1. Acid Base Titrations AP Chemistry Chapter 15

2. Titration Titrations are used to determine the amount of acid or base in a solution Titrant: the solution with known concentration delivered into the “unknown” Analyte: the substance of “unknown” concentration Equivalence point: the point at which all of the acid/base has been neutralized. Endpoint: the point at which the indicator changes color (ideally this is the same as the equivalence point) The pH is then plotted to track the titration!

3. Strong acid/base titrations NaOH + HCl  NaCl + H 2 O To compute the pH (or the [H+]) at any given point: 1. Determine the amount [H+] remaining…. 2. Divide by the total volume of the solution Since burets are graduated in mL, we can use a millimole (or mmole)… Molarity = moles solute Liters solution mmmm

4. Let’s look at a titration where an acid is being neutralized with a base in increments… 50.0 mL of 0.200 M HNO 3 with 0.10 M NaOH. Calculate pH at selected points where given quantities of NaOH are added. A. No NaOH added yet HNO 3  H + + NO 3 - 100% 0.200M[H + ] = 0.200 MpH = -log (0.200) = 0.699 B. 10.0 mL 0.100 M NaOH added 10.0 mL  0.100 mmole NaOH = 1.00 mmole NaOH 1 mL 1.00 mmole NaOH reacts completely with…. 50.0 mL  0.200 mmole HNO 3 = 10.00 mmole HNO 3 1 mL …only 1.00 mmole HNO 3 reacts, leaving 9.0 mmole HNO 3. [H + ] = 9.0 mmole HNO 3 = 0.15M 50.0 + 10.0 mL pH = -log (0.15) = 0.824

5. Calculate pH when 50.0 mL of 0.200 M HNO 3 titrated with ? mL 0.10 M NaOH. C. 20.0 mL (total) 0.100 M NaOH added 20.0 mL  0.100 mmole NaOH = 2.00 mmole NaOH 1 mL 2.00 mmole NaOH reacts completely with…. 50.0 mL  0.200 mmole HNO 3 = 10.00 mmole HNO 3 1 mL …only 2.00 mmole HNO 3 reacts, leaving 8.0 mmole HNO 3. [H + ] = 8.0 mmole HNO 3 = 0.114M 50.0 + 20.0 mL pH = -log (0.114) = 0.942

6. Calculate pH when 50.0 mL of 0.200 M HNO 3 titrated with ? mL 0.10 M NaOH. D. 50.0 mL (total) 0.100 M NaOH added 50.0 mL  0.100 mmole NaOH = 5.00 mmole NaOH 1 mL 5.00 mmole NaOH reacts completely with…. 50.0 mL  0.200 mmole HNO 3 = 10.00 mmole HNO 3 1 mL …only 5.00 mmole HNO 3 reacts, leaving 5.0 mmole HNO 3. [H + ] = 5.0 mmole HNO 3 = 0.05M 50.0 + 50.0 mL pH = -log (0.05) = 1.301

7. Calculate pH when 50.0 mL of 0.200 M HNO 3 titrated with ? mL 0.10 M NaOH. E. 100.0 mL (total) 0.100 M NaOH added 100.0 mL  0.100 mmole NaOH = 10.00 mmole NaOH 1 mL 10.00 mmole NaOH reacts completely with…. 50.0 mL  0.200 mmole HNO 3 = 10.00 mmole HNO 3 1 mL …all 10.00 mmole HNO 3 reacts, leaving 0 mmole HNO 3. [H + ] = 0 mmole HNO 3 = 0 M 50.0 + 100.0 mL pH = 7 The only source of [H + ]is from the water thus…

8. Calculate pH when 50.0 mL of 0.200 M HNO 3 titrated with ? mL 0.10 M NaOH. F. 150.0 mL (total) 0.100 M NaOH added 150.0 mL  0.100 mmole NaOH = 15.00 mmole NaOH 1 mL 15.00 mmole NaOH reacts completely with…. 50.0 mL  0.200 mmole HNO 3 = 10.00 mmole HNO 3 1 mL …all 10.00 mmole HNO 3 reacts, leaving 5.0 mmole OH- from the NaOH. [OH - ] = 5.0 mmole OH - = 0.025M 50.0 + 150.0 mL pOH = -log (0.025) = 1.602 pH = 12.40

9. Calculate pH when 50.0 mL of 0.200 M HNO 3 titrated with ? mL 0.10 M NaOH. G. 200.0 mL (total) 0.100 M NaOH added 200.0 mL  0.100 mmole NaOH = 20.00 mmole NaOH 1 mL 20.00 mmole NaOH reacts completely with…. 50.0 mL  0.200 mmole HNO 3 = 10.00 mmole HNO 3 1 mL …all 10.00 mmole HNO 3 reacts, leaving 10.0 mmole OH- from the NaOH. [OH - ] = 10.0 mmole OH - = 0.04M 50.0 + 200.0 mL pOH = -log (0.04) = 1.4 pH = 12.6

10. Gathering all of the information, the pH curve looks like this: The pH changes slowly at first… The closer to the equivalence point… …the more dramatic the pH change x

11. Weak titrated with strong In an acid/base titration in which one substance is strong and the other weak… The solution is not neutral at the equivalence point due to the hydrolysis of the salt. So we have to work with a series of buffer problems. Working first the stoichiometry problem, then the equilibrium problem.

12. Calculate pH when 50.0 mL of 0.100 M HC 2 H 3 O 2 is titrated with ? mL 0.10 M NaOH. A. No NaOH added yet.10 – x x x 1.8x10 -5 = x 2.10-x x=[H + ]=1.34x10 -3 M pH = 2.87 HC 2 H 3 O 2 H + + C 2 H 3 O 2 -  

13. B. 10.0 mL 0.100 M NaOH added 10.0 mL  0.100 mmole NaOH = 1.00 mmole NaOH 1 mL 1.00 mmole NaOH reacts completely with…. 50.0 mL  0.100 mmole HC 2 H 3 O 2 =5.00 mmole HC 2 H 3 O 2 1 mL …only 1.00 mmole HC 2 H 3 O 2 reacts, leaving 4.0 mmole HC 2 H 3 O 2 in 60 mL and 1.00 mmole C 2 H 3 O 2 - in 60 mL x = [H + ] = 7.2x10 -5 M pH = 4.14.0667 – x x.01667+x 1.8x10 -5 = x(.0167+x).0667-x HC 2 H 3 O 2 H + + C 2 H 3 O 2 -   Calculate pH when 50.0 mL of 0.100 M HC 2 H 3 O 2 is titrated with ? mL 0.10 M NaOH.

14. C. 25.0 mL 0.100 M NaOH added 25.0 mL  0.100 mmole NaOH = 2.5 mmole NaOH 1 mL 2.50 mmole NaOH reacts completely with…. 50.0 mL  0.100 mmole HC 2 H 3 O 2 =5.00 mmole HC 2 H 3 O 2 1 mL …only 2.50 mmole HC 2 H 3 O 2 reacts, leaving 2.5 mmole HC 2 H 3 O 2 in 75 mL and 2.5 mmole C 2 H 3 O 2 - in 75 mL x = [H + ] = 1.8x10 -5 M pH = 4.74.0333 – x x.0333+x 1.8x10 -5 = x(.0333+x).0333-x HC 2 H 3 O 2 H + + C 2 H 3 O 2 -   Calculate pH when 50.0 mL of 0.100 M HC 2 H 3 O 2 is titrated with ? mL 0.10 M NaOH. The the midway point since ½ of original acid is left.

15. D. 40.0 mL 0.100 M NaOH added 40.0 mL  0.100 mmole NaOH = 4.0 mmole NaOH 1 mL 4.0 mmole NaOH reacts completely with…. 50.0 mL  0.100 mmole HC 2 H 3 O 2 =5.00 mmole HC 2 H 3 O 2 1 mL …only 4.0 mmole HC 2 H 3 O 2 reacts, leaving 1.0 mmole HC 2 H 3 O 2 in 90 mL and 4.0 mmole C 2 H 3 O 2 - in 90 mL x = [H + ] = 4.5x10 -6 M pH = 5.35.0111 – x x.044+x 1.8x10 -5 = x(.044+x).0111-x HC 2 H 3 O 2 H + + C 2 H 3 O 2 -   Calculate pH when 50.0 mL of 0.100 M HC 2 H 3 O 2 is titrated with ? mL 0.10 M NaOH.

16. E. 50.0 mL 0.100 M NaOH added 50.0 mL  0.100 mmole NaOH = 5.0 mmole NaOH 1 mL 5.0 mmole NaOH reacts completely with…. 50.0 mL  0.100 mmole HC 2 H 3 O 2 =5.00 mmole HC 2 H 3 O 2 1 mL …all 5.0 mmole HC 2 H 3 O 2 reacts, leaving no H + from HC 2 H 3 O 2 in but 5.0 mmole C 2 H 3 O 2 - in 100 mL x = [OH - ] = 5.3x10 -6 M pOH = 5.28 so pH = 8.72.05 – x x x 5.6x10 -10 = x 2.05-x C 2 H 3 O 2 - + H 2 O OH - + HC 2 H 3 O 2   Calculate pH when 50.0 mL of 0.100 M HC 2 H 3 O 2 is titrated with ? mL 0.10 M NaOH. K b =10 -14 =5.6x10 -10 1.8x10 -5 This is the equivilance point Notice that the pH at the equivalence point of a strong base in a weak acid is always higher (more basic) than 7!

17. F. 60.0 mL 0.100 M NaOH added 60.0 mL  0.100 mmole NaOH = 6.0 mmole NaOH 1 mL 5.0 mmole NaOH reacts completely with…. 50.0 mL  0.100 mmole HC 2 H 3 O 2 =5.00 mmole HC 2 H 3 O 2 1 mL …all 5.0 mmole HC 2 H 3 O 2 reacts, leaving 5.0 mmole C 2 H 3 O 2 - in 110 mL and 1.0 mmol OH - from NaOH [OH - ] = 9.1x10 -3 M pOH = 2.04 so pH = 11.96 Calculate pH when 50.0 mL of 0.100 M HC 2 H 3 O 2 is titrated with ? mL 0.10 M NaOH. Now the major species are: Na +, C 2 H 3 O 2 -, OH -, H 2 O 2 bases, but C 2 H 3 O 2 - is weak

18. G. 75.0 mL 0.100 M NaOH added 75.0 mL  0.100 mmole NaOH = 7.5 mmole NaOH 1 mL 5.0 mmole NaOH reacts completely with…. 50.0 mL  0.100 mmole HC 2 H 3 O 2 =5.00 mmole HC 2 H 3 O 2 1 mL …all 5.0 mmole HC 2 H 3 O 2 reacts, leaving 5.0 mmole C 2 H 3 O 2 - in 125 mL and 2.5 mmol OH - from NaOH [OH - ] = 2.0x10 -2 MpOH = 1.07 so pH = 12.30 Calculate pH when 50.0 mL of 0.100 M HC 2 H 3 O 2 is titrated with ? mL 0.10 M NaOH. Again the major species are: Na +, C 2 H 3 O 2 -, OH -, H 2 O using the stronger of the 2 bases… pH is determined by the OH - excess

19. Equivalence point 8.72  4.74 was ½ way to the equivalence point. Here, pH = pKa. This is where pH changes least rapidly…a solution with this pH makes a good buffer. It is important to note that equivalence does not mean neutral.

20. The titration of a strong base with a strong acid.

21. The titration of a weak base with a strong acid has a different looking titration curve.

22. 15.9 If a 50.0 mL sample of 0.100 M HCN (K a =6.2x10 -10 ) is titrated with 0.10 M NaOH, calculate the pH of the solution. a. After 8.00 mL of 0.10 M NaOH has been added. 8.0 mL  0.100 mmole NaOH =.80 mmole NaOH 1 mL.8 mmole NaOH reacts completely with… 50.0 mL  0.100 mmole HCN =5.0 mmole HCN 1 mL …only.8 mmole HCN reacts (1:1 ratio) leaving 4.2 mmole H + from HCN and.8 mmole CN - in 58 mL x = [H + ] = 3.25x10 -9 M pH = 8.49.0724 – x x.0138+x 6.2x10 -10 = x(.0138+x).0724-x HCN H + + CN -  

23. 15.9 If a 50.0 mL sample of 0.100 M HCN (K a =6.2x10 -10 ) is titrated with 0.10 M NaOH, calculate the pH of the solution. B. At the ½ point of the titration 2.5 mmoleNaOH  1 mL = 25 mL NaOH.100 mmole NaOH Half of the 5.0 mmole HCN is 2.5 mmole HCN (also 2.5 mmole CN - ) in what volume? 50.0 mL  0.100 mmole HCN =5.0 mmole HCN 1 mL x = [H + ] = 6.2x10 -10 M pH = 9.21.0333 – x x.0333+x 6.2x10 -10 = x(.0333+x).0333-x HCN H + + CN -  

24. 15.9 If a 50.0 mL sample of 0.100 M HCN (K a =6.2x10 -10 ) is titrated with 0.10 M NaOH, calculate the pH of the solution. C. At the equivalence point 5.0 mmoleNaOH  1 mL = 50 mL NaOH.100 mmole NaOH All of the 5.0 mmole HCN reacts and leaves 5.0 mmole of the salt CN - in what volume of NaOH? (remember the 1:1 ration HCN:NaOH) 50.0 mL  0.100 mmole HCN =5.0 mmole HCN 1 mL x = [OH - ] = 8.9x10 -4 M pH = 10.96.05 – x x x 1.6x10 -5 = x 2.05-x CN- + H 2 O OH - + HCN   Remember Kb=Kw Ka

25. If we look back at the previous 2 examples, the equivalence point occurred in both when 50.0 mL of.10 M NaOH reacted with 50 mL of the weak acid. It is the amount of the acid, not its strength, that determines the equivalence point. The pH at the equivalence point is determined by the acid strength.

26. 15.10 What is the Ka value for an unknown weak acid if 2.00 mmole of the solid acid is dissolved in 100.0 mL water and the solution is titrated with 0.0500 M NaOH. After 20.0 mL NaOH has been added, the pH is 6.00. HA  H + + A - 20.0 mL  0.050 mmol NaOH = 1.00 mmol NaOH 1 mL 1.00 mmol 120 mL 8.33x10 -3 -x x 8.33x10 -3 +x Since pH = 6 …..[H+] = 1.0x10 -6 M Ka = (1.0x10 -6 )(8.33x10 -3 + 1.0x10 -6 ) (8.33x10 -3 – 1.0x10 -6 ) =1.0x10 -6

27. Determining Equivalence Point Either use a pH meter…monitor pH, plot the curve…the center of the vertical region is the equivalence point or… Use an indicator. The endpoint is indicated by a color change. Ideally the endpoint is the same as the equivalence point.

28. Acid-Base indicators Used to detect the equivalence point in an acid- base titration These work on LeChatelier’s Principle HIn  H + + In - clear pink Changes in equilibrium result in color changes. Indicators are usually weak organic bases or acids … acid form has different color than conjugate base form. Indicators change colors at different pH’s. We select an indicator that changes color at the expected equivalence point.

29. Equivalence point vs endpoint Ideally they are the same Equivalence…when all acid & base neutralized Endpoint…when indicator changes color

30. Hypothetical indicator, HIn with a Ka = 1.8x10 -8 HIn H + + In -   red blue Ka = [H + ][In - ] [HIn] Ka = [In - ] [H + ] [HIn] If we add indicator to [H+] =.01 M acidic solution… = 1.0x10 -8.01 = 1. 10 000 000 Predominant color is HIn (red) …as OH - added, equation shifts right (less HIn) As ratio gets closer to 1/10, color change can be detected by the eye.

31. 15.11 Bromthymol blue, an indicator with a Ka=1.0x10 -7, is yellow in its HIn form and blue in its In - form. Suppose we put a few drops of this indicator in a strongly acidic solution. If the solution is then titrated with NaOH, at what pH will the indicator color change first be visible? HIn + OH - H 2 O + In -   yellow blue Ka = 1.0x10 -7 = [H + ][In - ] [HIn] [H + ] 1 10 As ratio gets closer to 1/10, color change can be detected by the eye. [H + ] = 1.0x10 -6 pH = -log(1.0x10 -6 ) = 6 Or we can use the Henderson-Hasselbalch equation….

32. Henderson-Hasselbalch pH = pKa + log [base] [acid] In the previous problem, we assumed a color change is visible when we have a 1/10 ratio pH = pKa + log [1] [10] pH = pKa + -1 For bromthymol blue Ka=1.0x10 -7 or pKa = 7 pH = 7 + -1 = 6

33. HIn H + + In -   yellow blue When a basic solution is titrated, HIn exists initially as In - but as acid is added, more HIn is formed so the color change is visible when pH = pKa + log [base] [acid] pH = pKa + log [In-] [HIn] 10 1 So for the same bromthymol blue, titration of a base starts as blue and changes to yellow at…. pH = pKa + log [10] [1] = pKa + 1= 7 + 1 = 8 The useful range of bromthymol blue is pKa + 1 or pH range of 6 to 8.

34. We then choose an indicator so the indicator endpoint and titration equivalence point are close. Choose an indicator for the titration of 100.00 mL of 0.100 M HCl with 0.100 M NaOH. Both are strong so pH at equivalence pt is 7 Since we are titrating an acid, the indicator is in acid form initially so the change in color is observed as… pH = pKa + log [1] [10] 7 = pKa + -1pKa = 8 or Ka = 1.0x10 -8

35. Now continue chapter 15 with show for Solubility Equilibria