1. Aqueous Ionic Equilibrium

2. 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  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. Demo of a pH Buffer Assume 1 drop = 0.05mL When this 1 drop is dissolved in 30 mL of H 2 O What should the pH be when we add 1 drop of 0.1M HNO 3 to 30mL of deionized water ?

4. Why are pH buffers important? Life on Earth is water-based  Human 48-75% water  Plants as high as 95% water  H + and OH - are chemically and structurally reactive in cells The functioning of the cell is very pH dependent  Blood plasma pH = 7.4  pH < 6.9 fatal (acidosis)  pH > 7.9 fatal (alkalosis)  Fish die if the pH of the water goes below 5 or above 9 Life on Earth is water-based  Human 48-75% water  Plants as high as 95% water  H + and OH - are chemically and structurally reactive in cells The functioning of the cell is very pH dependent  Blood plasma pH = 7.4  pH < 6.9 fatal (acidosis)  pH > 7.9 fatal (alkalosis)  Fish die if the pH of the water goes below 5 or above 9

5. 5 Making an Acid Buffer

6. 6 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  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  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  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

7. 7 H2OH2O How Buffers Work HA + H3O+H3O+ A−A− A−A− Added H 3 O + new HA

8. 8 H2OH2O How Buffers Work HA + H3O+H3O+ A−A− Added HO − new A − A−A−

9. 9 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  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

10. 10 Common Ion Effect

11. 11 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 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 given K a for HC 2 H 3 O 2 = 1.8 x 10 -5

12. 12 [HA][A - ][H 3 O + ] initial 0.100 0 change equilibrium 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 -x0.100 + x x 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 given K a for HC 2 H 3 O 2 = 1.8 x 10 -5 HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 - + H 3 O +

13. 13 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 0.100 +x 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 given K a for HC 2 H 3 O 2 = 1.8 x 10 -5

14. 14 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 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 given K a for HC 2 H 3 O 2 = 1.8 x 10 -5

15. 15 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 x0.100 -x 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 given K a for HC 2 H 3 O 2 = 1.8 x 10 -5

16. 16 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 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 given K a for HC 2 H 3 O 2 = 1.8 x 10 -5

17. 17 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 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 given K a for HC 2 H 3 O 2 = 1.8 x 10 -5

18. 18 What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF?

19. 19 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

20. 20 [HA][A - ][H 3 O + ] initial 0.140.071 0 change equilibrium 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 +

21. 21 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 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

22. 22 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

23. 23 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

24. 24 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

25. 25 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

26. 26 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  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

27. 27 Deriving the Henderson-Hasselbalch Equation

28. 28 What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF? Solve for x assuming x << 0.14 we get … [HA][A - ][H 3 O + ] initial 0.140.071 ≈ 0 change -x-x+x+x+x+x equilibrium 0.140.0721.4E-3

29. What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and 0.071 M KF ? 29 Same answer we got solving with an ICE table in the x is small approximation! Solve this time with Henderson-Hasselbalch

30. 30 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 where K a =6.5x10 -5 for HC 7 H 5 O 2 ?

31. 31 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 where K a =6.5x10 -5 for HC 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 +

32. 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 where K a =6.5x10 -5 for HC 7 H 5 O 2 ? HC 7 H 5 O 2 + H 2 O C 7 H 5 O 2 - + H 3 O + [HA][A - ][H 3 O + ] initial 0.0500.150 ≈ 0 change -x-x+x+x+x+x equilibrium 0.050-x0.150-xx

33. 33 Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation?  It is not clear that the Henderson-Hasselbalch Equation is an approximation, but the way we use it, (by placing the initial concentration of [HA] o and [A - ] o into the equation assuming they are equilibrium values), makes it an approximation  For this reason 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  It is not clear that the Henderson-Hasselbalch Equation is an approximation, but the way we use it, (by placing the initial concentration of [HA] o and [A - ] o into the equation assuming they are equilibrium values), makes it an approximation  For this reason 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

34. 34 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 [H 3 O + ] using the new initial values of [HA] and [A − ]  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 [H 3 O + ] using the new initial values of [HA] and [A − ]

35. 35 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 Before0.100 0 mols added --0.010 mols After Not an ICE table

36. 36 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 Before0.100 ≈ 0 mols added --0.010 mols After 0.0900.110≈ 0 Assume OH - completely reactions with HA to make A -

37. 37 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

38. 38 [HA][A - ][H 3 O + ] initial 0.0900.1100 change equilibrium 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 +

39. 39 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 0.0900.110x 0.090  x 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

40. 40 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

41. 41 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 x0.090 -x

42. 42 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

43. 43 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

44. 44 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

45. 45 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

46. 46 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 Buffer Pure Water

47. 47 Basic Buffers  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) B: (aq) + H 2 O (l) H:B + (aq) + OH − (aq)

48. 48 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 −

49. 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  (2) the absolute concentrations of acid and base  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  (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 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 pH change after adding 0.010M NaOH

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 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 pH change after adding 0.010M NaOH

52. 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  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  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  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 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. 54 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. 55 The pK a of HCHO 2 is closest to the desired pH of the buffer, so it would give the most effective buffering range. 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

56. 56 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. 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]  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. 58 Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer

59. 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  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. 60 Titration

61. 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  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. 62 Titration Curve: Unknown Strong Base Added to Strong Acid

63. 63 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

64. 64 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)  What is the Initial pH before adding any NaOH?  HCHO 2(aq) + NaOH (aq)  NaCHO 2(aq) + H 2 O (aq)  What is the Initial pH before adding any NaOH? [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

65. 65 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  What is the pH after adding 5.0 mL 0.100M 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  What is the pH after adding 5.0 mL 0.100M NaOH? HCHO 2 CHO 2 - OH − mols Before2.50E-300 mols added --5.0E-4 mols After 2.00E-35.0E-4≈ 0

66. 66 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) : K b = 5.6 x 10 -11  initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 -3  at equivalence (25 mL of NaOH added  HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq) : K b = 5.6 x 10 -11  initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 -3  at equivalence (25 mL of NaOH added 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 [OH - ] = 1.7 x 10 -6 M

67. 67 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)  Added 30 mL NaOH (after equivalence point)  HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq)  Added 30 mL NaOH (after equivalence point) 5.0 x 10 -4 mol NaOH xs

68. 68 added 30.0 mL NaOH 0.00050 mol NaOH xs pH = 11.96 added 35.0 mL NaOH 0.00100 mol NaOH xs pH = 12.22 Adding NaOH to HCHO 2 added 12.5 mL NaOH 0.00125 mol HCHO 2 pH = 3.74 = pK a half-neutralization 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 15.0 mL NaOH 0.00100 mol HCHO 2 pH = 3.92 added 20.0 mL NaOH 0.00050 mol HCHO 2 pH = 4.34 added 40.0 mL NaOH 0.00150 mol NaOH xs pH = 12.36 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 50.0 mL NaOH 0.00250 mol NaOH xs pH = 12.52

69. 69 Titrating Weak Acid with a Strong Base  the initial pH is that of the weak acid solution  calculate like a weak acid equilibrium problem  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  the initial pH is that of the weak acid solution  calculate like a weak acid equilibrium problem  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

70. 70 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  [A − ] init = mol A − /total liters  calculate like a weak base equilibrium problem  beyond equivalence point, the OH is in excess  [OH − ] = mol MOH xs/total liters  [H 3 O + ][OH − ]=1 x 10 -14  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  [A − ] init = mol A − /total liters  calculate like a weak base equilibrium problem  beyond equivalence point, the OH is in excess  [OH − ] = mol MOH xs/total liters  [H 3 O + ][OH − ]=1 x 10 -14

71. 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 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

72. 72 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 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 Before0.004000≈ 0 mols added --0.00100 mols After ≈ 0 0.003000.00100

73. 73 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 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 Before0.004000≈ 0 mols added --0.00100 mols After 0.003000.00100≈ 0 Look up K a = 4.6 x 10 -4

74. 74 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 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 NO 2 - OH − mols Before0.004000≈ 0 mols added --0.00200 mols After ≈ 0 0.00200 at half-equivalence, moles KOH = ½ mole HNO 2 HNO 2 + H 2 O  NO 2 - + H 3 O +

75. 75 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 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 NO 2 - OH − mols Before0.004000≈ 0 mols added --0.00200 mols After 0.00200 ≈ 0 Look up K a = 4.6 x 10 -4 HNO 2 + H 2 O  NO 2 - + H 3 O +

76. 76 Titration Curve of a Weak Base with a Strong Acid

77. 77 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  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

78. 78 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  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

79. 79 Monitoring pH During a Titration

80. 80 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 ≈ 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  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 ≈ 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

81. 81 Phenolphthalein

82. 82 Methyl Red

83. 83 Monitoring a Titration with an Indicator  for most titrations, the titration curve shows a very large change in pH for very small additions of base 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  for most titrations, the titration curve shows a very large change in pH for very small additions of base 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

84. 84 Acid-Base Indicators

85. 85 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  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

86. 86 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  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

87. 87

88. 88 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)  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)

89. 89 Calculate the molar solubility of PbCl 2 in pure water at 25C 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

90. 90 Calculate the molar solubility of PbCl 2 in pure water at 25°C 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

91. 91 Determine the K sp of PbBr 2 if its molar solubility in water at 25°C is 1.05 x 10 -2 M

92. 92 Determine the K sp of PbBr 2 if its molar solubility in water at 25°C is 1.05 x 10 -2 M 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

93. 93 Determine the K sp of PbBr 2 if its molar solubility in water at 25°C is 1.05 x 10 -2 M 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 )

94. 94 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  in order to compare K sp s, the compounds must have the same dissociation stoichiometry  molar solubility is related to K sp  but you cannot always compare solubilities of compounds by comparing their K sp s  in order to compare K sp s, the compounds must have the same dissociation stoichiometry

95. 95 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 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

96. 96 Calculate the molar solubility of CaF 2 in 0.100 M NaF at 25°C 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

97. 97 Calculate the molar solubility of CaF 2 in 0.100 M NaF at 25°C 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 In the absence of NaF

98. 98 The Effect of pH on Solubility  For insoluble ionic hydroxides  M(OH) n (s) M n+ (aq) + nOH − (aq)  The lower the pH, the higher the solubility  Adding H + removes OH - by H + +OH -  H 2 O so shift equilibrium right  The higher the pH the lower the solubility  higher pH is like adding OH - shift equlibrium left)  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)  For insoluble ionic hydroxides  M(OH) n (s) M n+ (aq) + nOH − (aq)  The lower the pH, the higher the solubility  Adding H + removes OH - by H + +OH -  H 2 O so shift equilibrium right  The higher the pH the lower the solubility  higher pH is like adding OH - shift equlibrium left)  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)

99. 99 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  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

100. 100 precipitation occurs if Q > K sp a supersaturated solution will precipitate if a seed crystal is added

101. 101 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  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

102. 102 You have some seawater [Mg 2+ ]=0.059M and [Ca 2+ ] = 0.011M you want to selectively precipitate out the Mg 2+ with OH − can it be done? Assume K sp (Mg(OH) 2 )=2.06x10 -13 and K sp (Ca(OH) 2 )=4.68x10 -6 precipitating may just occur when Q = K sp Magnesium hydroxide is the most insoluble. What is the minimum [OH − ] necessary to just begin to precipitate Mg 2+ ?

103. 103 What is the [Mg 2+ ] when Ca 2+ just begins to precipitate from seawater (assume K sp Ca(OH) 2 = 4.68x10 -6 )? precipitating Mg 2+ begins when [OH − ] = 1.9 x 10 -6 M Ca 2+ precipitates for [OH - ] > 0.021M

104. 104 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 What is the [Mg 2+ ] when Ca 2+ just begins to precipitate from seawater (assume K sp Ca(OH) 2 = 4.68x10 -6 )? This is an example of selective precipitation, we can remove all of the Mg 2+ with OH - because its K sp is much smaller than that for Ca 2+

105. 105 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  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

106. 106 Qualitative Analysis

107. 107

108. 108 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  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

109. 109 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  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

110. 110 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  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

111. 111 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  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

112. 112 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  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

113. 113 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  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

114. Naming Complex Ions 114 Anionic LigandsNames Neutral LigandsNames Br - bromo NH 3 ammine F - fluoro H 2 Oaqua O 2- oxo NONitrosyl OH - Hydroxo COCarbonyl CN - cyano O 2 dioxygen C 2 O 4 2- oxalato N 2 dinitrogen CO 3 2- carbonato C 5 H 5 Npyridine CH 3 COO - acetato H 2 NCH 2 CH 2 NH 2 ethylenediamine NumberPrefixNumberPrefixNumberPrefix 1mono5penta (pentakis)9nona (ennea) 2di (bis)6hexa (hexakis)10deca 3tri (tris)7hepta11undeca 4tetra (tetrakis)8octa12dodeca Name of MetalName in an Anionic Complex IronFerratePlatinumPlatinate CopperCuprate AluminumAluminate LeadPlumbateManganeseManganate SilverArgenate GoldAurate TinStannate

115. Naming Complex Ions 115 Name cation then anion Name ligand first in alphabetical order then the metal ion Greek prefixes are used to designate the number of each type of ligand in the complex ion, e.g. di-, tri- and tetra-. After naming the ligands, name the central metal. If the complex ion is an anion, the name of the metal ends with the suffix –ate. [Cr(NH 3 ) 3 (H 2 O) 3 ]Cl 3 NH 3 is ammine H 2 O is aqua Cr is Chromium (III) triamminetriaquachromium(III) chloride 3 is tri [Pt(NH 3 ) 5 Cl]Br 3 pentaamminechloroplatinum(IV) bromide K 4 [Fe(CN) 6 ] potassium hexacyanoferrate(II)

116. 116 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)  diaquaSilver(I) and  diammineSilver(i)  generally H 2 O is not included, since its complex ion is always present in aqueous solution Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq)  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)  diaquaSilver(I) and  diammineSilver(i)  generally H 2 O is not included, since its complex ion is always present in aqueous solution Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq)

117. 117 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  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

118. 118 Formation Constants

119. 119 200.0 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 ) 4 2+ (aq)

120. 120 200.0 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. Since 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+x -4(6.7E-4-x) + 6.7E-4-x Equilibriumx ≃ 0.11 ≃ 6.7E-4 Cu 2+ (aq) + 4 NH 3 (aq) Cu(NH 3 ) 4 2+ (aq) Where x << 6.4E-4

121. 121 200.0 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” approximation [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 since 2.7 x 10 -13 << 6.7 x 10 -4, the approximation is valid What? Only 1 copper in 10 13 is Cu 2+ (aq)! NH 3 is a stronger ligand than H 2 O

122. 122 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 +  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 +

123. 123

124. 124 Solubility of Amphoteric Metal Hydroxides

125. 125 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)  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)

126. 126