1. Copyright  2011 Pearson Education, Inc. Chapter 16 Aqueous Ionic Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro

2. Copyright  2011 Pearson Education, Inc. ethylene glycol (aka 1,2–ethandiol) The Danger of Antifreeze Each year, thousands of pets and wildlife species die from consuming antifreeze Most brands of antifreeze contain ethylene glycol sweet taste initial effect drunkenness Metabolized in the liver to glycolic acid HOCH 2 COOH 2 glycolic acid (aka  -hydroxyethanoic acid) Tro: Chemistry: A Molecular Approach, 2/e

3. Copyright  2011 Pearson Education, Inc. Why Is Glycolic Acid Toxic? If present in high enough concentration in the bloodstream, glycolic acid overwhelms the buffering ability of the HCO 3 − in the blood, causing the blood pH to drop When the blood pH is low, its ability to carry O 2 is compromised acidosis HbH + (aq) + O 2 (g)  HbO 2 (aq) + H + (aq) One treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol 3Tro: Chemistry: A Molecular Approach, 2/e

4. Copyright  2011 Pearson Education, Inc. Buffers Buffers are solutions that resist changes in pH when an acid or base is added They act by neutralizing acid or base that is added to the buffered solution But just like everything else, there is a limit to what they can do, and eventually the pH changes Many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion blood has a mixture of H 2 CO 3 and HCO 3 − 4Tro: Chemistry: A Molecular Approach, 2/e

5. Copyright  2011 Pearson Education, Inc. Making an Acid Buffer 5Tro: Chemistry: A Molecular Approach, 2/e

6. Copyright  2011 Pearson Education, Inc. How Acid Buffers Work: Addition of Base HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) Buffers work by applying Le Châtelier’s Principle to weak acid equilibrium Buffer solutions contain significant amounts of the weak acid molecules, HA These molecules react with added base to neutralize it HA(aq) + OH − (aq) → A − (aq) + H 2 O(l) you can also think of the H 3 O + combining with the OH − to make H 2 O; the H 3 O + is then replaced by the shifting equilibrium 6Tro: Chemistry: A Molecular Approach, 2/e

7. Copyright  2011 Pearson Education, Inc. H2OH2O HA How Buffers Work HA  + H3O+H3O+ A−A− Added HO − new A − A−A− 7Tro: Chemistry: A Molecular Approach, 2/e

8. Copyright  2011 Pearson Education, Inc. How Acid Buffers Work: Addition of Acid HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) The buffer solution also contains significant amounts of the conjugate base anion, A − These ions combine with added acid to make more HA H + (aq) + A − (aq) → HA(aq) After the equilibrium shifts, the concentration of H 3 O + is kept constant 8Tro: Chemistry: A Molecular Approach, 2/e

9. Copyright  2011 Pearson Education, Inc. H2OH2O How Buffers Work HA  + H3O+H3O+ A−A− A−A− Added H 3 O + new HA 9Tro: Chemistry: A Molecular Approach, 2/e

10. Copyright  2011 Pearson Education, Inc. Common Ion Effect HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) Adding a salt containing the anion NaA, which is the conjugate base of the acid (the common ion), shifts the position of equilibrium to the left This causes the pH to be higher than the pH of the acid solution lowering the H 3 O + ion concentration 10Tro: Chemistry: A Molecular Approach, 2/e

11. Copyright  2011 Pearson Education, Inc. Common Ion Effect 11Tro: Chemistry: A Molecular Approach, 2/e

12. Copyright  2011 Pearson Education, Inc. Example 16.1: What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 ? 12 write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0 HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O + [HA][A − ][H 3 O + ] initial 0.100 ≈ 0 change equilibrium Tro: Chemistry: A Molecular Approach, 2/e

13. Copyright  2011 Pearson Education, Inc. [HA][A − ][H 3 O + ] initial 0.100 0 change equilibrium Example 16.1: What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 ? 13 represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x+x+x xx 0.100  x 0.100 + x x HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O + Tro: Chemistry: A Molecular Approach, 2/e

14. Copyright  2011 Pearson Education, Inc. Example 16.1: What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 ? 14 determine the value of K a because K a is very small, approximate the [HA] eq = [HA] init and [A − ] eq = [A − ] init, then solve for x [HA][A − ][H 3 O + ] initial 0.100 ≈ 0 change −x−x+x+x+x+x equilibrium 0.100 x 0.100  x 0.100 +x K a for HC 2 H 3 O 2 = 1.8 x 10 −5 Tro: Chemistry: A Molecular Approach, 2/e

15. Copyright  2011 Pearson Education, Inc. Example 16.1: What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 ? 15 check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init K a for HC 2 H 3 O 2 = 1.8 x 10 −5 the approximation is valid x = 1.8 x 10 −5 [HA][A − ][H 3 O + ] initial 0.100 ≈ 0 change −x−x+x+x+x+x equilibrium 0.100 x Tro: Chemistry: A Molecular Approach, 2/e

16. Copyright  2011 Pearson Education, Inc. Example 16.1: What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 ? 16 substitute x into the equilibrium concentration definitions and solve x = 1.8 x 10 −5 [HA][A − ][H 3 O + ] initial 0.100 ≈ 0 change −x−x+x+x+x+x equilibrium 0.100 1.8E-5 0.100 + x x 0.100  x Tro: Chemistry: A Molecular Approach, 2/e

17. Copyright  2011 Pearson Education, Inc. Example 16.1: What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 ? 17 substitute [H 3 O + ] into the formula for pH and solve [HA][A − ][H 3 O + ] initial 0.100 ≈ 0 change −x−x+x+x+x+x equilibrium 0.100 1.8E−5 Tro: Chemistry: A Molecular Approach, 2/e

18. Copyright  2011 Pearson Education, Inc. Example 16.1: What is the pH of a buffer that is 0.100 M HC 2 H 3 O 2 and 0.100 M NaC 2 H 3 O 2 ? 18 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values match [HA][A − ][H 3 O + ] initial 0.100 ≈ 0 change −x−x+x+x+x+x equilibrium 0.100 1.8E−5 K a for HC 2 H 3 O 2 = 1.8 x 10 −5 Tro: Chemistry: A Molecular Approach, 2/e

19. Copyright  2011 Pearson Education, Inc. Practice − What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 19Tro: Chemistry: A Molecular Approach, 2/e

20. Copyright  2011 Pearson Education, Inc. Practice − What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 20 write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0 HF + H 2 O  F  + H 3 O + [HA][A − ][H 3 O + ] initial0.140.071≈ 0 change equilibrium Tro: Chemistry: A Molecular Approach, 2/e

21. Copyright  2011 Pearson Education, Inc. [HA][A − ][H 3 O + ] initial 0.140.0710 change equilibrium Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 21 represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x+x+x xx 0.14  x 0.071 + x x HF + H 2 O  F  + H 3 O + Tro: Chemistry: A Molecular Approach, 2/e

22. Copyright  2011 Pearson Education, Inc. pK a for HF = 3.15K a for HF = 7.0 x 10 −4 Practice – What is the pH of a buffer that is 0.14 M and 0.071 M KF? 22 determine the value of K a because K a is very small, approximate the [HA] eq = [HA] init and [A − ] eq = [A − ] init solve for x [HA][A − ][H 3 O + ] initial 0.140.071≈ 0 change −x−x+x+x+x+x equilibrium 0.012 0.100 x 0.14  x 0.071 +x Tro: Chemistry: A Molecular Approach, 2/e

23. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 23 check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init K a for HF = 7.0 x 10 −4 the approximation is valid x = 1.4 x 10 −3 [HA][A − ][H 3 O + ] initial 0.140.071≈ 0 change −x−x+x+x+x+x equilibrium 0.140.071x Tro: Chemistry: A Molecular Approach, 2/e

24. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 24 substitute x into the equilibrium concentration definitions and solve x = 1.4 x 10 −3 [HA][A − ][H 3 O + ] initial 0.140.071≈ 0 change −x−x+x+x+x+x equilibrium 0.140.0721.4E-3 0.071 + x x 0.14  x Tro: Chemistry: A Molecular Approach, 2/e

25. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 25 substitute [H 3 O + ] into the formula for pH and solve [HA][A − ][H 3 O + ] initial 0.140.071≈ 0 change −x−x+x+x+x+x equilibrium 0.140.0721.4E−3 Tro: Chemistry: A Molecular Approach, 2/e

26. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 26 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values are close enough [HA][A − ][H 3 O + ] initial 0.140.071≈ 0 change −x−x+x+x+x+x equilibrium 0.140.0721.4E−3 K a for HF = 7.0 x 10 −4 Tro: Chemistry: A Molecular Approach, 2/e

27. Copyright  2011 Pearson Education, Inc. Henderson-Hasselbalch Equation Calculating the pH of a buffer solution can be simplified by using an equation derived from the K a expression called the Henderson- Hasselbalch Equation The equation calculates the pH of a buffer from the pK a and initial concentrations of the weak acid and salt of the conjugate base as long as the “x is small” approximation is valid 27Tro: Chemistry: A Molecular Approach, 2/e

28. Copyright  2011 Pearson Education, Inc. Deriving the Henderson-Hasselbalch Equation 28Tro: Chemistry: A Molecular Approach, 2/e

29. Copyright  2011 Pearson Education, Inc. Example 16.2: What is the pH of a buffer that is 0.050 M HC 7 H 5 O 2 and 0.150 M NaC 7 H 5 O 2 ? 29 assume the [HA] and [A − ] equilibrium concentrations are the same as the initial substitute into the Henderson- Hasselbalch Equation check the “x is small” approximation HC 7 H 5 O 2 + H 2 O  C 7 H 5 O 2  + H 3 O + K a for HC 7 H 5 O 2 = 6.5 x 10 −5 Tro: Chemistry: A Molecular Approach, 2/e

30. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 30Tro: Chemistry: A Molecular Approach, 2/e

31. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 31 find the pK a from the given K a assume the [HA] and [A − ] equilibrium concentrations are the same as the initial substitute into the Henderson- Hasselbalch equation check the “x is small” approximation HF + H 2 O  F  + H 3 O + Tro: Chemistry: A Molecular Approach, 2/e

32. Copyright  2011 Pearson Education, Inc. Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation? The Henderson-Hasselbalch equation is generally good enough when the “x is small” approximation is applicable Generally, the “x is small” approximation will work when both of the following are true: a)the initial concentrations of acid and salt are not very dilute b)the K a is fairly small For most problems, this means that the initial acid and salt concentrations should be over 100 to 1000x larger than the value of K a 32Tro: Chemistry: A Molecular Approach, 2/e

33. Copyright  2011 Pearson Education, Inc. How Much Does the pH of a Buffer Change When an Acid or Base Is Added? Though buffers do resist change in pH when acid or base is added to them, their pH does change Calculating the new pH after adding acid or base requires breaking the problem into two parts 1.a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other added acid reacts with the A − to make more HA added base reacts with the HA to make more A − 2.an equilibrium calculation of [H 3 O + ] using the new initial values of [HA] and [A − ] 33Tro: Chemistry: A Molecular Approach, 2/e

34. Copyright  2011 Pearson Education, Inc. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 34 If the added chemical is a base, write a reaction for OH − with HA. If the added chemical is an acid, write a reaction for H 3 O + with A −. construct a stoichiometry table for the reaction HC 2 H 3 O 2 + OH −  C 2 H 3 O 2  + H 2 O HAA−A− OH − mols before 0.100 0 mols added ──0.010 mols after Tro: Chemistry: A Molecular Approach, 2/e

35. Copyright  2011 Pearson Education, Inc. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 35 fill in the table – tracking the changes in the number of moles for each component HC 2 H 3 O 2 + OH −  C 2 H 3 O 2  + H 2 O HAA−A− OH − mols before 0.100 ≈ 0 mols added ──0.010 mols after 0.0900.110≈ 0 Tro: Chemistry: A Molecular Approach, 2/e

36. Copyright  2011 Pearson Education, Inc. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 36 If the added chemical is a base, write a reaction for OH − with HA. If the added chemical is an acid, write a reaction for it with A −. construct a stoichiometry table for the reaction enter the initial number of moles for each HC 2 H 3 O 2 + OH −  C 2 H 3 O 2  + H 2 O HAA–A– OH − mols before 0.100 0.010 mols change mols end new molarity Tro: Chemistry: A Molecular Approach, 2/e

37. Copyright  2011 Pearson Education, Inc. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 37 using the added chemical as the limiting reactant, determine how the moles of the other chemicals change add the change to the initial number of moles to find the moles after reaction divide by the liters of solution to find the new molarities HC 2 H 3 O 2 + OH −  C 2 H 3 O 2  + H 2 O HAA−A− OH − mols before 0.100 0.010 mols change mols end new molarity −0.010 +0.010 00.110 0.090 0.1100 Tro: Chemistry: A Molecular Approach, 2/e

38. Copyright  2011 Pearson Education, Inc. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 38 write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0, and using the new molarities of the [HA] and [A − ] HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O + [HA][A − ][H 3 O + ] initial 0.0900.110 ≈ 0≈ 0 change equilibrium Tro: Chemistry: A Molecular Approach, 2/e

39. Copyright  2011 Pearson Education, Inc. [HA][A − ][H 3 O + ] initial 0.0900.110≈ 0 change equilibrium Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 39 represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x+x+x xx 0.090  x 0.110 + x x HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O + Tro: Chemistry: A Molecular Approach, 2/e

40. Copyright  2011 Pearson Education, Inc. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 40 determine the value of K a because K a is very small, approximate the [HA] eq = [HA] init and [A − ] eq = [A − ] init solve for x [HA][A − ][H 3 O + ] initial 0.100 ≈ 0≈ 0 change −x−x+x+x+x+x equilibrium 0.0900.110 x 0.090  x 0.110 +x K a for HC 2 H 3 O 2 = 1.8 x 10 −5 Tro: Chemistry: A Molecular Approach, 2/e

41. Copyright  2011 Pearson Education, Inc. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 41 check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init K a for HC 2 H 3 O 2 = 1.8 x 10 −5 the approximation is valid x = 1.47 x 10 −5 [HA][A − ][H 3 O + ] initial 0.0900.110≈ 0 change −x−x+x+x+x+x equilibrium 0.0900.110 x Tro: Chemistry: A Molecular Approach, 2/e

42. Copyright  2011 Pearson Education, Inc. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 42 substitute x into the equilibrium concentration definitions and solve x = 1.47 x 10 −5 [HA][A - ][H 3 O + ] initial 0.0900.110 ≈ 0≈ 0 change −x−x+x+x+x+x equilibrium 0.0900.1101.5E-5 0.110 + x x 0.090  x Tro: Chemistry: A Molecular Approach, 2/e

43. Copyright  2011 Pearson Education, Inc. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 43 substitute [H 3 O + ] into the formula for pH and solve [HA][A − ][H 3 O + ] initial 0.0900.110 ≈ 0≈ 0 change −x−x+x+x+x+x equilibrium 0.0900.1101.5E−5 Tro: Chemistry: A Molecular Approach, 2/e

44. Copyright  2011 Pearson Education, Inc. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 44 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K a to the given K a the values match [HA][A − ][H 3 O + ] initial 0.0900.110 ≈ 0≈ 0 change −x−x+x+x+x+x equilibrium 0.0900.1101.5E−5 K a for HC 2 H 3 O 2 = 1.8 x 10 −5 Tro: Chemistry: A Molecular Approach, 2/e

45. Copyright  2011 Pearson Education, Inc. or, by using the Henderson-Hasselbalch Equation

46. Copyright  2011 Pearson Education, Inc. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 46 write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H 3 O + ] from water is ≈ 0, and using the new molarities of the [HA] and [A − ] fll in the ICE table HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O + [HA][A − ][H 3 O + ] initial 0.0900.110 ≈ 0≈ 0 change equilibrium +x+x+x+x xx 0.090  x 0.110 + x x Tro: Chemistry: A Molecular Approach, 2/e

47. Copyright  2011 Pearson Education, Inc. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 47 find the pK a from the given K a assume the [HA] and [A − ] equilibrium concentrations are the same as the initial HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O + K a for HC 2 H 3 O 2 = 1.8 x 10 −5 [HA][A − ][H 3 O + ] initial 0.0900.110≈ 0 change −x−x+x+x+x+x equilibrium 0.0900.110x Tro: Chemistry: A Molecular Approach, 2/e

48. Copyright  2011 Pearson Education, Inc. Example 16.3: What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? 48 substitute into the Henderson- Hasselbalch equation check the “x is small” approximation HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O + pK a for HC 2 H 3 O 2 = 4.745 Tro: Chemistry: A Molecular Approach, 2/e

49. Copyright  2011 Pearson Education, Inc. Example 16.3: Compare the effect on pH of adding 0.010 mol NaOH to a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L to adding 0.010 mol NaOH to 1.00 L of pure water 49 HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2  + H 3 O + pK a for HC 2 H 3 O 2 = 4.745 Tro: Chemistry: A Molecular Approach, 2/e

50. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a buffer that has 0.140 moles HF (pK a = 3.15) and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? (The “x is small” approximation is valid) 50Tro: Chemistry: A Molecular Approach, 2/e

51. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? 51 If the added chemical is a base, write a reaction for OH − with HA. If the added chemical is an acid, write a reaction for H 3 O + with A −. construct a stoichiometry table for the reaction F − + H 3 O +  HF + H 2 O F−F− H3O+H3O+ HF mols before 0.07100.140 mols added– 0.020 – mols after Tro: Chemistry: A Molecular Approach, 2/e

52. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? 52 fill in the table – tracking the changes in the number of moles for each component F − + H 3 O +  HF + H 2 O F−F− H3O+H3O+ HF mols before 0.07100.140 mols added– 0.020 – mols after 0.05100.160 Tro: Chemistry: A Molecular Approach, 2/e

53. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? 53 If the added chemical is a base, write a reaction for OH − with HA. If the added chemical is an acid, write a reaction for H 3 O + with A −. construct a stoichiometry table for the reaction enter the initial number of moles for each F − + H 3 O +  HF + H 2 O F−F− H3O+H3O+ HF mols before0.0710.0200.140 mols change mols after new molarity Tro: Chemistry: A Molecular Approach, 2/e

54. Copyright  2011 Pearson Education, Inc. F−F− H3O+H3O+ HF mols before0.0710.0200.140 mols change mols after new molarity Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? 54 using the added chemical as the limiting reactant, determine how the moles of the other chemicals change add the change to the initial number of moles to find the moles after reaction divide by the liters of solution to find the new molarities 0.16000.051 F − + H 3 O +  HF + H 2 O +0.020−0.020 0.16000.051 Tro: Chemistry: A Molecular Approach, 2/e

55. Copyright  2011 Pearson Education, Inc. Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is added? 55 write the reaction for the acid with water construct an ICE table. assume the [HA] and [A − ] equilibrium concentrations are the same as the initial substitute into the Henderson- Hasselbalch equation HF + H 2 O  F  + H 3 O + [HF][F − ][H 3 O + ] initial 0.1600.051≈ 0 change −x−x+x+x+x+x equilibrium 0.1600.051x Tro: Chemistry: A Molecular Approach, 2/e

56. Copyright  2011 Pearson Education, Inc. Basic Buffers B: (aq) + H 2 O (l)  H:B + (aq) + OH − (aq) Buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B + Cl − H 2 O (l) + NH 3 (aq)  NH 4 + (aq) + OH − (aq) 56Tro: Chemistry: A Molecular Approach, 2/e

57. Copyright  2011 Pearson Education, Inc. The Henderson-Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product B: + H 2 O  H:B + + OH − To apply the Henderson-Hasselbalch equation, the chemical equation of the basic buffer must be looked at like an acid reaction H:B + + H 2 O  B: + H 3 O + this does not affect the concentrations, just the way we are looking at the reaction Henderson-Hasselbalch Equation for Basic Buffers 57Tro: Chemistry: A Molecular Approach, 2/e

58. Copyright  2011 Pearson Education, Inc. Relationship between pK a and pK b Just as there is a relationship between the K a of a weak acid and K b of its conjugate base, there is also a relationship between the pK a of a weak acid and the pK b of its conjugate base K a  K b = K w = 1.0 x 10 −14 −log(K a  K b ) = −log(K w ) = 14 −log(K a ) + −log(K b ) = 14 pK a + pK b = 14 58Tro: Chemistry: A Molecular Approach, 2/e

59. Copyright  2011 Pearson Education, Inc. Example 16.4: What is the pH of a buffer that is 0.50 M NH 3 (pK b = 4.75) and 0.20 M NH 4 Cl? 59 find the pK a of the conjugate acid (NH 4 + ) from the given K b assume the [B] and [HB + ] equilibrium concentrations are the same as the initial substitute into the Henderson- Hasselbalch equation check the “x is small” approximation NH 3 + H 2 O  NH 4 + + OH − Tro: Chemistry: A Molecular Approach, 2/e

60. Copyright  2011 Pearson Education, Inc. The Henderson-Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product B: + H 2 O  H:B + + OH − We can rewrite the Henderson-Hasselbalch equation for the chemical equation of the basic buffer in terms of pOH Henderson-Hasselbalch Equation for Basic Buffers 60Tro: Chemistry: A Molecular Approach, 2/e

61. Copyright  2011 Pearson Education, Inc. Example 16.4: What is the pH of a buffer that is 0.50 M NH 3 (pK b = 4.75) and 0.20 M NH 4 Cl? 61 find the pK b if given K b assume the [B] and [HB + ] equilibrium concentrations are the same as the initial substitute into the Henderson-Hasselbalch equation base form, find pOH check the “x is small” approximation calculate pH from pOH NH 3 + H 2 O  NH 4 + + OH − Tro: Chemistry: A Molecular Approach, 2/e

62. Copyright  2011 Pearson Education, Inc. Buffering Effectiveness A good buffer should be able to neutralize moderate amounts of added acid or base However, there is a limit to how much can be added before the pH changes significantly The buffering capacity is the amount of acid or base a buffer can neutralize The buffering range is the pH range the buffer can be effective The effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base 62Tro: Chemistry: A Molecular Approach, 2/e

63. Copyright  2011 Pearson Education, Inc. HAA−A− OH − mols before0.180.0200 mols added ── 0.010 mols after 0.170.030≈ 0 Effect of Relative Amounts of Acid and Conjugate Base Buffer 1 0.100 mol HA & 0.100 mol A − Initial pH = 5.00 Buffer 2 0.18 mol HA & 0.020 mol A − Initial pH = 4.05 pK a (HA) = 5.00 after adding 0.010 mol NaOH pH = 5.09 HA + OH −  A  + H 2 O HAA−A− OH − mols before0.100 0 mols added ── 0.010 mols after 0.0900.110≈ 0 after adding 0.010 mol NaOH pH = 4.25 A buffer is most effective with equal concentrations of acid and base

64. Copyright  2011 Pearson Education, Inc. HAA−A− OH − mols before0.500.5000 mols added ── 0.010 mols after 0.490.51≈ 0 HAA−A− OH − mols before0.050 0 mols added ── 0.010 mols after 0.0400.060≈ 0 Effect of Absolute Concentrations of Acid and Conjugate Base Buffer 1 0.50 mol HA & 0.50 mol A − Initial pH = 5.00 Buffer 2 0.050 mol HA & 0.050 mol A − Initial pH = 5.00 pK a (HA) = 5.00 after adding 0.010 mol NaOH pH = 5.02 HA + OH −  A  + H 2 O after adding 0.010 mol NaOH pH = 5.18 A buffer is most effective when the concentrations of acid and base are largest

65. Copyright  2011 Pearson Education, Inc. Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer 65Tro: Chemistry: A Molecular Approach, 2/e

66. Copyright  2011 Pearson Education, Inc. Effectiveness of Buffers A buffer will be most effective when the [base]:[acid] = 1 equal concentrations of acid and base A buffer will be effective when 0.1 < [base]:[acid] < 10 A buffer will be most effective when the [acid] and the [base] are large 66Tro: Chemistry: A Molecular Approach, 2/e

67. Copyright  2011 Pearson Education, Inc. Buffering Range We have said that a buffer will be effective when 0.1 < [base]:[acid] < 10 Substituting into the Henderson-Hasselbalch equation we can calculate the maximum and minimum pH at which the buffer will be effective Lowest pHHighest pH Therefore, the effective pH range of a buffer is pK a ± 1 When choosing an acid to make a buffer, choose one whose is pK a closest to the pH of the buffer 67Tro: Chemistry: A Molecular Approach, 2/e

68. Copyright  2011 Pearson Education, Inc. Formic acid, HCHO 2 pK a = 3.74 Example 16.5a: Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25? Chlorous acid, HClO 2 pK a = 1.95 Nitrous acid, HNO 2 pK a = 3.34 Hypochlorous acid, HClOpK a = 7.54 The pK a of HCHO 2 is closest to the desired pH of the buffer, so it would give the most effective buffering range 68Tro: Chemistry: A Molecular Approach, 2/e

69. Copyright  2011 Pearson Education, Inc. Example 16.5b: What ratio of NaCHO 2 : HCHO 2 would be required to make a buffer with pH 4.25? 69 Formic acid, HCHO 2, pK a = 3.74 To make a buffer with pH 4.25, you would use 3.24 times as much NaCHO 2 as HCHO 2 Tro: Chemistry: A Molecular Approach, 2/e

70. Copyright  2011 Pearson Education, Inc. Practice – What ratio of NaC 7 H 5 O 2 : HC 7 H 5 O 2 would be required to make a buffer with pH 3.75? Benzoic acid, HC 7 H 5 O 2, pK a = 4.19 70Tro: Chemistry: A Molecular Approach, 2/e

71. Copyright  2011 Pearson Education, Inc. Practice – What ratio of NaC 7 H 5 O 2 : HC 7 H 5 O 2 would be required to make a buffer with pH 3.75? Benzoic acid, HC 7 H 5 O 2, pK a = 4.19 to make a buffer with pH 3.75, you would use 0.363 times as much NaC 7 H 5 O 2 as HC 7 H 5 O 2 71Tro: Chemistry: A Molecular Approach, 2/e

72. Copyright  2011 Pearson Education, Inc. Buffering Capacity Buffering capacity is the amount of acid or base that can be added to a buffer without causing a large change in pH The buffering capacity increases with increasing absolute concentration of the buffer components As the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves Buffers that need to work mainly with added acid generally have [base] > [acid] Buffers that need to work mainly with added base generally have [acid] > [base] 72Tro: Chemistry: A Molecular Approach, 2/e

73. Copyright  2011 Pearson Education, Inc. Titration In an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete when the reaction is complete we have reached the endpoint of the titration An indicator may be added to determine the endpoint an indicator is a chemical that changes color when the pH changes When the moles of H 3 O + = moles of OH −, the titration has reached its equivalence point 73Tro: Chemistry: A Molecular Approach, 2/e

74. Copyright  2011 Pearson Education, Inc. Titration 74Tro: Chemistry: A Molecular Approach, 2/e

75. Copyright  2011 Pearson Education, Inc. Titration Curve A plot of pH vs. amount of added titrant The inflection point of the curve is the equivalence point of the titration Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH The pH of the equivalence point depends on the pH of the salt solution equivalence point of neutral salt, pH = 7 equivalence point of acidic salt, pH < 7 equivalence point of basic salt, pH > 7 Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH 75Tro: Chemistry: A Molecular Approach, 2/e

76. Copyright  2011 Pearson Education, Inc. Titration Curve: Unknown Strong Base Added to Strong Acid 76Tro: Chemistry: A Molecular Approach, 2/e

77. Copyright  2011 Pearson Education, Inc. Before Equivalence (excess acid) After Equivalence (excess base) Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH Equivalence Point equal moles of HCl and NaOH pH = 7.00  Because the solutions are equal concentration, and 1:1 stoichiometry, the equivalence point is at equal volumes 77Tro: Chemistry: A Molecular Approach, 2/e

78. Copyright  2011 Pearson Education, Inc. HClNaClNaOH mols before2.50E-305.0E-4 mols change −5.0E-4+5.0E-4−5.0E-4 mols end 2.00E-35.0E-40 molarity, new 0.06670.0170 HClNaClNaOH mols before2.50E-305.0E-4 mols change mols end molarity, new HClNaClNaOH mols before2.50E-305.0E-4 mols change −5.0E-4+5.0E-4−5.0E-4 mols end 2.00E-35.0E-40 molarity, new 5.0 x 10 −4 mole NaOH added Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 Before equivalence point added 5.0 mL NaOH 78Tro: Chemistry: A Molecular Approach, 2/e

79. Copyright  2011 Pearson Education, Inc. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (aq) To reach equivalence, the added moles NaOH = initial moles of HCl = 2.50 x 10 −3 moles At equivalence, we have 0.00 mol HCl and 0.00 mol NaOH left over Because the NaCl is a neutral salt, the pH at equivalence = 7.00 79Tro: Chemistry: A Molecular Approach, 2/e

80. Copyright  2011 Pearson Education, Inc. HClNaClNaOH mols before2.50E-302.5E-3 mols change mols end molarity, new HClNaClNaOH mols before2.50E-302.5E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-30 molarity, new HClNaClNaOH mols before2.50E-302.5E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-30 molarity, new 00.0500 2.5 x 10 −3 mole NaOH added Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 At equivalence point added 25.0 mL NaOH 80Tro: Chemistry: A Molecular Approach, 2/e

81. Copyright  2011 Pearson Education, Inc. HClNaClNaOH mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new 00.0450.0091 HClNaClNaOH mols before2.50E-303.0E-3 mols change mols end molarity, new HClNaClNaOH mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 After equivalence point added 30.0 mL NaOH 3.0 x 10 −3 mole NaOH added

82. Copyright  2011 Pearson Education, Inc. added 5.0 mL NaOH 0.00200 mol HCl pH = 1.18 added 10.0 mL NaOH 0.00150 mol HCl pH = 1.37 added 25.0 mL NaOH equivalence point pH = 7.00 added 30.0 mL NaOH 0.00050 mol NaOH pH = 11.96 added 40.0 mL NaOH 0.00150 mol NaOH pH = 12.36 added 50.0 mL NaOH 0.00250 mol NaOH pH = 12.52 added 35.0 mL NaOH 0.00100 mol NaOH pH = 12.22 added 15.0 mL NaOH 0.00100 mol HCl pH = 1.60 added 20.0 mL NaOH 0.00050 mol HCl pH = 1.95 Adding 0.100 M NaOH to 0.100 M HCl 82 25.0 mL 0.100 M HCl 0.00250 mol HCl pH = 1.00 Tro: Chemistry: A Molecular Approach, 2/e

83. Copyright  2011 Pearson Education, Inc. Practice – Calculate the pH of the solution that results when 10.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO 3 83Tro: Chemistry: A Molecular Approach, 2/e

84. Copyright  2011 Pearson Education, Inc. HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-3 mols change −1.5E-3+1.5E-3−1.5E-3 mols end 1.1E-31.5E-30 molarity, new HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-3 mols change −1.5E-3+1.5E-3−1.5E-3 mols end 1.1E-31.5E-30 molarity, new 0.0180.0250 Practice – Calculate the pH of the solution that results when 10.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO 3 HNO 3 (aq) + NaOH(aq)  NaNO 3 (aq) + H 2 O(l) Initial pH = −log(0.250) = 0.60 Initial mol of HNO 3 = 0.0500 L x 0.25 mol/L=1.25 x 10 −2 Before equivalence point added 10.0 mL NaOH 84Tro: Chemistry: A Molecular Approach, 2/e

85. Copyright  2011 Pearson Education, Inc. Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M HNO 3 to reach equivalence 85Tro: Chemistry: A Molecular Approach, 2/e

86. Copyright  2011 Pearson Education, Inc. Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M HNO 3 to reach equivalence HNO 3 (aq) + NaOH(aq)  NaNO 3 (aq) + H 2 O(l) Initial pH = −log(0.250) = 0.60 Initial mol of HNO 3 = 0.0500 L x 0.25 mol/L=1.25 x 10 −2 At equivalence point: moles of NaOH = 1.25 x 10 −2 86Tro: Chemistry: A Molecular Approach, 2/e

87. Copyright  2011 Pearson Education, Inc. Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO 3 87Tro: Chemistry: A Molecular Approach, 2/e

88. Copyright  2011 Pearson Education, Inc. HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-2 mols change −1.25E-2+1.25E-2−1.25E-2 mols end 01.25E-20.0025 molarity, new 00.08330.017 HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-2 mols change −1.25E-2+1.25E-2−1.25E-2 mols end 01.25E-20.0025 molarity, new Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO 3 HNO 3 (aq) + NaOH(aq)  NaNO 3 (aq) + H 2 O(l) Initial pH = −log(0.250) = 0.60 Initial mol of HNO 3 = 0.0500 L x 0.25 mol/L=1.25 x 10 −2 After equivalence point added 100.0 mL NaOH

89. Copyright  2011 Pearson Education, Inc. HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-2 mols change −1.25E-2+1.25E-2−1.25E-2 mols end 01.25E-20.0025 molarity, new 00.08330.017 HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-2 mols change −1.25E-2+1.25E-2−1.25E-2 mols end 01.25E-20.0025 molarity, new Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO 3 HNO 3 (aq) + NaOH(aq)  NaNO 3 (aq) + H 2 O(l) Initial pH = −log(0.250) = 0.60 initial mol of HNO 3 = 0.0500 L x 0.25 mol/L=1.25 x 10 −2 After equivalence point added 100.0 mL NaOH 89Tro: Chemistry: A Molecular Approach, 2/e

90. Copyright  2011 Pearson Education, Inc. Titration of a Strong Base with a Strong Acid If the titration is run so that the acid is in the burette and the base is in the flask, the titration curve will be the reflection of the one just shown 90Tro: Chemistry: A Molecular Approach, 2/e

91. Copyright  2011 Pearson Education, Inc. Titration of a Weak Acid with a Strong Base Titrating a weak acid with a strong base results in differences in the titration curve at the equivalence point and excess acid region The initial pH is determined using the K a of the weak acid The pH in the excess acid region is determined as you would determine the pH of a buffer The pH at the equivalence point is determined using the K b of the conjugate base of the weak acid The pH after equivalence is dominated by the excess strong base the basicity from the conjugate base anion is negligible 91Tro: Chemistry: A Molecular Approach, 2/e

92. Copyright  2011 Pearson Education, Inc. Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH HCHO 2(aq) + NaOH (aq)  NaCHO 2(aq) + H 2 O (aq) Initial pH [HCHO 2 ][CHO 2 − ][H 3 O + ] initial 0.1000.000≈ 0 change −x−x+x+x+x+x equilibrium 0.100 − xxx K a = 1.8 x 10 −4 92Tro: Chemistry: A Molecular Approach, 2/e

93. Copyright  2011 Pearson Education, Inc. Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq) Initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 Before equivalence added 5.0 mL NaOH HAA−A− OH − mols before2.50E-300 mols added ––5.0E-4 mols after 2.00E-35.0E-4≈ 0 93Tro: Chemistry: A Molecular Approach, 2/e

94. Copyright  2011 Pearson Education, Inc. HAA−A− OH − mols before2.50E-300 mols added ––2.50E-3 mols after 02.50E-3≈ 0 [HCHO 2 ][CHO 2 − ][OH − ] initial 00.0500≈ 0 change +x+x−x−x+x+x equilibrium x5.00E-2-xx Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH added 25.0 mL NaOH CHO 2 − (aq) + H 2 O (l)  HCHO 2(aq) + OH − (aq) K b = 5.6 x 10 −11 [OH − ] = 1.7 x 10 −6 M HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq) Initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 At equivalence

95. Copyright  2011 Pearson Education, Inc. HAA−A− NaOH mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new HAA−A− NaOH mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new 00.0450.0091 HAA−A− NaOH mols before2.50E-303.0E-3 mols change mols end molarity, new Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH 95 added 30.0 mL NaOH 3.0 x 10 −3 mole NaOH added HCHO 2(aq) + NaOH (aq)  NaCHO 2(aq) + H 2 O (aq) Initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 After equivalence

96. Copyright  2011 Pearson Education, Inc. added 35.0 mL NaOH 0.00100 mol NaOH xs pH = 12.22 initial HCHO 2 solution 0.00250 mol HCHO 2 pH = 2.37 added 5.0 mL NaOH 0.00200 mol HCHO 2 pH = 3.14 added 10.0 mL NaOH 0.00150 mol HCHO 2 pH = 3.56 added 25.0 mL NaOH equivalence point 0.00250 mol CHO 2 − [CHO 2 − ] init = 0.0500 M [OH − ] eq = 1.7 x 10 −6 pH = 8.23 added 30.0 mL NaOH 0.00050 mol NaOH xs pH = 11.96 added 20.0 mL NaOH 0.00050 mol HCHO 2 pH = 4.34 added 15.0 mL NaOH 0.00100 mol HCHO 2 pH = 3.92 added 12.5 mL NaOH 0.00125 mol HCHO 2 pH = 3.74 = pK a half-neutralization Adding NaOH to HCHO 2 added 40.0 mL NaOH 0.00150 mol NaOH xs pH = 12.36 added 50.0 mL NaOH 0.00250 mol NaOH xs pH = 12.52 96Tro: Chemistry: A Molecular Approach, 2/e

97. Copyright  2011 Pearson Education, Inc. Titrating Weak Acid with a Strong Base The initial pH is that of the weak acid solution calculate like a weak acid equilibrium problem  e.g., 15.5 and 15.6 Before the equivalence point, the solution becomes a buffer calculate mol HA init and mol A − init using reaction stoichiometry calculate pH with Henderson-Hasselbalch using mol HA init and mol A − init Half-neutralization pH = pK a 97Tro: Chemistry: A Molecular Approach, 2/e

98. Copyright  2011 Pearson Education, Inc. Titrating Weak Acid with a Strong Base At the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established mol A − = original mole HA  calculate the volume of added base as you did in Example 4.8 [A − ] init = mol A − /total liters calculate like a weak base equilibrium problem  e.g., 15.14 Beyond equivalence point, the OH is in excess [OH − ] = mol MOH xs/total liters [H 3 O + ][OH − ]=1 x 10 −14 98Tro: Chemistry: A Molecular Approach, 2/e

99. Copyright  2011 Pearson Education, Inc. Example 16.7a: A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the volume of KOH at the equivalence point. 99 write an equation for the reaction for B with HA use stoichiometry to determine the volume of added B HNO 2 + KOH  NO 2  + H 2 O Tro: Chemistry: A Molecular Approach, 2/e

100. Copyright  2011 Pearson Education, Inc. Example 16.7b: A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH. 100 write an equation for the reaction for B with HA determine the moles of HA before & moles of added B make a stoichiometry table and determine the moles of HA in excess and moles A  made HNO 2 + KOH  NO 2  + H 2 O HNO 2 NO 2 − OH − mols before 0.004000≈ 0 mols added –– 0.00100 mols after ≈ 0≈ 0 0.00300 0.00100 Tro: Chemistry: A Molecular Approach, 2/e

101. Copyright  2011 Pearson Education, Inc. Example 16.7b: A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH. 101 write an equation for the reaction of HA with H 2 O determine K a and pK a for HA use the Henderson- Hasselbalch equation to determine the pH HNO 2 + H 2 O  NO 2  + H 3 O + HNO 2 NO 2 − OH − mols before 0.004000≈ 0 mols added –– 0.00100 mols after 0.003000.00100 ≈ 0≈ 0 Table 15.5 K a = 4.6 x 10 −4 Tro: Chemistry: A Molecular Approach, 2/e

102. Copyright  2011 Pearson Education, Inc. Example 16.7b: A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point. 102 write an equation for the reaction for B with HA determine the moles of HA before & moles of added B make a stoichiometry table and determine the moles of HA in excess and moles A  made HNO 2 + KOH  NO 2  + H 2 O HNO 2 NO 2 − OH − mols before 0.004000≈ 0 mols added –– 0.00200 mols after ≈ 0≈ 0 0.00200 at half-equivalence, moles KOH = ½ mole HNO 2 Tro: Chemistry: A Molecular Approach, 2/e

103. Copyright  2011 Pearson Education, Inc. Example 16.7b: A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point. 103 write an equation for the reaction of HA with H 2 O determine K a and pK a for HA use the Henderson- Hasselbalch equation to determine the pH HNO 2 + H 2 O  NO 2  + H 3 O + HNO 2 NO 2 − OH − mols before 0.004000≈ 0 mols added –– 0.00200 mols after 0.00200 ≈ 0≈ 0 Table 15.5 K a = 4.6 x 10 -4 Tro: Chemistry: A Molecular Approach, 2/e

104. Copyright  2011 Pearson Education, Inc. Titration Curve of a Weak Base with a Strong Acid 104Tro: Chemistry: A Molecular Approach, 2/e

105. Copyright  2011 Pearson Education, Inc. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the initial pH of the NH 3 solution. 105Tro: Chemistry: A Molecular Approach, 2/e

106. Copyright  2011 Pearson Education, Inc. Practice – Titration of 25.0 mL of 0.10 M NH 3 with 0.10 M HCl. Calculate the initial pH of the NH 3 (aq) 106 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial: NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH − (aq) [HCl][NH 4 + ][NH 3 ] initial 000.10 change +x+x+x+x−x−x equilibrium xx0.10−x pK b = 4.75 K b = 10 −4.75 = 1.8 x 10 −5 Tro: Chemistry: A Molecular Approach, 2/e

107. Copyright  2011 Pearson Education, Inc. Practice – Titration of 25.0 mL of 0.10 M NH 3 with 0.10 M HCl. Calculate the initial pH of the NH 3 (aq) 107 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial: NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH − (aq) [HCl][NH 4 + ][NH 3 ] initial 000.10 change +x+x+x+x−x−x equilibrium xx0.10−x pK b = 4.75 K b = 10 −4.75 = 1.8 x 10 −5 Tro: Chemistry: A Molecular Approach, 2/e

108. Copyright  2011 Pearson Education, Inc. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl. 108Tro: Chemistry: A Molecular Approach, 2/e

109. Copyright  2011 Pearson Education, Inc. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl. 109 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 Before equivalence: after adding 5.0 mL of HCl NH 3 NH 4 ClHCl mols before2.50E-305.0E-4 mols change −5.0E-4 mols end 2.00E-35.0E-40 molarity, new 0.06670.0170 NH 4 + (aq) + H 2 O (l)  NH 4 + (aq) + H 2 O (l) pK b = 4.75 pK a = 14.00 − 4.75 = 9.25 Tro: Chemistry: A Molecular Approach, 2/e

110. Copyright  2011 Pearson Education, Inc. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl. NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 Before equivalence: after adding 5.0 mL of HCl NH 3 NH 4 ClHCl mols before2.50E-305.0E-4 mols change −5.0E-4 mols end 2.00E-35.0E-40 molarity, new 0.06670.0170 110Tro: Chemistry: A Molecular Approach, 2/e

111. Copyright  2011 Pearson Education, Inc. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence. 111Tro: Chemistry: A Molecular Approach, 2/e

112. Copyright  2011 Pearson Education, Inc. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence. 112 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 At equivalence mol NH 3 = mol HCl = 2.50 x 10 −3 added 25.0 mL HCl NH 3 NH 4 ClHCl mols before2.50E-302.5E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-30 molarity, new 00.0500 Tro: Chemistry: A Molecular Approach, 2/e

113. Copyright  2011 Pearson Education, Inc. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence. 113 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) at equivalence [NH 4 Cl] = 0.050 M [NH 3 ][NH 4 + ][H 3 O + ] initial 00.050≈ 0 change +x+x−x−x+x+x equilibrium x0.050−xx NH 4 + (aq) + H 2 O (l)  NH 3(aq) + H 3 O + (aq) Tro: Chemistry: A Molecular Approach, 2/e

114. Copyright  2011 Pearson Education, Inc. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 30.0 mL of HCl. 114Tro: Chemistry: A Molecular Approach, 2/e

115. Copyright  2011 Pearson Education, Inc. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 30.0 mL of HCl. 115 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 After equivalence: after adding 30.0 mL HCl NH 3 NH 4 ClHCl mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new 00.0450.0091 when you mix a strong acid, HCl, with a weak acid, NH 4 +, you only need to consider the strong acid Tro: Chemistry: A Molecular Approach, 2/e

116. Copyright  2011 Pearson Education, Inc. Titration of a Polyprotic Acid If K a1 >> K a2, there will be two equivalence points in the titration the closer the K a ’s are to each other, the less distinguishable the equivalence points are titration of 25.0 mL of 0.100 M H 2 SO 3 with 0.100 M NaOH 116Tro: Chemistry: A Molecular Approach, 2/e

117. Copyright  2011 Pearson Education, Inc. Monitoring pH During a Titration The general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H 3 O + ] using a probe that specifically measures just H 3 O + The endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve If you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator 117Tro: Chemistry: A Molecular Approach, 2/e

118. Copyright  2011 Pearson Education, Inc. Monitoring pH During a Titration 118Tro: Chemistry: A Molecular Approach, 2/e

119. Copyright  2011 Pearson Education, Inc. Indicators Many dyes change color depending on the pH of the solution These dyes are weak acids, establishing an equilibrium with the H 2 O and H 3 O + in the solution HInd (aq) + H 2 O (l)  Ind  (aq) + H 3 O + (aq) The color of the solution depends on the relative concentrations of Ind  :HInd when Ind  :HInd ≈ 1, the color will be mix of the colors of Ind  and HInd when Ind  :HInd > 10, the color will be mix of the colors of Ind  when Ind  :HInd < 0.1, the color will be mix of the colors of HInd 119Tro: Chemistry: A Molecular Approach, 2/e

120. Copyright  2011 Pearson Education, Inc. Phenolphthalein 120Tro: Chemistry: A Molecular Approach, 2/e

121. Copyright  2011 Pearson Education, Inc. Methyl Red 121Tro: Chemistry: A Molecular Approach, 2/e

122. Copyright  2011 Pearson Education, Inc. Monitoring a Titration with an Indicator For most titrations, the titration curve shows a very large change in pH for very small additions of titrant near the equivalence point An indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH pK a of HInd ≈ pH at equivalence point 122Tro: Chemistry: A Molecular Approach, 2/e

123. Copyright  2011 Pearson Education, Inc. Acid-Base Indicators 123Tro: Chemistry: A Molecular Approach, 2/e

124. Copyright  2011 Pearson Education, Inc. Solubility Equilibria All ionic compounds dissolve in water to some degree however, many compounds have such low solubility in water that we classify them as insoluble We can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water 124Tro: Chemistry: A Molecular Approach, 2/e

125. Copyright  2011 Pearson Education, Inc. Solubility Product The equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, K sp For an ionic solid M n X m, the dissociation reaction is: M n X m (s)  nM m+ (aq) + mX n− (aq) The solubility product would be K sp = [M m+ ] n [X n− ] m For example, the dissociation reaction for PbCl 2 is PbCl 2 (s)  Pb 2+ (aq) + 2 Cl − (aq) And its equilibrium constant is K sp = [Pb 2+ ][Cl − ] 2 125Tro: Chemistry: A Molecular Approach, 2/e

126. Copyright  2011 Pearson Education, Inc. 126Tro: Chemistry: A Molecular Approach, 2/e

127. Copyright  2011 Pearson Education, Inc. Molar Solubility Solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature The molar solubility is the number of moles of solute that will dissolve in a liter of solution the molarity of the dissolved solute in a saturated solution for the general reaction M n X m (s)  nM m+ (aq) + mX n− (aq) 127Tro: Chemistry: A Molecular Approach, 2/e

128. Copyright  2011 Pearson Education, Inc. Example 16.8: Calculate the molar solubility of PbCl 2 in pure water at 25  C 128 write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid [Pb 2+ ][Cl − ] Initial00 Change+S+2S EquilibriumS2S PbCl 2 (s)  Pb 2+ (aq) + 2 Cl − (aq) K sp = [Pb 2+ ][Cl − ] 2 Tro: Chemistry: A Molecular Approach, 2/e

129. Copyright  2011 Pearson Education, Inc. Example 16.8: Calculate the molar solubility of PbCl 2 in pure water at 25  C 129 substitute into the K sp expression find the value of K sp from Table 16.2, plug into the equation, and solve for S [Pb 2+ ][Cl − ] Initial00 Change+S+2S EquilibriumS2S K sp = [Pb 2+ ][Cl − ] 2 K sp = (S)(2S) 2 Tro: Chemistry: A Molecular Approach, 2/e

130. Copyright  2011 Pearson Education, Inc. Practice – Determine the K sp of PbBr 2 if its molar solubility in water at 25  C is 1.05 x 10 −2 M 130Tro: Chemistry: A Molecular Approach, 2/e

131. Copyright  2011 Pearson Education, Inc. Practice – Determine the K sp of PbBr 2 if its molar solubility in water at 25  C is 1.05 x 10 −2 M 131 write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid [Pb 2+ ][Br − ] initial00 change+(1.05 x 10 −2 )+2(1.05 x 10 −2 ) equilibrium(1.05 x 10 −2 )(2.10 x 10 −2 ) PbBr 2 (s)  Pb 2+ (aq) + 2 Br − (aq) K sp = [Pb 2+ ][Br − ] 2 Tro: Chemistry: A Molecular Approach, 2/e

132. Copyright  2011 Pearson Education, Inc. Practice – Determine the K sp of PbBr 2 if its molar solubility in water at 25  C is 1.05 x 10 −2 M 132 substitute into the K sp expression plug into the equation and solve K sp = [Pb 2+ ][Br − ] 2 K sp = (1.05 x 10 −2 )(2.10 x 10 −2 ) 2 [Pb 2+ ][Br − ] initial00 change+(1.05 x 10 −2 )+2(1.05 x 10 −2 ) equilibrium(1.05 x 10 −2 )(2.10 x 10 −2 ) Tro: Chemistry: A Molecular Approach, 2/e

133. Copyright  2011 Pearson Education, Inc. K sp and Relative Solubility Molar solubility is related to K sp But you cannot always compare solubilities of compounds by comparing their K sp s To compare K sp s, the compounds must have the same dissociation stoichiometry 133Tro: Chemistry: A Molecular Approach, 2/e

134. Copyright  2011 Pearson Education, Inc. The Effect of Common Ion on Solubility Addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt For example, addition of NaCl to the solubility equilibrium of solid PbCl 2 decreases the solubility of PbCl 2 PbCl 2 (s)  Pb 2+ (aq) + 2 Cl − (aq) addition of Cl − shifts the equilibrium to the left 134Tro: Chemistry: A Molecular Approach, 2/e

135. Copyright  2011 Pearson Education, Inc. Example 16.10: Calculate the molar solubility of CaF 2 in 0.100 M NaF at 25  C 135 write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid [Ca 2+ ][F − ] initial00.100 change+S+2S equilibriumS0.100 + 2S CaF 2 (s)  Ca 2+ (aq) + 2 F − (aq) K sp = [Ca 2+ ][F − ] 2 Tro: Chemistry: A Molecular Approach, 2/e

136. Copyright  2011 Pearson Education, Inc. Example 16.10: Calculate the molar solubility of CaF 2 in 0.100 M NaF at 25  C 136 substitute into the K sp expression, assume S is small find the value of K sp from Table 16.2, plug into the equation, and solve for S [Ca 2+ ][F − ] initial00.100 change+S+2S equilibriumS0.100 + 2S K sp = [Ca 2+ ][F − ] 2 K sp = (S)(0.100 + 2S) 2 K sp = (S)(0.100) 2 Tro: Chemistry: A Molecular Approach, 2/e

137. Copyright  2011 Pearson Education, Inc. Practice – Determine the concentration of Ag + ions in seawater that has a [Cl − ] of 0.55 M K sp of AgCl = 1.77 x 10 −10 137Tro: Chemistry: A Molecular Approach, 2/e

138. Copyright  2011 Pearson Education, Inc. Practice – Determine the concentration of Ag + ions in seawater that has a [Cl − ] of 0.55 M write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid [Ag + ][Cl − ] initial00.55 change+S+S+S+S equilibriumS0.55 + S AgCl(s)  Ag + (aq) + Cl − (aq) K sp = [Ag + ][Cl − ] 138Tro: Chemistry: A Molecular Approach, 2/e

139. Copyright  2011 Pearson Education, Inc. Practice – Determine the concentration of Ag + ions in seawater that has a [Cl − ] of 0.55 M 139 substitute into the K sp expression, assume S is small find the value of K sp from Table 16.2, plug into the equation, and solve for S [Ag + ][Cl − ] Initial00.55 Change+S+S+S+S EquilibriumS0.55 + S K sp = [Ag + ][Cl − ] K sp = (S)(0.55 + S) K sp = (S)(0.55) Tro: Chemistry: A Molecular Approach, 2/e

140. Copyright  2011 Pearson Education, Inc. The Effect of pH on Solubility For insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide and the lower the pH, the higher the solubility higher pH = increased [OH − ] M(OH) n (s)  M n+ (aq) + nOH − (aq) For insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility M 2 (CO 3 ) n (s)  2 M n+ (aq) + nCO 3 2− (aq) H 3 O + (aq) + CO 3 2− (aq)  HCO 3 − (aq) + H 2 O(l) 140Tro: Chemistry: A Molecular Approach, 2/e

141. Copyright  2011 Pearson Education, Inc. Precipitation Precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound If we compare the reaction quotient, Q, for the current solution concentrations to the value of K sp, we can determine if precipitation will occur Q = K sp, the solution is saturated, no precipitation Q < K sp, the solution is unsaturated, no precipitation Q > K sp, the solution would be above saturation, the salt above saturation will precipitate Some solutions with Q > K sp will not precipitate unless disturbed – these are called supersaturated solutions 141Tro: Chemistry: A Molecular Approach, 2/e

142. Copyright  2011 Pearson Education, Inc. precipitation occurs if Q > K sp a supersaturated solution will precipitate if a seed crystal is added 142Tro: Chemistry: A Molecular Approach, 2/e

143. Copyright  2011 Pearson Education, Inc. Selective Precipitation A solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others A successful reagent can precipitate with more than one of the cations, as long as their K sp values are significantly different 143Tro: Chemistry: A Molecular Approach, 2/e

144. Copyright  2011 Pearson Education, Inc. Example 16.12: Will a precipitate form when we mix Pb(NO 3 ) 2(aq) with NaBr (aq) if the concentrations after mixing are 0.0150 M and 0.0350 M respectively? write the equation for the reaction determine the ion concentrations of the original salts determine the K sp for any “insoluble” product write the dissociation reaction for the insoluble product calculate Q, using the ion concentrations compare Q to K sp. If Q > K sp, precipitation Pb(NO 3 ) 2(aq) + 2 NaBr (aq) → PbBr 2(s) + 2 NaNO 3(aq) K sp of PbBr 2 = 4.67 x 10 –6 PbBr 2(s)  Pb 2+ (aq) + 2 Br − (aq) Pb(NO 3 ) 2 = 0.0150 M  Pb 2+ = 0.0150 M, NO 3 − = 2(0.0150 M) NaBr = 0.0350 M  Na + = 0.0350 M, Br − = 0.0350 M Q < K sp, so no precipitation 144Tro: Chemistry: A Molecular Approach, 2/e

145. Copyright  2011 Pearson Education, Inc. Practice – Will a precipitate form when we mix Ca(NO 3 ) 2(aq) with NaOH (aq) if the concentrations after mixing are both 0.0175 M? K sp of Ca(OH) 2 = 4.68 x 10 −6 145Tro: Chemistry: A Molecular Approach, 2/e

146. Copyright  2011 Pearson Education, Inc. Practice – Will a precipitate form when we mix Ca(NO 3 ) 2(aq) with NaOH (aq) if the concentrations after mixing are both 0.0175 M? write the equation for the reaction determine the ion concentrations of the original salts determine the K sp for any “insoluble” product write the dissociation reaction for the insoluble product calculate Q, using the ion concentrations compare Q to K sp. If Q > K sp, precipitation Ca(NO 3 ) 2(aq) + 2 NaOH (aq) → Ca(OH) 2(s) + 2 NaNO 3(aq) K sp of Ca(OH) 2 = 4.68 x 10 –6 Ca(OH) 2(s)  Ca 2+ (aq) + 2 OH − (aq) Ca(NO 3 ) 2 = 0.0175 M  Ca 2+ = 0.0175 M, NO 3 − = 2(0.0175 M) NaOH = 0.0175 M  Na + = 0.0175 M, OH − = 0.0175 M Q > K sp, so precipitation 146Tro: Chemistry: A Molecular Approach, 2/e

147. Copyright  2011 Pearson Education, Inc. 147 Example 16.13: What is the minimum [OH − ] necessary to just begin to precipitate Mg 2+ (with [0.059]) from seawater? precipitating may just occur when Q = K sp Mg(OH) 2(s)  Mg 2+ (aq) + 2 OH − (aq) Tro: Chemistry: A Molecular Approach, 2/e

148. Copyright  2011 Pearson Education, Inc. Practice – What is the minimum concentration of Ca(NO 3 ) 2(aq) that will precipitate Ca(OH) 2 from 0.0175 M NaOH (aq) ? K sp of Ca(OH) 2 = 4.68 x 10 −6 148Tro: Chemistry: A Molecular Approach, 2/e

149. Copyright  2011 Pearson Education, Inc. 149 Practice – What is the minimum concentration of Ca(NO 3 ) 2(aq) that will precipitate Ca(OH) 2 from 0.0175 M NaOH (aq) ? precipitating may just occur when Q = K sp [Ca(NO 3 ) 2 ] = [Ca 2+ ] = 0.0153 M Ca(NO 3 ) 2(aq) + 2 NaOH (aq) → Ca(OH) 2(s) + 2 NaNO 3(aq) Ca(OH) 2(s)  Ca 2+ (aq) + 2 OH − (aq) Tro: Chemistry: A Molecular Approach, 2/e

150. Copyright  2011 Pearson Education, Inc. 150 Example 16.14: What is the [Mg 2+ ] when Ca 2+ (with [0.011]) just begins to precipitate from seawater? precipitating may just occur when Q = K sp Ca(OH) 2(s)  Ca 2+ (aq) + 2 OH − (aq) Tro: Chemistry: A Molecular Approach, 2/e

151. Copyright  2011 Pearson Education, Inc. 151 Example 16.14: What is the [Mg 2+ ] when Ca 2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg 2+ begins when [OH − ] = 1.9 x 10 −6 M precipitating Ca 2+ begins when [OH − ] = 2.06 x 10 −2 M when Ca 2+ just begins to precipitate out, the [Mg 2+ ] has dropped from 0.059 M to 4.8 x 10 −10 M Mg(OH) 2(s)  Mg 2+ (aq) + 2 OH − (aq) Tro: Chemistry: A Molecular Approach, 2/e

152. Copyright  2011 Pearson Education, Inc. Practice – A solution is made by mixing Pb(NO 3 ) 2(aq) with AgNO 3(aq) so both compounds have a concentration of 0.0010 M. NaCl (s) is added to precipitate out both AgCl (s) and PbCl 2(aq). What is the [Ag + ] concentration when the Pb 2+ just begins to precipitate? 152Tro: Chemistry: A Molecular Approach, 2/e

153. Copyright  2011 Pearson Education, Inc. 153 Practice – What is the [Ag + ] concentration when the Pb 2+ (0.0010 M) just begins to precipitate? precipitating may just occur when Q = K sp AgNO 3(aq) + NaCl (aq) → AgCl (s) + NaNO 3(aq) AgCl (s)  Ag + (aq) + Cl − (aq) Tro: Chemistry: A Molecular Approach, 2/e

154. Copyright  2011 Pearson Education, Inc. 154 Practice – What is the [Ag + ] concentration when the Pb 2+ (0.0010 M) just begins to precipitate? precipitating may just occur when Q = K sp Pb(NO 3 ) 2(aq) + 2 NaCl (aq) → PbCl 2(s) + 2 NaNO 3(aq) PbCl 2(s)  Pb 2+ (aq) + 2 Cl − (aq) Tro: Chemistry: A Molecular Approach, 2/e

155. Copyright  2011 Pearson Education, Inc. 155 Practice – What is the [Ag + ] concentration when the Pb 2+ (0.0010 M) just begins to precipitate precipitating Ag + begins when [Cl − ] = 1.77 x 10 −7 M precipitating Pb 2+ begins when [Cl − ] = 1.08 x 10 −1 M when Pb 2+ just begins to precipitate out, the [Ag + ] has dropped from 0.0010 M to 1.6 x 10 −9 M AgCl (s)  Ag + (aq) + Cl − (aq) Tro: Chemistry: A Molecular Approach, 2/e

156. Copyright  2011 Pearson Education, Inc. 156 Qualitative Analysis An analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme wet chemistry A sample containing several ions is subjected to the addition of several precipitating agents Addition of each reagent causes one of the ions present to precipitate out Tro: Chemistry: A Molecular Approach, 2/e

157. Copyright  2011 Pearson Education, Inc. 157 Qualitative Analysis Tro: Chemistry: A Molecular Approach, 2/e

158. Copyright  2011 Pearson Education, Inc. 158Tro: Chemistry: A Molecular Approach, 2/e

159. Copyright  2011 Pearson Education, Inc. 159 Group 1 Group one cations are Ag +, Pb 2+ and Hg 2 2+ All these cations form compounds with Cl − that are insoluble in water as long as the concentration is large enough PbCl 2 may be borderline  molar solubility of PbCl 2 = 1.43 x 10 −2 M Precipitated by the addition of HCl Tro: Chemistry: A Molecular Approach, 2/e

160. Copyright  2011 Pearson Education, Inc. 160 Group 2 Group two cations are Cd 2+, Cu 2+, Bi 3+, Sn 4+, As 3+, Pb 2+, Sb 3+, and Hg 2+ All these cations form compounds with HS − and S 2− that are insoluble in water at low pH Precipitated by the addition of H 2 S in HCl Tro: Chemistry: A Molecular Approach, 2/e

161. Copyright  2011 Pearson Education, Inc. 161 Group 3 Group three cations are Fe 2+, Co 2+, Zn 2+, Mn 2+, Ni 2+ precipitated as sulfides; as well as Cr 3+, Fe 3+, and Al 3+ precipitated as hydroxides All these cations form compounds with S 2− that are insoluble in water at high pH Precipitated by the addition of H 2 S in NaOH Tro: Chemistry: A Molecular Approach, 2/e

162. Copyright  2011 Pearson Education, Inc. 162 Group 4 Group four cations are Mg 2+, Ca 2+, Ba 2+ All these cations form compounds with PO 4 3− that are insoluble in water at high pH Precipitated by the addition of (NH 4 ) 2 HPO 4 Tro: Chemistry: A Molecular Approach, 2/e

163. Copyright  2011 Pearson Education, Inc. 163 Group 5 Group five cations are Na +, K +, NH 4 + All these cations form compounds that are soluble in water – they do not precipitate Identified by the color of their flame Tro: Chemistry: A Molecular Approach, 2/e

164. Copyright  2011 Pearson Education, Inc. Complex Ion Formation Transition metals tend to be good Lewis acids They often bond to one or more H 2 O molecules to form a hydrated ion H 2 O is the Lewis base, donating electron pairs to form coordinate covalent bonds Ag + (aq) + 2 H 2 O(l)  Ag(H 2 O) 2 + (aq) Ions that form by combining a cation with several anions or neutral molecules are called complex ions e.g., Ag(H 2 O) 2 + The attached ions or molecules are called ligands e.g., H 2 O 164Tro: Chemistry: A Molecular Approach, 2/e

165. Copyright  2011 Pearson Education, Inc. Complex Ion Equilibria If a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand Ag(H 2 O) 2 + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) + 2 H 2 O (l) generally H 2 O is not included, because its complex ion is always present in aqueous solution Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) 165Tro: Chemistry: A Molecular Approach, 2/e

166. Copyright  2011 Pearson Education, Inc. Formation Constant The reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) The equilibrium constant for the formation reaction is called the formation constant, K f 166Tro: Chemistry: A Molecular Approach, 2/e

167. Copyright  2011 Pearson Education, Inc. Formation Constants 167Tro: Chemistry: A Molecular Approach, 2/e

168. Copyright  2011 Pearson Education, Inc. Example 16.15: 200.0 mL of 1.5 x 10 −3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Write the formation reaction and K f expression. Look up K f value determine the concentration of ions in the diluted solutions Cu 2+ (aq) + 4 NH 3 (aq)  Cu(NH 3 ) 2 2+ (aq) 168Tro: Chemistry: A Molecular Approach, 2/e

169. Copyright  2011 Pearson Education, Inc. Example 16.15: 200.0 mL of 1.5 x 10 −3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Create an ICE table. Because K f is large, assume all the Cu 2+ is converted into complex ion, then the system returns to equilibrium. [Cu 2+ ][NH 3 ][Cu(NH 3 ) 2 2+ ] initial6.7E-40.110 change≈−6.7E-4≈−4(6.7E-4)≈+6.7E-4 equilibriumx0.116.7E-4 Cu 2+ (aq) + 4 NH 3 (aq)  Cu(NH 3 ) 2 2+ (aq) 169Tro: Chemistry: A Molecular Approach, 2/e

170. Copyright  2011 Pearson Education, Inc. Example 16.15: 200.0 mL of 1.5 x 10 -3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Cu 2+ (aq) + 4 NH 3 (aq)  Cu(NH 3 ) 2 2+ (aq) substitute in and solve for x confirm the “x is small” approxima- tion [Cu 2+ ][NH 3 ][Cu(NH 3 ) 2 2+ ] initial6.7E-40.110 change≈−6.7E-4≈−4(6.7E-4)≈+6.7E-4 equilibriumx0.116.7E-4 2.7 x 10 −13 << 6.7 x 10 −4, so the approximation is valid 170Tro: Chemistry: A Molecular Approach, 2/e

171. Copyright  2011 Pearson Education, Inc. Practice – What is [HgI 4 2− ] when 125 mL of 0.0010 M KI is reacted with 75 mL of 0.0010 M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq) 171Tro: Chemistry: A Molecular Approach, 2/e

172. Copyright  2011 Pearson Education, Inc. Practice – What is [HgI 4 2− ] when 125 mL of 0.0010 M KI is reacted with 75 mL of 0.0010 M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq) Write the formation reaction and K f expression. Look up K f value determine the concentration of ions in the diluted solutions Hg 2+ (aq) + 4 I − (aq)  HgI 4 2− (aq) 172Tro: Chemistry: A Molecular Approach, 2/e

173. Copyright  2011 Pearson Education, Inc. Practice – What is [HgI 4 2− ] when 125 mL of 0.0010 M KI is reacted with 75 mL of 0.0010 M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq) Create an ICE table. Because K f is large, assume all the lim. rgt. is converted into complex ion, then the system returns to equilibrium. [Hg 2+ ][I − ][HgI 4 2− ] initial3.75E-46.25E-40 change ≈¼ ( −6.25E-4)≈−(6.25E-4)≈¼ ( +6.25E-4) equilibrium2.19E-4x1.56E-4 Hg 2+ (aq) + 4 I − (aq)  HgI 4 2− (aq) I − is the limiting reagent 173Tro: Chemistry: A Molecular Approach, 2/e

174. Copyright  2011 Pearson Education, Inc. Practice – What is [HgI 4 2− ] when 125 mL of 0.0010 M KI is reacted with 75 mL of 0.0010 M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq) substitute in and solve for x confirm the “x is small” approximation 2 x 10 −8 << 1.6 x 10 −4, so the approximation is valid Hg 2+ (aq) + 4 I − (aq)  HgI 4 2− (aq) [HgI 4 2− ] = 1.6 x 10 −4 [Hg 2+ ][I − ][HgI 4 2− ] initial3.75E-46.25E-40 change ≈¼ ( −6.25E-4)≈−(6.25E-4)≈¼ ( +6.25E-4) equilibrium2.19E-4x1.56E-4 174Tro: Chemistry: A Molecular Approach, 2/e

175. Copyright  2011 Pearson Education, Inc. 175 The Effect of Complex Ion Formation on Solubility The solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands AgCl (s)  Ag + (aq) + Cl − (aq) K sp = 1.77 x 10 −10 Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) K f = 1.7 x 10 7 Adding NH 3 to a solution in equilibrium with AgCl (s) increases the solubility of Ag + Tro: Chemistry: A Molecular Approach, 2/e

176. Copyright  2011 Pearson Education, Inc. 176Tro: Chemistry: A Molecular Approach, 2/e

177. Copyright  2011 Pearson Education, Inc. 177 Solubility of Amphoteric Metal Hydroxides Many metal hydroxides are insoluble All metal hydroxides become more soluble in acidic solution shifting the equilibrium to the right by removing OH − Some metal hydroxides also become more soluble in basic solution acting as a Lewis base forming a complex ion Substances that behave as both an acid and base are said to be amphoteric Some cations that form amphoteric hydroxides include Al 3+, Cr 3+, Zn 2+, Pb 2+, and Sb 2+ Tro: Chemistry: A Molecular Approach, 2/e

178. Copyright  2011 Pearson Education, Inc. 178 Al 3+ Al 3+ is hydrated in water to form an acidic solution Al(H 2 O) 6 3+ (aq) + H 2 O (l)  Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) Addition of OH − drives the equilibrium to the right and continues to remove H from the molecules Al(H 2 O) 5 (OH) 2+ (aq) + OH − (aq)  Al(H 2 O) 4 (OH) 2 + (aq) + H 2 O (l) Al(H 2 O) 4 (OH) 2 + (aq) + OH − (aq)  Al(H 2 O) 3 (OH) 3(s) + H 2 O (l) Tro: Chemistry: A Molecular Approach, 2/e

179. Copyright  2011 Pearson Education, Inc. 179Tro: Chemistry: A Molecular Approach, 2/e