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Chapter 16 Aqueous Ionic Equilibrium

Chemistry: A Molecular Approach , 1 st Ed. Nivaldo Tro. Chapter 16 Aqueous Ionic Equilibrium. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. 2008, Prentice Hall. Buffers. buffers are solutions that resist changes in pH when an acid or base is added

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Chapter 16 Aqueous Ionic Equilibrium

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  1. Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro Chapter 16Aqueous IonicEquilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall

  2. Buffers • buffers are solutions that resist changes in pH when an acid or base is added • they act by neutralizing the added acid or base • but just like everything else, there is a limit to what they can do, 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 Tro, Chemistry: A Molecular Approach

  3. Making an Acid Buffer Tro, Chemistry: A Molecular Approach

  4. How Acid Buffers WorkHA(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 • you can also think of the H3O+ combining with the OH− to make H2O; the H3O+ is then replaced by the shifting equilibrium • the buffer solutions also contain significant amounts of the conjugate base anion, A− - these ions combine with added acid to make more HA and keep the H3O+ constant Tro, Chemistry: A Molecular Approach

  5. Common Ion Effect HA(aq) + H2O(l) A−(aq) + H3O+(aq) • adding a salt containing the anion, NaA, that 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

  6. Common Ion Effect Tro, Chemistry: A Molecular Approach

  7. Ex 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

  8. Ex 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

  9. Ex 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

  10. Ex 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

  11. 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

  12. 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

  13. 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

  14. 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 0.14 x 0.071 +x Tro, Chemistry: A Molecular Approach

  15. 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

  16. 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

  17. 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

  18. 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 Ka 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

  19. 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 are added to them, their pH does change • calculating the new pH after adding acid or base requires breaking the problem into 2 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

  20. Ex 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

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

  22. Ex 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

  23. Ex 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

  24. Ex 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

  25. Ex 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

  26. Ex 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

  27. Ex 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

  28. Ex 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

  29. Ex 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

  30. Ex 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

  31. H2O(l) + NH3(aq) NH4+(aq) + OH−(aq) Basic BuffersB:(aq) + H2O(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− Tro, Chemistry: A Molecular Approach

  32. Buffer Solutions Consider HOAc/OAc- to see how buffers work. ACID USES UP ADDED OH-. We know that OAc- + H2O HOAc + OH-has Kb = 5.6 x 10-10 Therefore, the reverse reaction of the WEAK ACID with added OH- has Kreverse = 1/ Kb = 1.8 x 109 Kreverse is VERY LARGE, so HOAc completely uses up the OH- !!!!

  33. Buffer Solutions Consider HOAc/OAc- to see how buffers work. CONJUGATE BASE USES UP ADDED H+ HOAc + H2O OAc- + H3O+ has Ka = 1.8 x 10-5. Therefore, the reverse reaction of the WEAK BASE with added H+has Kreverse = 1/ Ka = 5.6 x 104 Kreverse is VERY LARGE, so OAc- completely uses up the H+ !

  34. Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? in 1 L HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 [HOAc] [OAc-] [H3O+] initial change equilib 0.700 0.600 0 -x +x +x 0.700 - x 0.600 + x x

  35. + [H O ](0.600) -5 3 K 1.8 x 10 = = a 0.700 Buffer Solutions Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? HOAc + H2O OAc- + H3O+ Ka = 1.8 x 10-5 [H3O+] = 2.1 x 10-5 and pH = 4.68 [HOAc] [OAc-] [H3O+] equilib 0.700 - x 0.600 + x x Assuming that x << 0.700 and 0.600, we have

  36. Adding an Acid to a Buffer Problem: What is the new pH when 1.00 mL of 1.00 M HCl is added to: a) 1.00 L of pure water (before HCl, pH = 7.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (a) Calculate [HCl] after adding 1.00 mL of HCl to 1.00 L of water M1 • V1 = M2 • V2 M2 = 1.00 x 10-3 M pH = 3.00

  37. Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to: a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b) Step 1 — do the stoichiometry H3O+ (from HCl) + OAc- (from buffer) ---> HOAc (from buffer) The reaction occurs completely because K is very large.

  38. Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to: a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 1—Stoichiometry [H3O+] [OAc-] [HOAc] Before rxn 0.00100 0.600 0.700 Change -0.00100 -0.00100 +0.00100 After rxn 0 0.599 0.701

  39. Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to; a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc- [HOAc] [OAc-] [H3O+] Before rxn 0.701 0.599 0 Change -x +x +x After rxn 0.701-x 0.599+x x

  40. Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) [HOAc] [OAc-] [H3O+] After rxn 0.710-x 0.599+x x Because [H3O+] = 2.1 x 10-5 M BEFORE adding HCl, we again neglect x relative to 0.701 and 0.599. Solution to Part (b): Step 2—Equilibrium HOAc + H2O H3O+ + OAc-

  41. [HOAc] 0.701 + -5 [ H O ] • K • ( 1 . 8 x 10 ) = = 3 a - 0.599 [OAc ] Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to a) 1.00 L of pure water (after HCl, pH = 3.00) b) 1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) [H3O+] = 2.1 x 10-5 M ------> pH = 4.68 The pH has not changed significantly upon adding HCl to the buffer! Solution to Part (b): Step 2—Equilibrium HOAc + H2O OAc- +H3O+

  42. REVIEW PROBLEMS • Calculate the pH of 0.500 L of a buffer solution composed of 0.10 M sodium acetate and 0.15 M acetic acid, before and after adding 1.0 grams of sodium hydroxide solid (no volume change).

  43. [C2H3O2-] [H+] [HC2H3O2] x(0.10) (0.15) Sample Problem Calculate the pH of 0.500 L of a buffer solution composed of 0.10 M sodium acetate and 0.15 M acetic acid. HC2H3O2 <---> H+ + C2H3O2- 0.15 0 0.10 - x + x + x 0.15 x 0.10 Ka = = = 1.8 x 10-5 X = 2.7 x 10-5 M = [H+] pH = 4.57

  44. .050 .500 .075 .500 Sample Problem Calculate the pH after adding 1.0 grams of sodium hydroxide solid. HC2H3O (.500L)(.15M) = .075 mole NaC2H3O (.500L)(.10M) = .050 mole NaOH (1.0g)(40.0g/mole) = .025 mole HC2H3O2 + OH- ---> HOH + C2H3O2- 0.075 0.025 0.050 - 0.025 - 0.025 + 0.025 0.050 0 0 .075 [C2H3O2-] = = 0.15 M [HC2H3O2] = = 0.10

  45. [C2H3O2-][H+] [HC2H3O2] x(0.15) (0.10) Sample Problem Calculate the pH after adding 1.0 grams of sodium hydroxide solid. HC2H3O2 <---> H+ + C2H3O2- 0.10 0.15 - x + x + x 0.15 x 0.10 Ka = = = 1.8 x 10-5 X = 1.2 x 10-5 M = [H+] pH = 4.92

  46. Ex. 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, HClO2 pKa = 1.95 Nitrous Acid, HNO2 pKa = 3.34 Formic Acid, HCHO2 pKa = 3.74 Hypochlorous Acid, HClO pKa = 7.54 Tro, Chemistry: A Molecular Approach

  47. Ex. 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, HClO2 pKa = 1.95 Nitrous Acid, HNO2 pKa = 3.34 Formic Acid, HCHO2 pKa = 3.74 Hypochlorous Acid, HClO pKa = 7.54 The pKa of HCHO2 is closest to the desired pH of the buffer, so it would give the most effective buffering range. Tro, Chemistry: A Molecular Approach

  48. Buffering Capacity • buffering capacity is the amount of acid or base that can be added to a buffer without destroying its effectiveness • 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] Tro, Chemistry: A Molecular Approach

  49. Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer Tro, Chemistry: A Molecular Approach

  50. 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 Tro, Chemistry: A Molecular Approach

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