1. A titration curve is a graph which shows how the pH of an acid solution changes as a base is added to it, or how the pH of a base solution changes as an acid is added to it. Strong Acid-Strong Base Titration Curves

2. Here, we’ll consider the addition a strong base to a solution which is initially a strong acid. Titration Curve for the Addition of a Strong Base to a Strong Acid

3. The base is in the buret and the acid is in the beaker below it 0.0 Strong Base 0.100 M NaOH Strong Acid 0.100 M HCl

4. The strong base we’ll use in our example is 0.100 M NaOH 0.0 Strong Base 0.100 M NaOH Strong Acid 0.100 M HCl

5. And the strong acid we’ll use is 0.100 M HCl. We have added 25.0 mL of 0.100 M HCl to the beaker. 0.0 Strong Base 0.100 M NaOH Strong Acid 0.100 M HCl 25.0 mL

6. A pH meter will be used to monitor the pH of the mixture in the beaker below the buret. 0.0 Strong Base 0.100 M NaOH Strong Acid 0.100 M HCl pH Meter 25.0 mL

7. What we’ll do is draw a graph of the pH in the beaker versus the Volume of NaOH added to the HCl in the beaker. 0.0 0.100 M HCl 0.100 M NaOH Graph of pH VS. Volume of 0.100 M NaOH Added 25.0 mL

8. W e’ll start with the pH before any NaOH is added at all. 0.0 0.100 M HCl 0.100 M NaOH 25.0 mL

9. At this point, we have zero moles of NaOH present and (25.0 mL or 0.0250 Litres) times 0.100 moles per litre, equals 0.0025 moles of HCl. 0.0 0.100 M HCl 0.100 M NaOH 25.0 mL 0.0000 mol0.0025 mol

10. Because all we have in the beaker is 0.100 M HCl, the hydronium ion concentration is 0.100 molar 0.0 0.100 M HCl 0.100 M NaOH 25.0 mL

11. and the pH is equal to 1.00 0.0 0.100 M HCl 0.100 M NaOH 25.0 mL 1.00

12. Now we’ll open the stopcock and slowly add 10.0 mL of NaOH to the beaker. Watch what happens to the pH (click). 0.0 0.100 M NaOH 1.00 1.37 10.0 0.100 M HCl 25.0 mL 1.00 1.37

13. As a result of adding 10.0 mL of NaOH, the pH rose to 1.37. 0.0 10.0 1.37 0.100 M NaOH

14. It can be shown that when 10.0 mL of 0.100 M NaOH is added to 25.00 mL of 0.100 M HCl, that we’ve added 0.001 moles of NaOH to 0.0025 moles of HCl. 0.0 10.0 0.100 M NaOH 0.0010 mol0.0025 mol 1.37

15. So the acid, HCl is in excess. This explains why the pH is still quite low at this point. 0.0 10.0 0.100 M NaOH 0.0010 mol0.0025 mol HCl is in Excess 1.37

16. Now, we’ll add more NaOH to the beaker until we get to a volume of 24.0 mL of NaOH added. Watch what happens to the pH (click) 0.0 10.0 0.100 M NaOH 1.37 24.0 2.69

17. The pH at 24.0 mL has gone up to 2.69. Notice it is starting to increase at a faster rate. 0.0 10.0 0.100 M NaOH 2.69 24.0

18. Calculations show that we have now added 0.0024 moles of NaOH to 0.0025 moles of HCl. The acid is still in excess, but not by as much as it was before. Therefore the pH is higher. 0.0 10.0 0.100 M NaOH 2.69 24.0 0.0024 mol0.0025 mol

19. Now we’ll slowly add just one more mL of NaOH, to give us a total volume of 25.0 mL of NaOH added. Notice how quickly the pH rises here. (Click) 0.0 10.0 2.69 24.0 25.0 0.100 M NaOH 7.00

20. The pH at this point has gone all the way up to 7.00. 0.0 10.0 7.00 24.0 25.0 0.100 M NaOH pH = 7.00

21. Remember if we assume we’re at 25°C, a pH of 7, means the solution is now neutral. 0.0 10.0 7.00 24.0 25.0 0.100 M NaOH pH = 7.00 Neutral

22. This is a special point in this titration. It’s called the equivalence point. 0.0 10.0 7.00 24.0 25.0 0.100 M NaOH Equivalence Point

23. At the equivalence point we added 25.0 mL or 0.0250 L of 0.100 M NaOH, which is 0.0025 moles, to 0.0025 moles of HCl. The equivalence point is defined as the point at which just enough base is added to neutralize all of the acid. 0.0 10.0 7.00 24.0 25.0 0.100 M NaOH 0.0025 mol Equivalence Point

24. The NaOH and HCl will completely react with each other, with no excess of either reactant (click).The coefficients in the balanced equation are all 1 So we see that 0.0025 moles of the salt NaCl is formed in this neutralization. 0.0 10.0 7.00 24.0 25.0 0.100 M NaOH 0.0025 mol 0.000 mol

25. Like all salts formed from the neutralization of a strong acid with a strong base, NaCl is a neutral salt. 0.0 10.0 7.00 24.0 25.0 0.100 M NaOH Strong BaseStrong Acid Neutral Salt

26. At the equivalence point, there is no strong base or strong acid remaining, 0.0 10.0 7.00 24.0 25.0 0.100 M NaOH 0.00 mol Equivalence Point

27. All there is present is water and a neutral salt. 0.0 10.0 7.00 24.0 25.0 0.100 M NaOH 0.00 mol Neutral Salt 0.0025 mol

28. And that’s why the pH at the equivalence point of all strong acid – strong base titrations is 7.00. 0.0 10.0 7.00 24.0 25.0 0.100 M NaOH 0.00 mol Neutral Salt 0.0025 mol

29. 11.29 After the equivalence point, we’ll slowly add just one more mL of NaOH to give us a total of 26 mL of NaOH added. Notice how quickly the pH increases here. (click) 0.0 10.0 7.00 24.0 25.0 11.29 26.0 0.100 M NaOH

30. 11.29 Notice that it went from 7.00, all the way up to 11.29 with the addition of just 1 mL of NaOH. We can see that small changes in the volume of NaOH make big changes in pH when we’re close to the equivalence point. 0.0 10.0 24.0 25.0 11.29 0.100 M NaOH 26.0 Equivalence Point

31. 11.29 At the 26 mL mark, we have added 0.0026 moles of NaOH to 0.0025 moles of HCl. So the base, NaOH is now in excess. Therefore, the pH is above 7. 0.0 10.0 24.0 25.0 11.29 0.100 M NaOH 26.0 0.0026 mol0.0025 mol Equivalence Point

32. 11.29 Now we’ll add NaOH until the volume of NaOH added is 50 mL. Notice the pH continues to go up, but not as quickly as it did when we were close to the equivalence point. 0.0 10.0 24.0 25.0 11.29 26.0 12.52 50.0

33. 11.29 The pH after adding 50 mL of NaOH to 25 mL of HCl is 12.52. 0.0 10.0 24.0 25.0 12.52 26.0 50.0

34. 11.29 At the 50 mL mark, we’ve added 0.005 moles of NaOH to 0.0025 moles of HCl. The base, NaOH is in excess enough to make the solution quite basic, hence it has a high pH. 0.0 10.0 24.0 25.0 12.52 26.0 50.0 0.0050 mol 0.0025 mol

35. Now, we’ll take a look at a strong acid-strong base titration curve and review its main features.

36. Because we are starting with a strong acid, the acidity typically starts out high, so the pH starts out with a low value. For example in 0.10 M HCl, the pH is equal to 1. pH starts with a low value

37. When we add the base, the pH goes up gradually, but when we get near the equivalence point, the slope of the titration curve shows a sharp increase. Near the equivalence point, there is a sharp increase in the slope of the curve

38. In a strong acid-strong base titration curve, there is a long section of the graph that is almost vertical. Long “almost vertical” section

39. Right in the center of the almost vertical portion of all titration curves is where the equivalence point is represented. Equivalence Point

40. It is important to remember that the pH at the equivalence point of ANY strong acid- strong base titration is always equal to 7.00. Equivalence Point The pH at the Equivalence Point of a Strong Acid-Strong Base Titration Curve is ALWAYS Equal to 7.00

41. This is because at the equivalence point the strong acid and strong base completely neutralize each other, Equivalence Point The pH at the Equivalence Point of a Strong Acid-Strong Base Titration Curve is ALWAYS Equal to 7.00 HA + BOH  H 2 O + BA Neutralize Each Other

42. And all were left with is water and a neutral salt. So the mixture is now neutral and the pH is equal to 7.00 Equivalence Point The pH at the Equivalence Point of a Strong Acid-Strong Base Titration Curve is ALWAYS Equal to 7.00 HA + BOH  H 2 O + BA Water and a Neutral Salt

43. After the equivalence point, when the base is in excess, the curve resembles a reverse reflection of what it was before the equivalence point. Notice the curve has a sort of S shape.

44. The best indicator for the titration depicted by this curve has a pH range that is somewhere on this almost vertical straight section

45. So any indicator that has a pH range somewhere between 3.6 and 10.4, would be suitable for this titration.

46. So theoretically any of these indicators would be suitable to use for this titration.

47. We see that the volumes corresponding to these two points have very little difference in value, so even a drop of two of base would be enough to change the colour of any of these indicators.

48. Although any of these indicators are suitable, the best indicators are ones which have the pH of the actual equivalence point within their range.

49. Bromthymol blue, Phenol Red, or Neutral red would all be excellent indicators for all strong acid-strong base titrations, because the pH at the equivalence point of all of strong acid-strong base titrations is 7.00, and 7 is within the ranges for these 3 indicators

50. The volume of base required to reach the equivalence point of any titration can be determined by drawing a straight vertical line down from the equivalence point. Equivalence Point

51. The value on the x axis where this line hits the axis is the volume of base required. We can see that in this titration, it is 25 mL. Equivalence Point