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Chapter 16 Aqueous Ionic Equilibrium

Chemistry: A Molecular Approach , 2nd Ed. Nivaldo Tro. Chapter 16 Aqueous Ionic Equilibrium. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. The Danger of Antifreeze. Each year, thousands of pets and wildlife species die from consuming antifreeze

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Chapter 16 Aqueous Ionic Equilibrium

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  1. Chemistry: A Molecular Approach, 2nd Ed.Nivaldo Tro Chapter 16Aqueous IonicEquilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

  2. 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 • HOCH2COOH glycolic acid (aka a-hydroxyethanoic acid) ethylene glycol (aka 1,2–ethandiol) Tro: Chemistry: A Molecular Approach, 2/e

  3. Why Is Glycolic Acid Toxic? • If present in high enough concentration in the bloodstream, glycolic acid overwhelms the buffering ability of the HCO3− in the blood, causing the blood pH to drop • When the blood pH is low, its ability to carry O2 is compromised • acidosis HbH+(aq) + O2(g)  HbO2(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 Tro: Chemistry: A Molecular Approach, 2/e

  4. 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 H2CO3 and HCO3− Tro: Chemistry: A Molecular Approach, 2/e

  5. Making an Acid Buffer Tro: Chemistry: A Molecular Approach, 2/e

  6. How Acid Buffers Work:Addition of BaseHA(aq) + H2O(l) A−(aq) + H3O+(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) + H2O(l) • you can also think of the H3O+ combining with the OH− to make H2O; the H3O+ is then replaced by the shifting equilibrium Tro: Chemistry: A Molecular Approach, 2/e

  7. H2O How Buffers Work new A− A− HA A− HA  H3O+ + Added HO− Tro: Chemistry: A Molecular Approach, 2/e

  8. How Acid Buffers Work:Addition of AcidHA(aq) + H2O(l) A−(aq) + H3O+(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 H3O+ is kept constant Tro: Chemistry: A Molecular Approach, 2/e

  9. H2O How Buffers Work new HA HA HA A− A−  H3O+ + Added H3O+ Tro: Chemistry: A Molecular Approach, 2/e

  10. Common Ion Effect HA(aq) + H2O(l) A−(aq) + H3O+(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 H3O+ ion concentration Tro: Chemistry: A Molecular Approach, 2/e

  11. Common Ion Effect Tro: Chemistry: A Molecular Approach, 2/e

  12. Example 16.1: What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? HC2H3O2 + H2O  C2H3O2 + H3O+ Tro: Chemistry: A Molecular Approach, 2/e

  13. Example 16.1: What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? HC2H3O2 + H2O  C2H3O2 + H3O+ x +x +x x 0.100 x 0.100 + x Tro: Chemistry: A Molecular Approach, 2/e

  14. Example 16.1: What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10−5 0.100 x 0.100 +x Tro: Chemistry: A Molecular Approach, 2/e

  15. Example 16.1: What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10−5 x = 1.8 x 10−5 the approximation is valid Tro: Chemistry: A Molecular Approach, 2/e

  16. Example 16.1: What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? 0.100 + x x 0.100 x x = 1.8 x 10−5 Tro: Chemistry: A Molecular Approach, 2/e

  17. Example 16.1: What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Tro: Chemistry: A Molecular Approach, 2/e

  18. Example 16.1: What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10−5 the values match Tro: Chemistry: A Molecular Approach, 2/e

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

  20. Practice − What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? HF + H2O  F + H3O+ Tro: Chemistry: A Molecular Approach, 2/e

  21. Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? HF + H2O  F + H3O+ x +x +x x 0.14 x 0.071 + x Tro: Chemistry: A Molecular Approach, 2/e

  22. Practice – What is the pH of a buffer that is 0.14 M and 0.071 M KF? Ka for HF = 7.0 x 10−4 pKa for HF = 3.15 0.14 x 0.071 +x Tro: Chemistry: A Molecular Approach, 2/e

  23. Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Ka for HF = 7.0 x 10−4 x = 1.4 x 10−3 the approximation is valid Tro: Chemistry: A Molecular Approach, 2/e

  24. Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? 0.14 x 0.071 + x x x = 1.4 x 10−3 Tro: Chemistry: A Molecular Approach, 2/e

  25. Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Tro: Chemistry: A Molecular Approach, 2/e

  26. Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Ka for HF = 7.0 x 10−4 the values are close enough Tro: Chemistry: A Molecular Approach, 2/e

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

  28. Deriving the Henderson-Hasselbalch Equation Tro: Chemistry: A Molecular Approach, 2/e

  29. Example 16.2: What is the pH of a buffer that is 0.050 M HC7H5O2 and 0.150 M NaC7H5O2? HC7H5O2 + H2O  C7H5O2 + H3O+ Ka for HC7H5O2 = 6.5 x 10−5 Tro: Chemistry: A Molecular Approach, 2/e

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

  31. Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? HF + H2O  F + H3O+ Tro: Chemistry: A Molecular Approach, 2/e

  32. 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: • the initial concentrations of acid and salt are not very dilute • the Ka 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 Ka Tro: Chemistry: A Molecular Approach, 2/e

  33. 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 • 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− • an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−] Tro: Chemistry: A Molecular Approach, 2/e

  34. Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + OH− C2H3O2 + H2O Tro: Chemistry: A Molecular Approach, 2/e

  35. Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + OH− C2H3O2 + H2O Tro: Chemistry: A Molecular Approach, 2/e

  36. Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + OH− C2H3O2 + H2O Tro: Chemistry: A Molecular Approach, 2/e

  37. Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + OH− C2H3O2 + H2O −0.010 +0.010 −0.010 0.110 0 0.090 0.090 0.110 0 Tro: Chemistry: A Molecular Approach, 2/e

  38. Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + H2O  C2H3O2 + H3O+ Tro: Chemistry: A Molecular Approach, 2/e

  39. Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + H2O  C2H3O2 + H3O+ x +x +x x 0.090 x 0.110 + x Tro: Chemistry: A Molecular Approach, 2/e

  40. Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? Ka for HC2H3O2 = 1.8 x 10−5 0.110 +x 0.090 x Tro: Chemistry: A Molecular Approach, 2/e

  41. Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? Ka for HC2H3O2 = 1.8 x 10−5 x = 1.47 x 10−5 the approximation is valid Tro: Chemistry: A Molecular Approach, 2/e

  42. Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? 0.090 x x 0.110 + x x = 1.47 x 10−5 Tro: Chemistry: A Molecular Approach, 2/e

  43. Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? Tro: Chemistry: A Molecular Approach, 2/e

  44. Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? Ka for HC2H3O2 = 1.8 x 10−5 the values match Tro: Chemistry: A Molecular Approach, 2/e

  45. or, by using the Henderson-Hasselbalch Equation

  46. Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + H2O  C2H3O2 + H3O+ x +x +x x 0.090 x 0.110 + x Tro: Chemistry: A Molecular Approach, 2/e

  47. Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + H2O  C2H3O2 + H3O+ Ka for HC2H3O2 = 1.8 x 10−5 Tro: Chemistry: A Molecular Approach, 2/e

  48. Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it? HC2H3O2 + H2O  C2H3O2 + H3O+ pKa for HC2H3O2 = 4.745 Tro: Chemistry: A Molecular Approach, 2/e

  49. Example 16.3: Compare the effect on pH of adding 0.010 mol NaOH to a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L to adding 0.010 mol NaOH to 1.00 L of pure water HC2H3O2 + H2O  C2H3O2 + H3O+ pKa for HC2H3O2 = 4.745 Tro: Chemistry: A Molecular Approach, 2/e

  50. Practice – What is the pH of a buffer that has 0.140 moles HF (pKa = 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) Tro: Chemistry: A Molecular Approach, 2/e

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