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Chapter 16 Aqueous Ionic Equilibria

Chemistry II. Chapter 16 Aqueous Ionic Equilibria. Buffers: Solutions that Resist pH Change. buffers are solutions that resist changes in pH when an acid or base is added they act by neutralizing the added acid or base

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Chapter 16 Aqueous Ionic Equilibria

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  1. Chemistry II Chapter 16Aqueous Ionic Equilibria

  2. Buffers: Solutions that Resist pH Change • 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

  3. Making an Acid Buffer

  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 weak acid molecules, HA – reacts with added base to neutralize it • the buffer solutions also contain significant amounts of conjugate base anion, A− - combine with added acid to make HA and keep the H3O+ constant

  5. H2O How Buffers Work new HA HA HA A− A− ⇌ H3O+ + Added H3O+

  6. H2O How Buffers Work new A− A− HA A− HA ⇌ H3O+ + Added HO−

  7. pH of a Buffer Solution • Consider a solution of a weak acid (HA) and its conjugate base (A-): • Acetic acid/Sodium acetate mixture, CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO-(aq) • Ionization of CH3COOH is suppressed by CH3COO- since CH3COO- in this system shifts equilibrium left

  8. 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 equilibrium to the left • this causes the pH to be higher than the pH of the acid solution • lowering the H3O+ ion concentration

  9. pH of a Buffer Solution • Use ICE table technique

  10. Ex 16.1 - What is the pH of a buffer that is 0.100 M CH3COOH and 0.100 M CH3COONa? CH3COOH + H2O ⇌CH3COO- + H3O+

  11. Ex 16.1 - What is the pH of a buffer that is 0.100 M CH3COOH and 0.100 M CH3COONa? CH3COOH + H2O ⇌CH3COO- + H3O+ x +x +x x 0.100 x 0.100 + x

  12. Ex 16.1 - What is the pH of a buffer that is 0.100 M CH3COOH and 0.100 M CH3COONa? Ka for CH3COOH = 1.8 x 10-5 0.100 x 0.100 +x

  13. Ex 16.1 - What is the pH of a buffer that is 0.100 M CH3COOH and 0.100 M CH3COONa? Ka for CH3COOH = 1.8 x 10-5 x = 1.8 x 10-5 the approximation is valid

  14. Ex 16.1 - What is the pH of a buffer that is 0.100 M CH3COOH and 0.100 M CH3COONa? Alternatively solve (0.100 + x)x = 1.8 x 10-5 0.100 – x Rearrange: x2 + 0.100018x -1.8 x 10-6 = 0 Use quadratic formula: x = -b ± √b2 – 4ac= 1.8 x 10-5 and -0.10 2a

  15. Ex 16.1 - What is the pH of a buffer that is 0.100 M CH3COOH and 0.100 M CH3COONa? 0.100 x 0.100 + x x x = 1.8 x 10-5

  16. Ex 16.1 - What is the pH of a buffer that is 0.100 M CH3COOH and 0.100 M CH3COONa?

  17. Henderson-Hasselbalch Equation • simplified buffer pH calculation – use the Henderson-Hasselbalch Equation • calculates pH of buffer from Ka and initial concentrations of the weak acid (HA) and salt of the conjugate base (A-) • as long as the “x is small” approximation is valid

  18. Deriving the Henderson-Hasselbalch Equation

  19. Ex 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

  20. 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 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 [H3O+] using the new initial values of [HA] and [A−]

  21. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa in 1.00 L that has 0.010 mol NaOH added to it? CH3COOH + OH− ⇌CH3COO + H2O Part 1: stoichiometry Original pH = 4.74

  22. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa in 1.00 L that has 0.010 mol NaOH added to it? CH3COOH + OH− ⇌CH3COO + H2O Part 1: stoichiometry

  23. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa in 1.00 L that has 0.010 mol NaOH added to it? CH3COOH + OH− ⇌CH3COO + H2O -0.010 +0.010 -0.010 0.110 0 0.090 0.090 0.110 Part 1: stoichiometry

  24. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa in 1.00 L that has 0.010 mol NaOH added to it? CH3COOH + OH− ⇌CH3COO + H2O Part 2: Equilibrium

  25. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa in 1.00 L that has 0.010 mol NaOH added to it? CH3COOH + OH− ⇌CH3COO + H2O x +x +x x 0.090 x 0.110 + x Part 2: Equilibrium

  26. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa 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 Part 2: Equilibrium

  27. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa in 1.00 L that has 0.010 mol NaOH added to it? Ka for HCH3COH = 1.8 x 10-5 x = 1.47 x 10-5 the approximation is valid

  28. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa 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

  29. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa in 1.00 L that has 0.010 mol NaOH added to it? Compared to 4.74 without addition of base

  30. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa in 1.00 L that has 0.010 mol NaOH added to it? Ka for HC2H3O2 = 1.8 x 10-5 the values match

  31. 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? • Compare to adding 0.010 mol NaOH to 1.0 L pure water [OH-] = 0.010 mol / 1.0 L = 0.010 M pOH = -log[OH-] = 2.00 pH + pOH = 14.00 pH = 14.00 - pOH = 12.00 vs 4.83 for buffered soln.

  32. Using Henderson-Hasselbalch equation:

  33. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa in 1.00 L that has 0.010 mol NaOH added to it? CH3COOH + H2O ⇌CH3COO + H3O+ Ka for CH3COOH = 1.8 x 10-5

  34. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH3COOH and 0.100 mol CH3COONa in 1.00 L that has 0.010 mol NaOH added to it? CH3COOH + H2O ⇌CH3COO + H3O+ pKa for CH3COOH = 4.745

  35. 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

  36. Buffering Effectiveness • Consider 2 buffers both with pKa = 5.00 • Calculate change in pH based on addition of 0.010 mol NaOH for two different 1.0 L solutions Both solutions have 0.20 mol total acid and conjugate base Solution 1: equal amounts (0.10 mols HA and 0.10 mols A-) Solution 2: more acid (0.18 mol HA and 0.020 mol A-)

  37. Effect of Relative Amounts of Acid and Conjugate Base a buffer is most effective with equal concentrations of acid and 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 pKa (HA) = 5.00 HA + OH− ⇌A + H2O after adding 0.010 mol NaOH pH = 5.09 after adding 0.010 mol NaOH pH = 4.25

  38. Effect of Absolute Concentrations of Acid and Conjugate Base a buffer is most effective when the concentrations of acid and base are largest 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 pKa (HA) = 5.00 HA + OH− ⇌A + H2O after adding 0.010 mol NaOH pH = 5.02 after adding 0.010 mol NaOH pH = 5.18

  39. Effectiveness of Buffers • a buffer will be most effective when the [base]:[acid] = 1 • equal concentrations of acid and base • effective when 0.1 < [base]:[acid] < 10 • a buffer will be most effective when the [acid] and the [base] are large

  40. Buffering Range • we have said that a buffer will be effective when 0.1 < [base]:[acid] < 10 • substituting into the Henderson-Hasselbalch we can calculate the maximum and minimum pH at which the buffer will be effective Lowest pH Highest pH therefore, the effective pH range of a buffer is pKa ± 1 when choosing an acid to make a buffer, choose one whose is pKa is closest to the pH of the buffer

  41. 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

  42. 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.

  43. Ex. 16.5b – What ratio of NaCHO2 : HCHO2 would be required to make a buffer with pH 4.25? Formic Acid, HCHO2, pKa = 3.74 to make the buffer with pH 4.25, you would use 3.24 times as much NaCHO2 as HCHO2

  44. 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]

  45. 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 H3O+ = moles of OH−, the titration has reached its equivalence point

  46. Titration

  47. 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

  48. Titration Curve:Unknown Strong Base Added to Strong Acid

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