1. 1 Chapter 15 Aqueous Equilibrium AP Chemistry Unit 12

2. The Common Ion Effect Consider a solution of acetic acid. If we add acetate ion as a second solute (i.e., sodium acetate), the pH of the solution increases: LeChâtelier’s principle: What happens to [H 3 O + ] when the equilibrium shifts to the left?

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.”

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. K a for HF is 6.8  10 −4. [H 3 O + ] [F − ] [HF] K a = = 6.8  10 -4

5. The Common-Ion Effect Because HCl, a strong acid, is also present, the initial [H 3 O + ] is not 0, but rather 0.10 M. [HF], M[H 3 O + ], M[F − ], M Initially0.200.100 Change−x−x+x+x+x+x At Equilibrium 0.20 − x  0.200.10 + x  0.10 x HF (aq) + H 2 O (l) H 3 O + (aq) + F − (aq)

6. The Common-Ion Effect = x 1.4  10 −3 = x (0.10) (x) (0.20) 6.8  10 −4 = (0.20) (6.8  10 −4 ) (0.10)

7. The Common-Ion Effect Therefore, [F − ] = x = 1.4  10 −3 [H 3 O + ] = 0.10 + x = 0.10 + 1.4  10 −3 = 0.10 M So,pH = −log (0.10) pH = 1.00

8. Example 15.16 Calculate the pH of an aqueous solution that is both 1.00 M CH 3 COOH and 1.00 M CH 3 COONa.

9. 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.

10. Buffer Solutions The acid component of the buffer neutralizes small added amounts of OH –, forming the weaker conjugate base which does not affect pH much: HA + OH –  H 2 O + A – The base component neutralizes small added amounts of H 3 O +, forming the weaker conjugate acid which does not affect pH much. A – + H 3 O +  H 2 O + HA Pure water does not buffer at all …

11. 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.

12. Buffers Similarly, if acid is added, the F − reacts with it to form HF and water.

13. 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 −

14. Buffer Calculations Rearranging slightly, this becomes [A − ] [HA] K a = [H 3 O + ] Taking the negative log of both sides, we get [A − ] [HA] −log K a = −log [H 3 O + ] + − log pKapKa pH acid base

15. Buffer Calculations So pK a = pH − log [base] [acid] Rearranging, this becomes pH = pK a + log [base] [acid] This is the Henderson–Hasselbalch equation. To use this equation, the ratio [conjugate base]/[weak acid] must have a value between 0.10–10 and both concentrations must exceed Ka by a factor of 100 or more.

16. Buffer Capacity and Buffer Range There is a limit to the ability of a buffer solution to neutralize added acid or base. This buffer capacity is reached before either buffer component has been consumed. In general, the more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize. As a rule, a buffer is most effective if the concentrations of the buffer acid and its conjugate base are equal or nearly so. Therefore, a buffer is most effective when the desired pH of the buffer is very near pK a of the weak acid of the buffer.

17. 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.

18. 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.

19. Example 15.17 A buffer solution is 0.24 M NH 3 and 0.20 M NH 4 Cl. (a) What is the pH of this buffer? (b) If 0.0050 mol NaOH is added to 0.500 L of this solution, what will be the pH? Example 15.18 What concentration of acetate ion in 0.500 M CH 3 COOH(aq) produces a buffer solution with pH = 5.00?

20. Acid–Base Indicators An acid–base indicator is a weak acid or base. The acid form (HA) of the indicator has one color, the conjugate base (A – ) has a different color. One of the “colors” may be colorless. In an acidic solution, [H 3 O + ] is high. Because H 3 O + is a common ion, it suppresses the ionization of the indicator acid, and we see the color of HA. In a basic solution, [OH – ] is high, and it reacts with HA, forming the color of A –. Acid–base indicators are often used for applications in which a precise pH reading isn’t necessary.

21. Different indicators have different values of K a, so they exhibit color changes at different values of pH

22. Example 15.19 A Conceptual Example Explain the series of color changes of thymol blue indicator produced by the actions pictured in Figure 15.14: (a) A few drops of thymol blue are added to HCl(aq). (b) Some sodium acetate is added to solution (a). (c) A small quantity of sodium hydroxide is added to solution (b). (d) An additional, larger quantity of sodium hydroxide is added to solution (c).

23. Neutralization Reactions At the equivalence point in an acid–base titration, the acid and base have been brought together in precise stoichiometric proportions. The endpoint is the point in the titration at which the indicator changes color. Ideally, the indicator is selected so that the endpoint and the equivalence point are very close together. The endpoint and the equivalence point for a neutralization titration can be best matched by plotting a titration curve, a graph of pH versus volume of titrant.

24. Titration Curve, Strong Acid with Strong Base Bromphenol blue, bromthymol blue, and phenolphthalein all change color at very nearly 20.0 mL At about what volume would we see a color change if we used methyl violet as the indicator?

25. Example 15.20 Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH: H 3 O + + Cl – + Na + + OH –  Na + + Cl – + 2 H 2 O (a) before the addition of any NaOH (b) after the addition of 10.00 mL of 0.500 M NaOH (c) after the addition of 20.00 mL of 0.500 M NaOH (d) after the addition of 20.20 mL of 0.500 M NaOH

26. Titration Curve, Weak Acid with Strong Base The equivalence-point pH is NOT 7.00 here. Why not?? Bromphenol blue was ok for the strong acid/strong base titration, but it changes color far too early to be useful here.

27. 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.

28. 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.

29. Example 15.21 Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH 3 COOH with 0.500 M NaOH: CH 3 COOH + Na + + OH –  Na + + CH 3 COO – + H 2 O (a) before the addition of any NaOH (b) after the addition of 8.00 mL of 0.500 M NaOH (c) after the addition of 10.00 mL of 0.500 M NaOH (d) after the addition of 20.00 mL of 0.100 M NaOH (e) after the addition of 21.00 mL of 0.100 M NaOH

30. Titration of a Weak Base with a Strong Acid The pH at the equivalence point in these titrations is < 7. Methyl red is the indicator of choice.

31. Titrations of Polyprotic Acids When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation.

32. Example 15.22 A Conceptual Example This titration curve shown in Figure 15.18 involves 1.0 M solutions of an acid and a base. Identify the type of titration it represents.

33. Solubility product constant, K sp : the equilibrium constant expression for the dissolving of a slightly soluble solid. The Solubility Product Constant, K sp K sp = [ Ba 2+ ][ SO 4 2– ] BaSO 4 (s) Ba 2+ (aq) + SO 4 2– (aq) Many important ionic compounds are only slightly soluble in water (we used to call them “insoluble”) An equation can represent the equilibrium between the compound and the ions present in a saturated aqueous solution:

35. Example 16.1 Write a solubility product constant expression for equilibrium in a saturated aqueous solution of the slightly soluble salts (a) iron(III) phosphate, FePO 4, and (b) chromium(III) hydroxide, Cr(OH) 3.

36. K sp is an equilibrium constant Molar solubility is the number of moles of compound that will dissolve per liter of solution. Molar solubility is related to the value of K sp, but molar solubility and K sp are not the same thing. In fact, “smaller K sp ” doesn’t always mean “lower molar solubility.” Solubility depends on both K sp and the form of the equilibrium constant expression. K sp and Molar Solubility

37. Example 16.2 At 20 °C, a saturated aqueous solution of silver carbonate contains 32 mg of Ag 2 CO 3 per liter of solution. Calculate K sp for Ag 2 CO 3 at 20 °C. The balanced equation is Ag 2 CO 3 (s) 2 Ag + (aq) + CO 3 2 – (aq) K sp = ? Example 16.3 From the K sp value for silver sulfate, calculate its molar solubility at 25 °C. Ag 2 SO 4 (s) 2 Ag + (aq) + SO 4 2 – (aq) K sp = 1.4 x 10 – 5 at 25 °C

38. 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)

39. Example 16.5 Calculate the molar solubility of Ag 2 SO 4 in 1.00 M Na 2 SO 4 (aq).

40. 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.

41. Example 16.11 What is the molar solubility of Mg(OH) 2 (s) in a buffer solution having [OH – ] = 1.0 x 10 – 5 M, that is, pH = 9.00? Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH – (aq) K sp = 1.8 x 10 –11 Example 16.12 A Conceptual Example Without doing detailed calculations, determine in which of the following solutions Mg(OH) 2 (s) is most soluble: (a) 1.00 M NH 3 (b) 1.00 M NH 3 /1.00 M NH 4 + (c) 1.00 M NH 4 Cl.

42. Factors Affecting Solubility Complex Ions –Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.

43. Factors Affecting Solubility Complex Ions –The formation of these complex ions increases the solubility of these salts.

44. 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+.

45. Q ip can then be compared to K sp. Precipitation should occur if Q ip > K sp. Precipitation cannot occur if Q ip < K sp. A solution is just saturated if Q ip = K sp. In applying the precipitation criteria, the effect of dilution when solutions are mixed must be considered. Q ip is the ion product reaction quotient and is based on initial conditions of the reaction. Q ip and Q c : new look, same great taste! Will Precipitation Occur?

46. Example 16.6 If 1.00 mg of Na 2 CrO 4 is added to 225 mL of 0.00015 M AgNO 3, will a precipitate form? Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2 – (aq) K sp = 1.1 x 10 – 12

47. Example 16.7 A Conceptual Example Pictured here is the result of adding a few drops of concentrated KI(aq) to a dilute solution of Pb(NO 3 ) 2. What is the solid that first appears? Explain why it then disappears. Example 16.8 If 0.100 L of 0.0015 M MgCl 2 and 0.200 L of 0.025 M NaF are mixed, should a precipitate of MgF 2 form? MgF 2 (s) Mg 2+ (aq) + 2 F – (aq) K sp = 3.7 x 10 –8

48. Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture.