1. EQUILIBRIUM Part 1 Common Ion Effect

2. COMMON ION EFFECT Whenever a weak electrolyte and a strong electrolyte share the same solution, the strong electrolyte dissociates completely and effects the solubility of the weak electrolyte. The weak electrolyte dissolves less than it would when it is by itself.

3. COMMON ION EFFECT Keep in mind how to identify the strength of an acid. The larger the K a the stronger the acid. It means that the acid dissociates more into the ions. K a = [H + ][A - ] [HA]

4. EXAMPLE: What is the pH of a solution of 0.30 mol acetic acid and 0.30 mol sodium acetate to added to enough water to make 1 L solution? What should we start to create? Think of other equilibrium or acid-base problems. Correct, a Table!

5. EXAMPLE: Identify the strong and weak electrolytes. Identify the source of H +, so can find pH. Sodium Acetate is a strong electrolyte, so need to find the Ka of acetic acid, Ka= 1.8 x10 -5, and all H + will come from scetic acid. Common ion is CH 3 COO -.

6. EXAMPLE: [CH 3 COOH] [H + ] [CH 3 COO - ] Initial [ ]0.300 Change in [ ] -x+x Equilibrium [ ] 0.30 – x x.30 +x What are the initial concentrations (M) and the change in concentrations (variable)?

7. EXAMPLE: [CH 3 COOH] [H + ] [CH 3 COO - ] Initial [ ]0.300 Change in [ ] -x+x Equilibrium [ ] 0.30 – x x.30 +x How can we use this information?

8. EXAMPLE: [CH 3 COOH] [H + ] [CH 3 COO - ] Initial [ ]0.300 Change in [ ] -x+x Equilibrium [ ] 0.30 – x x.30 +x How can we use this information?

9. CALCULATING PH FROM K A K a = [H + ][CH 3 COO - ] [CH 3 COOH] K a = (x) (0.30 + x) = 1.8 x 10 -5 0.30 - x We can assume that x is negligible compared to 0.30 since K a is small.

10. CALCULATING PH FROM K A K a = (x) (0.30) = 1.8 x 10 -5 0.30 x = 1.8 x 10 -5 [H + ] = 1.8 x 10 -5 M pH = -log(1.8 x 10 -5 ) = 4.74 This would have been pH = 2.64 if no common ion.

11. CALCULATING CONCENTRATION WHEN A COMMON ION IS INVOLVED Calculate the fluoride ion concentration and pH of a solution that is 0.20 M HF and 0.10 M HCl.

12. CALCULATING CONCENTRATION WHEN A COMMON ION IS INVOLVED Write the reactions in order to find the common ion. HF   H + + F - HCl  H + + Cl - The common ion is H +. HF is a weak acid so it does not completely dissociate, like HCl does.

13. CALCULATING CONCENTRATION WHEN A COMMON ION IS INVOLVED What do we need to create?

14. CALCULATING CONCENTRATION WHEN A COMMON ION IS INVOLVED A Table! How did we get this info? [HF] [H + ] [F - ] Initial [ ]0.200.100 Change in [ ] -x+x Equilibrium [ ] 0.20 – x 0.10 +x x

15. CALCULATING CONCENTRATION WHEN A COMMON ION IS INVOLVED What do we need to find x? Where can we find Ka? [HF] [H + ] [F - ] Initial [ ]0.200.100 Change in [ ] -x+x Equilibrium [ ] 0.20 – x 0.10 +x x

16. CALCULATING CONCENTRATION WHEN A COMMON ION IS INVOLVED Find Ka in Appendix D in Text. [HF] [H + ] [F - ] Initial [ ]0.200.100 Change in [ ] -x+x Equilibrium [ ] 0.20 – x 0.10 +x x

18. CALCULATING CONCENTRATION WHEN A COMMON ION IS INVOLVED K a = (.10 + x) (x) = 6.8 x 10 -4 0.20 - x Simplifies if we assume x is relatively small compared to.10 or.20. Wy can we assume this? Ka = (.10 ) (x) = 6.8 x 10 -4.20

19. CALCULATING CONCENTRATION WHEN A COMMON ION IS INVOLVED Ka = (.10 ) (x) = 6.8 x 10 -4.20 x = (.20 ) (6.8 x 10 -4 ) = 1.4 x 10 -3 = [F - ].10

20. CALCULATING CONCENTRATION WHEN A COMMON ION IS INVOLVED Then we can use x to find the concentration (molarity) of H + ions. [H + ] =.10 +.0014 = (roughly).10 pH = 1.00

22. BUFFERED SOLUTIONS Solutions that contain a weak conjugate acid-base pair are able to resist drastic changes in pH when small amounts of strong acids or strong bases are added are called BUFFERS.

23. BUFFERS Buffers have an acid to neutralize OH - and a base to neutralize H +. The acid and base must not consume each other. So the weak acid or weak base must be paired with a salt of the acid or base.

24. BUFFERS Examples of buffers: CH 3 COOH and CH 3 COONa to get CH 3 COO - Or NH 4 Cl and NH 3 to get NH 4 + By choosing appropriate components and adjusting relative concentrations, a solution can be buffered at a pH.

25. BUFFERS We need to follow some manipulation of formulas to see how this works. HX is an acid MX is a salt of the acid where M is most likely an alkali metal

26. BUFFERS HX is an acid MX is a salt of the acid where M is most likely an alkali metal So HX (aq)  H + (aq) + X - (aq) The HX is the acid, H + is from the acid dissociating, and the X - is from the acid and salt disscociating.

27. BUFFERS HX (aq)  H + (aq) + X - (aq) The Ka is Ka = [H+] [X-] [HX] So [H+] = Ka [HX] [X-]

28. BUFFERS [H + ] = Ka [HX] [X - ] [H + ] is dependent on the Ka and the relationship of [HX] and [X - ]

29. BUFFERS [H + ] = Ka [HX] [X - ] If a base is added ([OH - ]) then: OH - (aq) + HX(aq)  H 2 O(l) + X - (aq) And HX decreases and X - increases. Presence of HX counteracts the addition of base and pH increase is small.

30. BUFFERS [H + ] = Ka [HX] [X - ] If an acid is added ([H + ] or [H 3 O + ]) then: H 3 O + (aq) + X - (aq)  H 2 O(l) + HX(aq) And HX increases and X - decreases. Presence of HX counteracts the addition of acid and pH decrease is small.

31. CALCULATING PH OF BUFFERS

32. -log[H + ] = -logKa - log [HX] [X - ] pH = pKa - log [HX] [X - ] pH = pKa + log [X - ] [HX]

33. HENDERSON-HASSELBALCH EQUATION pH = pKa + log [X - ] [HX] pH = pKa + log [base] [acid] Base and acid refer to the equilibrium concentration of the conjugate acid-base pair.

34. BUFFER CAPACITY Amount of acid or base that a buffer can neutralize before the pH begins to change at a greater degree. pH of a 1L solution of 1 M CH 3 COOH and 1 M CH 3 COONa has same pH as a 1L solution of.1 M CH 3 COOH and.1 M CH 3 COONa, just is a greater buffer.

35. PH RANGE If the concentrations of weak acid and conjugate base pair is the same, then pH = pKa. So try to select a buffer based on pKa that is close to desired pH. Range that is good is pH = pKa + 1.

36. STRONG ACID OR STRONG BASES AND BUFFERS When a strong acid or base is added to a weak acid buffer, the strong acid or strong base is consumed. We need to calculate the new values of [HX] and [X - ], then use that with Ka to calculate the [H + ] and then pH.

37. CALCULATING PH OF BUFFER EXAMPLE: What is the pH of a buffer that is 0.12 M lactic acid (HC 3 H 5 O 3 ) and 0.10 M sodium lactate (NaC 3 H 5 O 3 )? Lactic Acid Ka = 1.4 x 10 -4 What should we start to create? Think of other equilibrium or acid-base problems. Correct, a Table!

38. CALCULATING PH OF BUFFER EXAMPLE: Identify the strong and weak electrolytes. Identify the source of H +, so can find pH. Sodium Lactate is a strong electrolyte, so need to find the Ka of lactic acid, Ka= 1.4 x10 -4, and all H + will come from lactic acid. Common ion is C 3 H 5 O 3 -.

39. CALCULATING PH OF BUFFER EXAMPLE: [HC 3 H 5 O 3 ] [H + ] [C3H5O3-][C3H5O3-] Initial [ ]0.1200.10 Change in [ ] -x+x Equilibrium [ ] 0.12 – x x.10 +x What are the initial concentrations (M) and the change in concentrations (variable)?

40. CALCULATING PH OF BUFFER EXAMPLE: How can we use this information? [HC 3 H 5 O 3 ] [H + ] [C3H5O3-][C3H5O3-] Initial [ ]0.1200.10 Change in [ ] -x+x Equilibrium [ ] 0.12 – x x.10 +x

41. CALCULATING PH FROM K A K a = [H + ][C 3 H 5 O 3 - ] [HC 3 H 5 O 3 ] K a = (x) (0.10 + x) = 1.4 x 10 -4 0.12 - x We can assume that x is negligible compared to 0.10 since K a is small.

42. CALCULATING PH FROM K A K a = (x) (0.10) = 1.4 x 10 -4 0.12 x = 1.8 x 10 -4 (0.12 / 0.10) [H + ] = 1.7 x 10 -4 M pH = -log(1.7 x 10 -4 ) = 3.77 Or: pH = pKa + log [base] = 3.85 + (-0.08) [acid] still 3.77

44. EXAMPLE: PREPARING A BUFFER How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer with a pH of 9.00? (assume the addition of NH 4 Cl does not change the volume of the solution)

45. EXAMPLE: PREPARING A BUFFER Identify what we are dealing with. NH 4 Cl, NH 4 +, Cl -, NH 3, H 2 O Cl - is a spectator ion. NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) K b = 1.8 x 10 -5

46. EXAMPLE: PREPARING A BUFFER Identify what we are dealing with. NH 4 Cl, NH 4 +, Cl -, NH 3, H 2 O Cl - is a spectator ion. NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) K b = 1.8 x 10 -5

47. EXAMPLE: PREPARING A BUFFER NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) K b = 1.8 x 10 -5 = [NH 4 + ] [OH - ] [NH 3 ] pOH = 14.00- pH = 14.00 – 9.00 = 5.00 [OH - ] = 1.0 x10 -5

48. EXAMPLE: PREPARING A BUFFER NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) K b = 1.8 x 10 -5 = [NH 4 + ] [OH - ] [NH 3 ] pOH = 14.00- pH = 14.00 – 9.00 = 5.00 [OH - ] = 1.0 x10 -5

49. EXAMPLE: PREPARING A BUFFER When we look at the K b and notice that it is small and the common ion is already present, then we can use the initial concentration for NH 3 of 0.10 M.

50. EXAMPLE: PREPARING A BUFFER Then we can solve for [NH 4 ] K b = 1.8 x 10 -5 = [NH 4 + ] [OH - ] [NH 3 ] [NH 4 ] = 1.8 x10 -5 [NH 3 ] [OH - ]

51. EXAMPLE: PREPARING A BUFFER [NH 4 ] = (1.8 x10 -5 )( 0.10 M) (1.0 x10 -5 M) =.18 M So the answer is: (2.0 L).18 mol NH 4 =0.36 mol NH 4 Cl L

53. EXAMPLE: CALCULATING PH CHANGES IN BUFFERS A buffer that is made by adding 0.300 mol CH 3 COOH and 0.300 mol CH 3 COONa to enough water to make 1.000 L of solution has a pH of 4.74. (1) calculate the pH of the solution after 5.0 mL of 4.0 M NaOH (aq) is added. (2) Then compare the pH of a solution of adding 5.0 mL of 4.0 M NaOH (aq) added to 1.000 L H 2 O.

54. EXAMPLE: CALCULATING PH CHANGES IN BUFFERS What should we make? A variation of the table for the reaction: CH 3 COOH + OH -  CH3COO - + H 2 O(l) CH 3 COOH OH - CH 3 COO -

55. EXAMPLE: CALCULATING PH CHANGES IN BUFFERS CH 3 COOH + OH -  CH3COO - + H 2 O(l) Water does not show up because of species. We also focus on mol not [ ] CH 3 COOH OH - CH 3 COO - Buffer before 0.300 mol0 Addition.020 mol Buffer after 0.280 mol 0.320 mol

56. EXAMPLE: CALCULATING PH CHANGES IN BUFFERS Now look at the equilibrium of: CH 3 COOH  CH 3 COO - + H + And we can use the info in the table to help us find the concentrations (molarity) keeping in mind the volume has changed due to addition of the base

57. EXAMPLE: CALCULATING PH CHANGES IN BUFFERS The concentrations are [CH 3 COOH] = 0.280 mol / 1.005 L [CH 3 COO - ] = 0.320 mol / 1.005 L [CH 3 COOH] = 0.279 M [CH 3 COO - ] = 0.318 M

58. EXAMPLE: CALCULATING PH CHANGES IN BUFFERS With these concentrations: [CH 3 COOH] = 0.279 M [CH 3 COO - ] = 0.318 M Use Henderson-Hasselbalch Equation: pH = pKa + log [X - ] [HX]

59. EXAMPLE: CALCULATING PH CHANGES IN BUFFERS Use Henderson-Hasselbalch Equation: pH = 4.74 + log.318 M = 4.80.279 M Remember that the pKa is the same as the pH when the acid and conjugate base concentrations are equal.

60. EXAMPLE: CALCULATING PH CHANGES IN BUFFERS Now if we compare that change in pH to the change in pH with the same amount of base added to just water, we will see what a difference a buffer makes.

61. EXAMPLE: CALCULATING PH CHANGES IN BUFFERS We would still have the OH - added to the water. The concentration would then be 0.20 mol OH - / 1.005 L = 0.020 M OH - Then the pH can be found by: pH = 14.00 – pOH pH= 14.00 – (-log 0.020 ) pH = 14.00 - 1.70 pH = 12.3

62. EXAMPLE: CALCULATING PH CHANGES IN BUFFERS In the first example with the buffer, the pH increased by.06 Without the buffer, the pH increased from 7.00 to 12.3 (increase of 5.30)