1. Chapter 16 Aqueous Ionic Equilibrium 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro

2. 2 The Danger of Antifreeze each year, thousands of pets and wildlife 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 if present in high enough concentration in the bloodstream, it overwhelms the buffering ability of HCO 3 −, causing the blood pH to drop when the blood pH is low, it ability to carry O 2 is compromised – acidosis the treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol ethylene glycol (aka 1,2–ethandiol)

3. Tro, Chemistry: A Molecular Approach 3 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

4. Tro, Chemistry: A Molecular Approach 4 Making an Acid Buffer

5. Tro, Chemistry: A Molecular Approach 5 How Acid Buffers Work 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 – 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 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 H 3 O + constant

6. Tro, Chemistry: A Molecular Approach 6 H2OH2O How Buffers Work HA  + H3O+H3O+ A−A− A−A− Added H 3 O + new HA

7. Tro, Chemistry: A Molecular Approach 7 H2OH2O HA How Buffers Work HA  + H3O+H3O+ A−A− Added HO − new A − A−A−

8. Tro, Chemistry: A Molecular Approach 8 Common Ion Effect HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (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 H 3 O + ion concentration

9. Tro, Chemistry: A Molecular Approach 9 Common Ion Effect

10. Tro, Chemistry: A Molecular Approach 10 Ex 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 ? 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

11. Tro, Chemistry: A Molecular Approach 11 [HA][A - ][H 3 O + ] initial 0.100 0 change equilibrium Ex 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 ? 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 +

12. Tro, Chemistry: A Molecular Approach 12 determine the value of K a since 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 change -x-x+x+x+x+x equilibrium 0.100 x 0.100  x Ex 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 ? 0.100 +x K a for HC 2 H 3 O 2 = 1.8 x 10 -5

13. Tro, Chemistry: A Molecular Approach 13 Ex 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 ? K a for HC 2 H 3 O 2 = 1.8 x 10 -5 check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init 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

14. Tro, Chemistry: A Molecular Approach 14 Ex 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 ? x = 1.8 x 10 -5 substitute x into the equilibrium concentration definitions and solve [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

15. Tro, Chemistry: A Molecular Approach 15 Ex 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 ? 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

16. Tro, Chemistry: A Molecular Approach 16 Ex 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 ? 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

17. Tro, Chemistry: A Molecular Approach 17 Practice - What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF?

18. Tro, Chemistry: A Molecular Approach 18 Practice - What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 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 + ] initial 0.140.071 ≈ 0 change equilibrium

19. Tro, Chemistry: A Molecular Approach 19 [HA][A - ][H 3 O + ] initial 0.140.071 0 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? 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 +

20. Tro, Chemistry: A Molecular Approach 20 determine the value of K a since 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 Practice - What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 0.071 +x K a for HF = 7.0 x 10 -4

21. Tro, Chemistry: A Molecular Approach 21 Practice - What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? K a for HF = 7.0 x 10 -4 check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init the approximation is valid x = 1.4 x 10 -3 [HA][A 2 - ][H 3 O + ] initial 0.140.071 ≈ 0 change -x-x+x+x+x+x equilibrium 0.140.071 x

22. Tro, Chemistry: A Molecular Approach 22 Practice - What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? x = 1.4 x 10 -3 substitute x into the equilibrium concentration definitions and solve [HA][A 2 - ][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

23. Tro, Chemistry: A Molecular Approach 23 Practice - What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 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

24. Tro, Chemistry: A Molecular Approach 24 Practice - What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 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

25. Tro, Chemistry: A Molecular Approach 25 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 K a and initial concentrations of the weak acid and salt of the conjugate base – as long as the “x is small” approximation is valid

26. Tro, Chemistry: A Molecular Approach 26 Deriving the Henderson-Hasselbalch Equation

27. Tro, Chemistry: A Molecular Approach 27 Ex 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 ? 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

28. Tro, Chemistry: A Molecular Approach 28 Practice - What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF?

29. Tro, Chemistry: A Molecular Approach 29 Practice - What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? 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 +

30. Tro, Chemistry: A Molecular Approach 30 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 1000x larger than the value of K a

31. Tro, Chemistry: A Molecular Approach 31 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 [H 3 O + ] using the new initial values of [HA] and [A − ]

32. Tro, Chemistry: A Molecular Approach 32 Ex 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? 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 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

33. Tro, Chemistry: A Molecular Approach 33 Ex 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? 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

34. Tro, Chemistry: A Molecular Approach 34 Ex 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? 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

35. Tro, Chemistry: A Molecular Approach 35 Ex 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? 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 0.110

36. Tro, Chemistry: A Molecular Approach 36 Ex 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? 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 change equilibrium

37. Tro, Chemistry: A Molecular Approach 37 [HA][A - ][H 3 O + ] initial 0.0900.110 0 change equilibrium Ex 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? 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 +

38. Tro, Chemistry: A Molecular Approach 38 determine the value of K a since 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 change -x-x+x+x+x+x equilibrium 0.0900.110 x 0.090  x Ex 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? 0.110 +x K a for HC 2 H 3 O 2 = 1.8 x 10 -5

39. Tro, Chemistry: A Molecular Approach 39 Ex 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? K a for HC 2 H 3 O 2 = 1.8 x 10 -5 check if the approximation is valid by seeing if x < 5% of [HC 2 H 3 O 2 ] init 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

40. Tro, Chemistry: A Molecular Approach 40 Ex 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? x = 1.47 x 10 -5 substitute x into the equilibrium concentration definitions and solve [HA][A - ][H 3 O + ] initial 0.0900.110 ≈ 0 change -x-x+x+x+x+x equilibrium 0.0900.1101.5E-5 0.110 + x x 0.090  x

41. Tro, Chemistry: A Molecular Approach 41 Ex 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? substitute [H 3 O + ] into the formula for pH and solve [HA][A - ][H 3 O + ] initial 0.0900.110 ≈ 0 change -x-x+x+x+x+x equilibrium 0.0900.1101.5E-5

42. Tro, Chemistry: A Molecular Approach 42 Ex 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? 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 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

43. Tro, Chemistry: A Molecular Approach 43 Ex 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? 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

44. Tro, Chemistry: A Molecular Approach 44 Ex 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? 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

45. Tro, Chemistry: A Molecular Approach 45 Ex 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? 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

46. Tro, Chemistry: A Molecular Approach 46 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)

47. Tro, Chemistry: A Molecular Approach 47 Ex 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? 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 −

48. Tro, Chemistry: A Molecular Approach 48 Ex 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? 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 −

49. Tro, Chemistry: A Molecular Approach 49 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

50. 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 12 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

51. 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 12 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

52. Tro, Chemistry: A Molecular Approach 52 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

53. 53 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 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 is closest to the pH of the buffer

54. Tro, Chemistry: A Molecular Approach 54 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, HClO 2 pK a = 1.95 Nitrous Acid, HNO 2 pK a = 3.34 Formic Acid, HCHO 2 pK a = 3.74 Hypochlorous Acid, HClOpK a = 7.54

55. Tro, Chemistry: A Molecular Approach 55 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, HClO 2 pK a = 1.95 Nitrous Acid, HNO 2 pK a = 3.34 Formic Acid, HCHO 2 pK a = 3.74 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.

56. Tro, Chemistry: A Molecular Approach 56 Ex. 16.5b – What ratio of NaCHO 2 : HCHO 2 would be required to make a buffer with pH 4.25? Formic Acid, HCHO 2, pK a = 3.74 to make the buffer with pH 4.25, you would use 3.24 times as much NaCHO 2 as HCHO 2

57. Tro, Chemistry: A Molecular Approach 57 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]

58. Tro, Chemistry: A Molecular Approach 58 Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer

59. Tro, Chemistry: A Molecular Approach 59 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

60. Tro, Chemistry: A Molecular Approach 60 Titration

61. Tro, Chemistry: A Molecular Approach 61 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

62. Tro, Chemistry: A Molecular Approach 62 Titration Curve: Unknown Strong Base Added to Strong Acid

63. Tro, Chemistry: A Molecular Approach 63

64. Tro, Chemistry: A Molecular Approach 64 Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (aq) 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 5.0 x 10 -4 mol NaOH 2.00 x 10 -3 mol HCl

65. Tro, Chemistry: A Molecular Approach 65 Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (aq) at equivalence, 0.00 mol HCl and 0.00 mol NaOH pH at equivalence = 7.00 after equivalence point added 30.0 mL NaOH 5.0 x 10 -4 mol NaOH xs

66. Tro, Chemistry: A Molecular Approach 66 Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (aq) at equivalence, 0.00 mol HCl and 0.00 mol NaOH pH at equivalence = 7.00 after equivalence point added 30.0 mL NaOH 5.0 x 10 -4 mol NaOH xs

67. Tro, Chemistry: A Molecular Approach 67 added 30.0 mL NaOH 0.00050 mol NaOH pH = 11.96 added 35.0 mL NaOH 0.00100 mol NaOH pH = 12.22 Adding NaOH to HCl 25.0 mL 0.100 M HCl 0.00250 mol HCl pH = 1.00 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 15.0 mL NaOH 0.00100 mol HCl pH = 1.60 added 20.0 mL NaOH 0.00050 mol HCl pH = 1.95 added 25.0 mL NaOH equivalence point pH = 7.00 added 40.0 mL NaOH 0.00150 mol NaOH pH = 12.36 added 50.0 mL NaOH 0.00250 mol NaOH pH = 12.52

68. Tro, Chemistry: A Molecular Approach 68 Titration of 25.0 mL of 0.100 M HCl with 0.100 M NaOH The 1st derivative of the curve is maximum at the equivalence point Since the solutions are equal concentration, the equivalence point is at equal volumes

69. Tro, Chemistry: A Molecular Approach 69 After about pH 3, there is practically no HCl left, it has all been reacted and become NaCl + H 2 O

70. Tro, Chemistry: A Molecular Approach 70 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

71. Tro, Chemistry: A Molecular Approach 71 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

72. 72 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 at equivalence added 25.0 mL NaOH 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 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

73. Tro, Chemistry: A Molecular Approach 73 Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq) after equivalence point added 30.0 mL NaOH 5.0 x 10 -4 mol NaOH xs