and base, 50.0 mL + 50.0 mL = 100.0 mL = 0.1000 L. Thus, the concentration of CH3COO– is The CH3COO– ion is a weak base:

45. Sample Exercise 17.8Calculating the pH at the Equivalence Point Continued The Kb for CH3COO– can be calculated from the Ka value of its conjugate acid, Kb = Kw/Ka = (1.01014)/(1.8105) = 5.61010. Using the Kb expression, we have Making the approximation that and then solving for x, we have x = [OH–] = 5.3 106M, which gives pOH = 5.28 pH = 8.72.

46. 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.

47. 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.

48. BaSO4(s) Ba2+(aq) + SO42(aq) Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO4 in water: © 2012 Pearson Education, Inc.

49. Solubility Products The equilibrium constant expression for this equilibrium is Ksp = [Ba2+] [SO42] where the equilibrium constant, Ksp, is called the solubility product. © 2012 Pearson Education, Inc.