Analytical chemistry ocw

1. STK1094 Analytical Chemistry 1 Dayang Norafizan binti Awang Chee Faculty of Resource Science and Technology Universiti Malaysia Sarawak This OpenCourseWare@UNIMAS and its related course materials are licensed under a Creative Commons Attribution- NonCommercial - ShareAlike 4.0 International License.

2. Learning Objectives At the end of this lesson, students should be able to: Describe the difference between an “endpoint” and an “equivalence point” in an acid-base titration Identify the equivalence point in an acid-base titration from the pH titration curve Illustrate titration of a weak acid with a strong base .

3. Defining Terms Standard solution: A reagent of a known concentration which used in the titrimetric analysis Titration: This is performed by adding a standard solution from burette or other liquid-dispensing device to a solution of the analyte until the point at which the reaction is believe to be completed Equivalence point: Occurs in a titration at the point in which the amount of added titrant is chemically equivalent to the amount of analyte in a sample Back-titration : This is a process in which an excess of the standard titrant is added, and the amount of the excess is determined by back titration with a second standard titrant. In this instance, the equivalent point corresponds with the amount of initial titrant is chemically equivalent to the amount of analyte plus the amount of back titrant.

4. Defining Terms End point: The point in titration when a physical change occurs that is associated with the condition of chemical equivalence Indicators are used to give an observable physical change (end point) or at near the equivalence point by adding them to the analyte . The difference between end point and equivalence point should be very small and this difference is referred to as titration error. Titration error, E t E t = V ep – V eq V ep is the actual volume used to get to the end point. V eq is the theoretical value of reagent required to reach the end point.

5. 5 Acid-Base Titrations A quick and accurate method for determining acidic or basic substances in many samples . The titrant is typically a strong acid or base . The sample species can be either a strong or weak acid or base.

6. 6 Acid-Base Titration Types of acid-base titrations: 1)strong acid – strong base titration 2)weak acid – strong base titration 3)strong acid – weak base titration 4) polyprotic acid – strong base titration 5) polybasic base – strong acid titration

7. 7 Strong Acid - Strong Base Titration In strong acid – strong base titration , there are three regions of the titration curve that represent different kinds of calculations : before equivalence point at equivalence point after equivalence point

8. Titration curve Strong acid titrated with a strong base: The net reaction is H 3 O + + OH - → 2H 2 O Before the equivalence point, acid is present in excess pH is determined by the concentration of excess HCl [ ] =   8

9. Strong acid titrated with a strong base At equivalence point, moles of acid and moles of base are equal. At equivalence point, [H 3 O + ] = [OH - ] pKw = 14 = pH + pOH pH = 7 So, the equivalence point for strong acid/base is always a pH=7 9

10. Strong acid titrated with a strong base Overtitration Pass the equivalence point, we don’t have any acid remaining. All that we are doing is diluting our titrant. ] = pH = 14-pOH Eg . Construct a titration curve for the titration of 100 mL 0.1 M HCl with 0.1 M NaOH Volume of NaOH needed to reach eq. point Moles HCl = moles NaOH V NaOH = 100.0 mL 2) Before addition of NaOH pH = - log [0.1] = 1 3) After addition of 10mL NaOH ] = = 0.082 M, pH= 1.09   10

11. 11 mL titrant Total mL [H 3 O + ] pH 100 0.10 1.00 10 110 0.082 1.09 20 120 0.067 1.17 30 130 0.054 1.28 40 140 0.043 1.37 50 150 0.033 1.48 60 160 0.025 1.60 70 170 0.018 1.74 80 180 0.011 1.96 90 190 0.0053 2.28

12. Titration curve 4) At equivalence point Equivalence point, moles of HCl = moles of NaOH Since neither is in excess, pH is determined by K w K w = 1.00 x 10 -14 = [H 3 O + ][OH - ] = [H 3 O + ] 2 [H 3 O + ] = 1.00 x 10 -7 pH= 7 Note that for the first 90 mL of titration, pH = 2.28 At eq. point, the pH value jump of 4.72 pH unit 12

13. Titration curve 5) Overtitration Account for the dilution of titrant 10 mL overtitration [OH - ] = moles excess NaOH = MV NaOH - MV HCl V total = (0.1M)(110mL)-(0.1M)(100mL) 210 mL = 0.0048M pOH = 2.32 pH = 14-2.32 = 11.68 13

14. 14 mL titrant Total volume [OH - ] pH 110 210 0.0048 11.68 120 220 0.0091 11.96 130 230 0.013 12.11 140 240 0.017 12.23 150 250 0.020 12.30 160 260 0.023 12.36 170 270 0.026 12.41 180 280 0.029 12.46 190 290 0.031 12.49 200 300 0.033 12.52

15. Titration curve Titration of a strong base with a strong acid: If we plot pOH rather than pH, the result still look identical. Typically, we still plot pH verses mL titrant, so the curve is inverted. 15

16. Titration of weak acids & weak bases with strong titrant Must concerned with conjugate acid/base pairs & their equilibrium 16

17. Titration of weak acids & weak bases with strong titrant Before titration: If the sample is weak acid , then use = [H 3 O + ]=[A - ] Calculate the pH value   17 If the sample is weak base , then use = [OH - ]=[HA] Calculate the pH value = 14 - pOH  

18. Titration of weak acids & weak bases with strong titrant Before equivalence point: Equilibrium expression used is the Henderson- Hasselbalch equation Starting with an acid pH= Starting with base pH= )   18

19. Titration of weak acids & weak bases with strong titrant At equivalence point All sample is converted to its conjugate form If the sample was an acid-solve the pH using K B relationship If the sample was a base-solve the pH using K A relationship + = 14   19

20. Titration of weak acids & weak bases with strong titrant Overtitration : Identical to strong acid/strong base example. Need to account for the amount of excess titrant & how much it has been diluted. Eg . 100 mL solution of 0.1 M benzoic acid is titrated with 0.1 M NaOH . Construct a titration curve. For benzoic acid K a =6.31 x 10 -5 p K a =4.20 Volume of NaOH needed to reach eq. point Moles C 5 H 6 COOH = moles NaOH V NaOH = 100.0 mL 20

21. Titration of weak acids & weak bases with strong titrant 2) Before titration: = ]= [ ] Assume [A-] is negligible compared to [HA] = = = = 0.025 M pH= 2.60   21

22. Titration of weak acids & weak bases with strong titrant 3) After addition of 10mL NaOH Henderson- Hasselbalch equation pH = pKa + log [C 5 H 6 COO-] [C 5 H 6 COOH] [C 5 H 6 COOH] = moles unreacted C 5 H 6 C OOH = MV C5H6COOH - MV NaOH V total V total = (0.1M)(100mL)-(0.1M)(10mL) 110 mL = 0.082M [C 5 H 6 COO - ] = moles NaOH added = MV NaOH V total V total = (0.1M)(10mL) 110 mL = 0.009M pH = 4.2 + log (0.009/0.082) = 3.24 Calculate other point by repeating this process 22

23. Titration of weak acids & weak bases with strong titrant 23 Note: 50mL titrant pH = pK A Also, there was only a change of 1.91 in the pH from 10mL to 90mL titrant . mL titrant pH 2.60 10 3.24 20 3.60 30 3.83 40 4.02 50 4.20 60 4.38 70 4.57 80 4.80 90 5.15 mL titrant

24. Titration of weak acids & weak bases with strong titrant 4) At equivalence point 100mL titrant: All acid has been converted to its conjugate base – benzoate Use K B relationship. = = / = n benzoic acid = n NaOH Predominate ion in solution is A - , which is a weak base [A-] = moles acid/ total volume = 0.05M We have diluted the sample & the total volume at this point is 200 mL. We can assume that [benzoic acid] is negligible compared to [benzoate].   24

25. Titration of weak acids & weak bases with strong titrant 4) At equivalence point C 5 H 6 COO - ( aq ) + H 2 O (l) OH - ( aq ) + C 5 H 6 COOH =   25 mL titrant

26. Titration of weak acids or weak bases with strong titrant 5) Overtitration Need to account for the dilution of titrant. Eg : 10 mL excess. 26

27. Titration of weak acids or weak bases with strong titrant 27 mL titrant Total volume [OH - ] pH 110 210 0.0048 11.68 120 220 0.0091 11.96 130 230 0.013 12.11 140 240 0.017 12.23 150 250 0.020 12.30

28. © 2009, Prentice-Hall, Inc. Titrations of Polyprotic Acids When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation. Titration curve for the reaction of 50.0 mL of 0.10 M H 3 PO 3 with 0.10 M NaOH

29. Titrations of Polybasic Base 8- 29

30. Self-Reflection What is the difference between end-point and equivalence point? How to build the titration curve for strong acid/strong base with weak acid/weak base and vice versa?