1. Chapter 16 Aqueous Ionic Equilibria Chemistry II

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 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 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 H 3 O + constant

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

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

7. pH of a Buffer Solution Consider a solution of a weak acid (HA) and its conjugate base (A - ): Acetic acid/Sodium acetate mixture, CH 3 COOH(aq) + H 2 O(l) ⇌ H 3 O + (aq) + CH 3 COO - (aq) Ionization of CH 3 COOH is suppressed by CH 3 COO - since CH 3 COO - in this system shifts equilibrium left

8. 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 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. 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 CH 3 COOH and 0.100 M CH 3 COONa? 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 CH 3 COOH + H 2 O ⇌ CH 3 COO - + H 3 O + [HA][A - ][H 3 O + ] initial 0.100 ≈ 0 change equilibrium

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 CH 3 COOH and 0.100 M CH 3 COONa? 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 CH 3 COOH + H 2 O ⇌ CH 3 COO - + H 3 O +

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 CH 3 COOH and 0.100 M CH 3 COONa? 0.100 +x K a for CH 3 COOH = 1.8 x 10 -5

13. Ex 16.1 - What is the pH of a buffer that is 0.100 M CH 3 COOH and 0.100 M CH 3 COONa? K a for CH 3 COOH = 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. Ex 16.1 - What is the pH of a buffer that is 0.100 M CH 3 COOH and 0.100 M CH 3 COONa? Alternatively solve (0.100 + x)x = 1.8 x 10 -5 0.100 – x Rearrange: x 2 + 0.100018x -1.8 x 10 -6 = 0 Use quadratic formula: x = -b ± √b 2 – 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 CH 3 COOH and 0.100 M CH 3 COONa? 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

16. Ex 16.1 - What is the pH of a buffer that is 0.100 M CH 3 COOH and 0.100 M CH 3 COONa? 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

17. Henderson-Hasselbalch Equation simplified buffer pH calculation – use the Henderson-Hasselbalch Equation calculates pH of buffer from K a 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 HC 7 H 5 O 2 and 0.150 M NaC 7 H 5 O 2 ? Assume [HA] eq = [HA] init and [A − ] eq = [A − ] init (since K a is small) 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

21. 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 − ]

22. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH 3 COOH and 0.100 mol CH 3 COONa 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 CH 3 COOH + OH − ⇌ CH 3 COO  + H 2 O HAA-A- OH − mols Before 0.100 0 mols added -- 0.010 mols After Part 1: stoichiometry Original pH = 4.74

23. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH 3 COOH and 0.100 mol CH 3 COONa 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 CH 3 COOH + OH − ⇌ CH 3 COO  + H 2 O HAA-A- OH − mols Before 0.100 ≈ 0 mols added -- 0.010 mols After 0.0900.110≈ 0 Part 1: stoichiometry

24. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH 3 COOH and 0.100 mol CH 3 COONa 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 CH 3 COOH + OH − ⇌ CH 3 COO  + 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 Part 1: stoichiometry

25. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH 3 COOH and 0.100 mol CH 3 COONa in 1.00 L that has 0.010 mol NaOH added to it? Write the reaction for the ionization of 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 − ] CH 3 COOH + OH − ⇌ CH 3 COO  + H 2 O [HA][A - ][H 3 O + ] initial 0.0900.110 ≈ 0 change equilibrium Part 2: Equilibrium

26. [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 CH 3 COOH and 0.100 mol CH 3 COONa 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 CH 3 COOH + OH − ⇌ CH 3 COO  + H 2 O Part 2: Equilibrium

27. 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 CH 3 COOH and 0.100 mol CH 3 COONa 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 Part 2: Equilibrium

28. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH 3 COOH and 0.100 mol CH 3 COONa in 1.00 L that has 0.010 mol NaOH added to it? K a for HCH 3 COH = 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

29. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH 3 COOH and 0.100 mol CH 3 COONa 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

30. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH 3 COOH and 0.100 mol CH 3 COONa 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 Compared to 4.74 without addition of base

31. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH 3 COOH and 0.100 mol CH 3 COONa 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

32. 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? 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 = 12.00vs 4.83 for buffered soln.

33. Using Henderson-Hasselbalch equation:

34. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH 3 COOH and 0.100 mol CH 3 COONa 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 CH 3 COOH + H 2 O ⇌ CH 3 COO  + H 3 O + K a for CH 3 COOH = 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

35. Ex 16.3 - What is the pH of a buffer that has 0.100 mol CH 3 COOH and 0.100 mol CH 3 COONa 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 CH 3 COOH + H 2 O ⇌ CH 3 COO  + H 3 O + pK a for CH 3 COOH = 4.745

36. 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) weak baseConjugate acid Need K a for HH eqn. pK a + pK b = 14

37. 37 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 −

38. 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

39. Buffering Effectiveness Consider 2 buffers both with pK a = 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 - )

40. 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

41. HAA-A- OH − mols Before0.50 0 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

42. 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

43. 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

44. 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

45. 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.

46. 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

47. 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]

49. 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

50. Titration

51. 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

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

54. 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

55. 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

56. 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

57. 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

58. 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

59. After about pH 3, there is practically no HCl left, it has all been reacted and become NaCl + H 2 O

60. 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

61. 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

62. 62 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

63. 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

64. 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

65. Tro, Chemistry: A Molecular Approach 65 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

67. Tro, Chemistry: A Molecular Approach 67 Titration of 25.0 mL of 0.100 M HCHO 2 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 pH at equivalence = 8.23

68. Tro, Chemistry: A Molecular Approach 68 at pH 3.74, the [HCHO 2 ] = [CHO 2  ]; the acid is half neutralized half-neutralization occurs when pH = pK a

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  e.g., 15.5 and 15.6 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. Tro, Chemistry: A Molecular Approach 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  calculate the volume of added base like Ex 4.8 [A − ] init = mol A − /total liters calculate like a weak base equilibrium problem  e.g., 15.14 beyond equivalence point, the OH is in excess [OH − ] = mol MOH xs/total liters [H 3 O + ][OH − ]=1 x 10 -14

71. Ex 16.7a – 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. Ex 16.7b – 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 Before 0.004000≈ 0 mols added -- 0.00100 mols After ≈ 0 0.00300 0.00100

73. Ex 16.7b – 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 Before 0.004000≈ 0 mols added -- 0.00100 mols After 0.003000.00100 ≈ 0 Table 15.5 K a = 4.6 x 10 -4

74. Tro, Chemistry: A Molecular Approach 74 Ex 16.7b – 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 + KOH  NO 2  + H 2 O HNO 2 NO 2 - OH − mols Before 0.004000≈ 0 mols added -- 0.00200 mols After ≈ 0 0.00200 at half-equivalence, moles KOH = ½ mole HNO 2

75. Ex 16.7b – 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 + H 2 O  NO 2  + H 3 O + HNO 2 NO 2 - OH − mols Before 0.004000≈ 0 mols added -- 0.00200 mols After 0.00200 ≈ 0 Table 15.5 K a = 4.6 x 10 -4

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

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 titration of 25.0 mL of 0.100 M H 2 SO 3 with 0.100 M NaOH

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

79. Monitoring pH During a Titration

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 when 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. Methyl Red

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

84. Acid-Base Indicators

86. 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

87. 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

88. Solubility Product e.g., the dissociation reaction for PbCl 2 is PbCl 2 (s) ⇌ Pb 2+ (aq) + 2 Cl − (aq) and its equilibrium constant (solubility product) is K sp = [Pb 2+ ][Cl − ] 2

90. 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)

91. Ex 16.8 – Calculate the molar solubility of PbCl 2 in pure water 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 [Pb 2+ ][Cl − ] Initial00 Change+S+2S EquilibriumS2S PbCl 2 (s) ⇌ Pb 2+ (aq) + 2 Cl − (aq) K sp = [Pb 2+ ][Cl − ] 2

92. Ex 16.8 – 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

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

94. 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

95. Practice – 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 )

96. K sp and Relative Solubility molar solubility is related to K sp but you cannot always compare solubilities of 2 compounds by comparing their K sp s in order to compare K sp s, the compounds must have the same dissociation stoichiometry (same values of n and m)

97. 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 Cl − shifts the equilibrium to the left

98. Ex 16.10 – 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

99. Ex 16.10 – 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

100. The Effect of pH on Solubility for insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide and the lower the pH, the higher the solubility higher pH = increased [OH − ] M(OH) n (s) ⇌ M n+ (aq) + nOH − (aq) e.g. Mg(OH) 2 (s) ⇌ Mg 2+ (aq) + 2OH − (aq) Shifts left with inc. OH -

101. The Effect of pH on Solubility 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)

102. 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

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

104. 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

105. Ex 16.13 What is the minimum [OH − ] necessary to just begin to precipitate Mg 2+ (with [0.059]) from seawater? precipitating may just occur when Q = K sp

106. Ex 16.14 What is the [Mg 2+ ] when Ca 2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg 2+ begins when [OH − ] = 1.9 x 10 -6 M

107. Ex 16.14 What is the [Mg 2+ ] when Ca 2+ (with [0.011]) just begins to precipitate from seawater? 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

108. 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

109. Qualitative Analysis

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111. Tro, Chemistry: A Molecular Approach 111 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

112. 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

113. 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

114. 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

115. 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

116. 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

117. 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) 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)

118. 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

119. Formation Constants

120. Ex 16.15 – 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 ) 2 2+ (aq)

121. Ex 16.15 – 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-4-4(6.7E-4)+ 6.7E-4 Equilibriumx0.116.7E-4 Cu 2+ (aq) + 4 NH 3 (aq) ⇌ Cu(NH 3 ) 2 2+ (aq)

122. Ex 16.15 – 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

123. 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 +

124. 124

125. Solubility of Amphoteric Metal Hydroxides many metal hydroxides are insoluble all metal hydroxides become more soluble in acidic solution shifting the equilibrium to the right by removing OH − Act as a base and react with H 3 O + e.g. Al(OH) 3 dissolved in acid Al(OH) 3 (s) + 3H 3 O + →Al 3+ (aq) + 6 H 2 O(l)

126. Solubility of Amphoteric Metal Hydroxides some metal hydroxides also become more soluble in basic solution acting as a acid Al(OH) 3 (s) + OH - →Al(OH 4 ) - (aq) substances that behave as both an acid and base are said to be amphoteric (e.g. Cr 3+, Zn 2+, Pb 2+, and Sn 2+ ) Therefore Al(OH) 3 is soluble at high pH and at low pH but insoluble in a pH-neutral solution

127. 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(H 2 O) 3 (OH) 3(s) + OH − (aq) ⇌ Al(H 2 O) 2 (OH) 4(s) + H 2 O (l) Where Al(H 2 O) 3 (OH) 3(s) = Al(OH) 3(s)