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

2. Aqueous Equilibria 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. CH 3 COOH(aq) + H 2 O(l)H 3 O + (aq) + CH 3 COO  (aq) © 2012 Pearson Education, Inc.

3. Aqueous Equilibria 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. Aqueous Equilibria Exercise 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?

5. Aqueous Equilibria The Common-Ion Effect Exercise 2 : Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. K a for HF is 6.8  10  4. [H 3 O + ] [F  ] [HF] K a = = 6.8  10  4 © 2012 Pearson Education, Inc.

6. Aqueous Equilibria 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: CH 3 COOH and CH 3 COO - ion solution or NH 4 + ion and NH 3 solution or HF and F - ions solution. © 2012 Pearson Education, Inc.

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. © 2012 Pearson Education, Inc.

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

9. 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.

10. 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.

11. 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.

12. Aqueous Equilibria Henderson–Hasselbalch Equation Example 3 What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate? K a for lactic acid is 1.4  10  4. © 2012 Pearson Education, Inc.

13. 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.

14. Aqueous Equilibria Exercise 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.)

15. 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.

16. 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 the Henderson–Hasselbalch equation to determine the new pH of the solution. © 2012 Pearson Education, Inc.

17. Aqueous Equilibria Calculating pH Changes in Buffers Example 5 A buffer is made by adding 0.300 mol HC 2 H 3 O 2 and 0.300 mol NaC 2 H 3 O 2 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.

18. 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.

19. 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.

20. Aqueous Equilibria 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.

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

22. Aqueous Equilibria 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.

23. Aqueous Equilibria 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.

24. Aqueous Equilibria Exercise 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.

25. 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.

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27. 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.

28. 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.

29. Aqueous Equilibria Exercise 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 ).

30. Aqueous Equilibria Exercise 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.

31. 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.

32. 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.

33. Aqueous Equilibria Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s)Ba 2+ (aq) + SO 4 2  (aq) © 2012 Pearson Education, Inc.

34. Aqueous Equilibria Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2  ] where the equilibrium constant, K sp, is called the solubility product. © 2012 Pearson Education, Inc.

35. Aqueous Equilibria Solubility Products K sp is not the same as solubility. Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M). © 2012 Pearson Education, Inc.

36. Aqueous Equilibria Exercise 9 Calculating K sp from Solubility Solid silver chromate is added to pure water at 25  C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4 (s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3  10  4 M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving Ag + or CrO 4 2– ions in the solution, calculate K sp for this compound.

37. Aqueous Equilibria Exercise 10 Calculating Solubility from K sp The K sp for CaF 2 is 3.9  10  11 at 25  C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter.

38. Aqueous Equilibria Factors Affecting Solubility The Common-Ion Effect –If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease: BaSO 4 (s)Ba 2+ (aq) + SO 4 2  (aq) © 2012 Pearson Education, Inc.

39. Aqueous Equilibria Exercise 11 Calculating the Effect of a Common Ion on Solubility Calculate the molar solubility of CaF 2 at 25  C in a solution that is (a) 0.010 M in Ca(NO 3 ) 2, (b) 0.010 M in NaF.

40. Aqueous Equilibria Factors Affecting Solubility pH –If a substance has a basic anion, it will be more soluble in an acidic solution. –Substances with acidic cations are more soluble in basic solutions. © 2012 Pearson Education, Inc.

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42. Aqueous Equilibria Exercise 12 Predicting the Effect of Acid on Solubility Which of these substances are more soluble in acidic solution than in basic solution: (a) Ni(OH) 2 (s), (b) CaCO 3 (s), (c) BaF 2 (s), (d) AgCl(s)?

43. Aqueous Equilibria Factors Affecting Solubility Complex Ions –Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent. © 2012 Pearson Education, Inc.

44. Aqueous Equilibria Factors Affecting Solubility Complex Ions –The formation of these complex ions increases the solubility of these salts. © 2012 Pearson Education, Inc.

45. Aqueous Equilibria Factors Affecting Solubility Amphoterism –Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. –Examples of such cations are Al 3+, Zn 2+, and Sn 2+. © 2012 Pearson Education, Inc.

46. Aqueous Equilibria Will a Precipitate Form? In a solution, –If Q = K sp, the system is at equilibrium and the solution is saturated. –If Q < K sp, more solid can dissolve until Q = K sp. –If Q > K sp, the salt will precipitate until Q = K sp. © 2012 Pearson Education, Inc.

47. Aqueous Equilibria Exercise 13 Predicting Whether a Precipitate Forms Does a precipitate form when 0.10 L of 8.0  10  3 M Pb(NO 3 ) 2 is added to 0.40 L of 5.0  10  3 M Na 2 SO 4 ?

48. Aqueous Equilibria Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture. © 2012 Pearson Education, Inc.