1. Titration Curves What’s in the beaker

2. Always ask… What’s in the beaker? When I start the titration, there is only water and whatever it is I’m titrating (and maybe indicator, but we’ll ignore that). Let’s titrate HCl with NaOH

3. 10.00 mL of 0.100 M HCl diluted with 50.0 mL of water What’s in the beaker? Cl - H2OH2O H2OH2O H2OH2O H+H+ H+H+ H+H+

4. I know what’s in the beaker… Now I ask: what reactions are possible? Cl - H2OH2O H2OH2O H2OH2O H+H+ H+H+ H+H+ “What reactions are possible”? H + + H 2 O  H + + H 2 O Cl - + H 2 O  HCl + OH - H 2 O + H 2 O  H 3 O + + OH -

5. Which reactions matter? NONE! Cl - H2OH2O H2OH2O H2OH2O H+H+ H+H+ H+H+ “What reactions are possible”? H + + H 2 O  H + + H 2 O Cl - + H 2 O  HCl + OH - H 2 O + H 2 O  H 3 O + + OH -

6. 10.00 mL of 0.100 M HCl diluted with 50.0 mL of water What’s the pH? 0.100 M * 10.0 mL = [HCl]* 60.0 mL [HCl] = 1.66 7 x10 -2 M = [H + ] = [Cl - ] (it’s a strong acid) pH = - log (1.66 7 x10 -2 M) = 1.778

7. 10.00 mL of 0.100 M HCl diluted with 50.0 mL of water. Then add 2.00 mL of 0.100 M NaOH. What’s in the beaker? Cl - H2OH2O H2OH2O H2OH2O H+H+ Na + H+H+

8. I know what’s in the beaker… Now I ask: what reactions are possible? Cl - H2OH2O H2OH2O H2OH2O Na + H+H+ H+H+ “What reactions are possible”? H + + H 2 O  H + + H 2 O Cl - + H 2 O  HCl + OH - Na + + H 2 O  NaOH + H + H 2 O + H 2 O  H 3 O + + OH -

9. Which reactions matter? NONE! Cl - H2OH2O H2OH2O H2OH2O H+H+ H+H+ H+H+ “What reactions are possible”? H + + H 2 O  H + + H 2 O Cl - + H 2 O  HCl + OH - Na+ + H2O  NaOH + H+ H 2 O + H 2 O  H 3 O + + OH -

10. 10.00 mL of 0.100 M HCl diluted with 50.0 mL of water. Then add 2.00 mL of 0.100 M NaOH. What’s the pH? I like an ICE chart here, just to keep track of things! Moles is better for my ICE chart because of dilution issues. 0.100 M * 0.0100 L = 0.00100 moles HCl initial 0.100 M NaOH * 0.00200 L = 0.0002 mol NaOH

11. Always start with a balanced equation. Which one? H++OH -  H2OH2O I0.001 mol0.0002 mol- C-0.0002 mol - End0.0008 mol0 mol

12. So…. 0.0008 mol H + = 1.29x10 -2 M 0.062 L pH = - log(1.29x10 -2 M) = 1.889 Notice, not a lot of change.

13. 10.00 mL of 0.100 M HCl diluted with 50.0 mL of water. Then add 5.00 mL of 0.100 M NaOH. What’s the pH? I like an ICE chart here, just to keep track of things! Moles is better for my ICE chart because of dilution issues. 0.100 M * 0.0100 L = 0.00100 moles HCl initial 0.100 M NaOH * 0.00500 L = 0.0005 mol NaOH

14. Always start with a balanced equation. Which one? H++OH -  H2OH2O I0.001 mol0.0005 mol- C-0.0005 mol - End0.0005 mol0 mol

15. So…. 0.0005 mol H + = 7.69x10 -3 M 0.065 L pH = - log(7.69x10 -3 M) = 2.114 Notice, not a lot of change.

16. 10.00 mL of 0.100 M HCl diluted with 50.0 mL of water. Then add 10.00 mL of 0.100 M NaOH. What’s the pH? I like an ICE chart here, just to keep track of things! Moles is better for my ICE chart because of dilution issues. 0.100 M * 0.0100 L = 0.00100 moles HCl initial 0.100 M NaOH * 0.0100 L = 0.00100 mol NaOH

17. Always start with a balanced equation. Which one? H++OH -  H2OH2O I0.001 mol - C-0.001 mol - End0.00 mol0 mol

18. So…. Nothing left….so is [H + ] = 0? Suddenly, one of my reactions is relevant: H 2 O + H 2 O  H 3 O+ + OH- The pH is 7 at equivalence because of the K w reaction.

19. 10.00 mL of 0.100 M HCl diluted with 50.0 mL of water. Then add 12.00 mL of 0.100 M NaOH. What’s in the beaker? Cl - H2OH2O H2OH2O H2OH2O Na + OH - Na +

20. I know what’s in the beaker… Now I ask: what reactions are possible? “What reactions are possible”? OH - + H 2 O  OH - + H 2 O Cl - + H 2 O  HCl + OH - Na + + H 2 O  NaOH + H + H 2 O + H 2 O  H 3 O + + OH - Cl - H2OH2O H2OH2O H2OH2O Na +

21. 10.00 mL of 0.100 M HCl diluted with 50.0 mL of water. Then add 12.00 mL of 0.100 M NaOH. What’s the pH? I like an ICE chart here, just to keep track of things! Moles is better for my ICE chart because of dilution issues. 0.100 M * 0.0100 L = 0.00100 moles HCl initial 0.100 M NaOH * 0.01200 L = 0.0012 mol NaOH

22. Always start with a balanced equation. Which one? H++OH -  H2OH2O I0.001 mol0.0012 mol- C-0.001 mol - End0.00 mol0.0002 mol

23. So…. 0.0002 mol OH - = 2.778x10 -3 M OH - 0.072 L pOH = - log(2.778x10 -3 M) = 2.556 pH = 14 – 2.556 = 11.444

24. 10.00 mL of 0.100 M HCl diluted with 50.0 mL of water. Then add 15.00 mL of 0.100 M NaOH. What’s the pH? I like an ICE chart here, just to keep track of things! Moles is better for my ICE chart because of dilution issues. 0.100 M * 0.0100 L = 0.00100 moles HCl initial 0.100 M NaOH * 0.01500 L = 0.0015 mol NaOH

25. Always start with a balanced equation. Which one? H++OH -  H2OH2O I0.001 mol0.0015 mol- C-0.001 mol - End0.00 mol0.0005 mol

26. So…. 0.0005 mol OH - = 6.667x10 -3 M OH - 0.075 L pOH = - log(6.667x10 -3 M) = 2.176 pH = 14 – 2.176 = 11.824

27. 10.00 mL of 0.100 M HOAc diluted with 50.0 mL of water What’s in the beaker? H2OH2O HOAc H2OH2O H2OH2O

28. I know what’s in the beaker… Now I ask: what reactions are possible? “What reactions are possible”? HOAc + H 2 O  OAc - + H 3 O + H 2 O + H 2 O  H 3 O + + OH - H2OH2O HOAc H2OH2O H2OH2O

29. Which reactions matter? HOAc + H 2 O  OAc- + H 3 O+ H2OH2O HOAc H2OH2O H2OH2O

30. 10.00 mL of 0.100 M HOAc diluted with 50.0 mL of water What’s the pH? 0.100 M * 10.0 mL = [HOAc]* 60.0 mL [HOAc] = 1.66 7 x10 -2 M (initial concentration) ICE-ICE-BABY-ICE-ICE

31. Always start with a balanced equation. Which one? HOAc+H2OH2O  OAc - +H3O+H3O+ I 1.66 7 x10 -2 M -00 C-x-+x E 1.66 7 x10 -2 -x -xx

32. So…. K a = 1.80x10 -5 = (x)(x) (1.66 7 x10 -2 M –X) Assume x<<1.667x10 -2 K a = 1.80x10 -5 = (x)(x) (1.66 7 x10 -2 M) 3.000x10 -7 = x 2 X= 5.48x10 -4 M (works…barely)

33. pH = - log (5.48x10 -4 ) = 3.26 HOAc+H2OH2O  OAc - +H3O+H3O+ I 1.66 7 x10 -2 M -00 C - 5.48x10 -4 M- 5.48x10 -4 M E 1.612x10 -2 -x-5.48x10 -4 M

34. 10.00 mL of 0.100 M HOAc diluted with 50.0 mL of water. Add 2.00 mL NaOH. What’s in the beaker? OAc - H2OH2O HOAc H2OH2O OAc - H2OH2O Na + HOAc Na + HOAc Why? NaOH + HOAc  NaOAc + H 2 O

35. I know what’s in the beaker… Now I ask: what reactions are possible? “What reactions are possible”? HOAc + H 2 O  OAc - + H 3 O + H 2 O + H 2 O  H 3 O + + OH - OAc - + H 2 O  HOAc + OH - (but this is the same as the first one) OAc - H2OH2O HOAc H2OH2O OAc - H2OH2O H+H+ HOAc H+H+

36. Which reactions matter? HOAc + H 2 O  OAc- + H 3 O+ OAc - H2OH2O HOAc H2OH2O OAc - H2OH2O H+H+ HOAc H+H+

37. 10.00 mL of 0.100 M HOAc diluted with 50.0 mL of water. Add 2.00 mL of NaOH. What’s the pH? 0.100 M * 0.010 L = 0.001 mol HOAc 0.100 M * 0.002 L = 0.0002 mol NaOH DOUBLE ICE- DOUBLE ICE-BABY- DOUBLE ICE- DOUBLE ICE

38. Always start with a balanced equation. Which one? HOAc+NaOH  NaOAc+H2OH2O I0.0010 mol 0.0002 mol0- C-0.0002 mol +0.0002- E0.0008 mol 0 mol0.0002

39. On to the 2 nd ICE chart HOAc+H2OH2O  OAc - +H3O+H3O+ I- C-x-+x E-

40. On to the 2 nd ICE chart HOAc+H2OH2O  OAc - +H3O+H3O+ I 0.0008 mol 0.062 L -0.0002 0.062 C-x-+x E-

41. THIS IS ACTUALLY A BUFFER!!!

42. So….H-H equation K a = 1.80x10 -5 = pK a = - log (1.80x10 -5 ) = 4.74 pH = 4.74 + log [OAc - ] [HOAc] pH = 4.74 + log (0.0002) = 4.14 (0.0008)

43. 10.00 mL of 0.100 M HOAc diluted with 50.0 mL of water. Add 5.00 mL NaOH. What’s in the beaker? OAc - H2OH2O HOAc H2OH2O OAc - H2OH2O Na + HOAc Na + HOAc Why? NaOH + HOAc  NaOAc + H 2 O

44. I know what’s in the beaker… Now I ask: what reactions are possible? “What reactions are possible”? HOAc + H 2 O  OAc - + H 3 O + H 2 O + H 2 O  H 3 O + + OH - OAc - + H 2 O  HOAc + OH - (but this is the same as the first one) OAc - H2OH2O HOAc H2OH2O OAc - H2OH2O Na + HOAc Na + HOAc

45. Which reactions matter? HOAc + H 2 O  OAc- + H 3 O+ OAc - H2OH2O HOAc H2OH2O OAc - H2OH2O Na + HOAc Na + HOAc

46. 10.00 mL of 0.100 M HOAc diluted with 50.0 mL of water. Add 5.00 mL of NaOH. What’s the pH? 0.100 M * 0.010 L = 0.001 mol HOAc 0.100 M * 0.005 L = 0.0005 mol NaOH DOUBLE ICE- DOUBLE ICE-BABY- DOUBLE ICE- DOUBLE ICE

47. Always start with a balanced equation. Which one? HOAc+NaOH  NaOAc+H2OH2O I0.0010 mol 0.0005 mol0- C-0.0005 mol +0.0005- E0.0005 mol 0 mol0.0005

48. On to the 2 nd ICE chart HOAc+H2OH2O  OAc - +H3O+H3O+ I0.0005- C-x-+x E-

49. So….H-H equation K a = 1.80x10 -5 = pK a = - log (1.80x10 -5 ) = 4.74 pH = 4.74 + log [OAc - ] [HOAc] pH = 4.74 + log (0.0005) = 4.74 (0.0005)

50. 10.00 mL of 0.100 M HOAc diluted with 50.0 mL of water. Add 10.00 mL NaOH. What’s in the beaker? OAc - H2OH2O H2OH2O H2OH2O Na + OAc- Na + OAc - Na+ Why? NaOH + HOAc  NaOAc + H 2 O

51. I know what’s in the beaker… Now I ask: what reactions are possible? “What reactions are possible”? H 2 O + H 2 O  H 3 O + + OH - OAc - + H 2 O  HOAc + OH - No more HOAc, no more K a OAc - H2OH2O Na+ H2OH2O OAc - H2OH2O Na + OAc- Na + OAc Na+

52. Which reactions matter? OAc - + H 2 O  HOAc + OH-

53. 10.00 mL of 0.100 M HOAc diluted with 50.0 mL of water. Add 10.00 mL of NaOH. What’s the pH? 0.100 M * 0.010 L = 0.001 mol HOAc 0.100 M * 0.010 L = 0.001 mol NaOH DOUBLE ICE- DOUBLE ICE-BABY- DOUBLE ICE- DOUBLE ICE

54. Always start with a balanced equation. Which one? HOAc+NaOH  NaOAc+H2OH2O I0.0010 mol 0.0005 mol0- C-0.0010 mol +0.001- E00 mol0.001

55. On to the 2 nd ICE chart OAc+H2OH2O  HOAc+OH - I0.001 mol 0.070 L -0.0 C-x-+x E 6.9x10 -2 -x -xx

56. So…. K b = K w = 1.0x10 -14 = 5.56x10 -10 K a 1.80x10 -5 5.56x10 -10 = (x)(x) 0.069-X Assume x<<0.069x10 -2 5.56x10 -10 = (x)(x) 0.069 3.836x10 -11 = x 2 X= 6.19x10 -6 M

57. pOH = - log 6.19x10 -6 = 5.2 pH = 14 – 5.2 = 8.8 OAc+H2OH2O  HOAc+OH - I0.001 mol 0.070 L -0.0 C -6.19x10 -6 M-+6.19x10 -6 M E 6.9x10 -2 -6.19x10 -6 M

58. So, for titrations in general… In general, titrations do not have endpoints at pH=7. That only occurs for a strong acid and a strong base. If either the acid or base is weak, then there is a conjugate acid and/or conjugate base floating around at equivalence!