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Acid Base Titrations

Acid Base Titrations. AP Chemistry Chapter 15. 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

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Acid Base Titrations

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  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 + H2O 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 m m

  4. 100% HNO3 H+ + NO3- 50.0 mL of 0.200 M HNO3 with 0.10 M NaOH. Calculate pH at selected points where given quantities of NaOH are added. Let’s look at a titration where an acid is being neutralized with a base in increments… A. No NaOH added yet [H+] = 0.200 M pH = -log (0.200) = 0.699 0.200M 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 HNO3 = 10.00 mmole HNO3 1 mL …only 1.00 mmole HNO3 reacts, leaving 9.0 mmole HNO3. [H+] = 9.0 mmole HNO3 = 0.15M 50.0 + 10.0 mL pH = -log (0.15) = 0.824

  5. Calculate pH when 50.0 mL of 0.200 M HNO3 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 HNO3 = 10.00 mmole HNO3 1 mL …only 2.00 mmole HNO3 reacts, leaving 8.0 mmole HNO3. [H+] = 8.0 mmole HNO3 = 0.114M 50.0 + 20.0 mL pH = -log (0.114) = 0.942

  6. Calculate pH when 50.0 mL of 0.200 M HNO3 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 HNO3 = 10.00 mmole HNO3 1 mL …only 5.00 mmole HNO3 reacts, leaving 5.0 mmole HNO3. [H+] = 5.0 mmole HNO3 = 0.05M 50.0 + 50.0 mL pH = -log (0.05) = 1.301

  7. Calculate pH when 50.0 mL of 0.200 M HNO3 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 HNO3 = 10.00 mmole HNO3 1 mL …all 10.00 mmole HNO3 reacts, leaving 0 mmole HNO3. [H+] = 0 mmole HNO3 = 0 M 50.0 + 100.0 mL The only source of [H+]is from the water thus… pH = 7

  8. Calculate pH when 50.0 mL of 0.200 M HNO3 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 HNO3 = 10.00 mmole HNO3 1 mL …all 10.00 mmole HNO3 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 HNO3 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 HNO3 = 10.00 mmole HNO3 1 mL …all 10.00 mmole HNO3 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 closer to the equivalence point… The pH changes slowly at first… x …the more dramatic the pH change

  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. HC2H3O2 H+ + C2H3O2-  Calculate pH when 50.0 mL of 0.100 M HC2H3O2 is titrated with ? mL 0.10 M NaOH. A. No NaOH added yet 1.8x10-5= x2 .10-x .10 – x x x x=[H+]=1.34x10-3 M pH = 2.87

  13. HC2H3O2 H+ + C2H3O2-  Calculate pH when 50.0 mL of 0.100 M HC2H3O2 is titrated with ? mL 0.10 M NaOH. 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 HC2H3O2 =5.00 mmole HC2H3O2 1 mL …only 1.00 mmole HC2H3O2 reacts, leaving 4.0 mmole HC2H3O2 in 60 mL and 1.00 mmole C2H3O2- in 60 mL .0667 – x x .01667+x 1.8x10-5= x(.0167+x) .0667-x x = [H+] = 7.2x10-5 M pH = 4.14

  14. HC2H3O2 H+ + C2H3O2-  Calculate pH when 50.0 mL of 0.100 M HC2H3O2 is titrated with ? mL 0.10 M NaOH. 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 HC2H3O2 =5.00 mmole HC2H3O2 1 mL …only 2.50 mmole HC2H3O2 reacts, leaving 2.5 mmole HC2H3O2 in 75 mL and 2.5 mmole C2H3O2- in 75 mL The the midway point since ½ of original acid is left. .0333 – x x .0333+x 1.8x10-5= x(.0333+x) .0333-x x = [H+] = 1.8x10-5 M pH = 4.74

  15. HC2H3O2 H+ + C2H3O2-  Calculate pH when 50.0 mL of 0.100 M HC2H3O2 is titrated with ? mL 0.10 M NaOH. 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 HC2H3O2 =5.00 mmole HC2H3O2 1 mL …only 4.0 mmole HC2H3O2 reacts, leaving 1.0 mmole HC2H3O2 in 90 mL and 4.0 mmole C2H3O2- in 90 mL .0111 – x x .044+x 1.8x10-5= x(.044+x) .0111-x x = [H+] = 4.5x10-6 M pH = 5.35

  16. C2H3O2- + H2O OH- + HC2H3O2  Calculate pH when 50.0 mL of 0.100 M HC2H3O2 is titrated with ? mL 0.10 M NaOH. 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 HC2H3O2 =5.00 mmole HC2H3O2 1 mL …all 5.0 mmole HC2H3O2 reacts, leaving no H+ from HC2H3O2 in but 5.0 mmole C2H3O2- in 100 mL Kb=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! .05 – x x x 5.6x10-10= x2 .05-x x = [OH-] = 5.3x10-6 M pOH = 5.28 so pH = 8.72

  17. Calculate pH when 50.0 mL of 0.100 M HC2H3O2 is titrated with ? mL 0.10 M NaOH. 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 HC2H3O2 =5.00 mmole HC2H3O2 1 mL …all 5.0 mmole HC2H3O2 reacts, leaving 5.0 mmole C2H3O2- in 110 mL and 1.0 mmol OH- from NaOH Now the major species are: Na+, C2H3O2-, OH-, H2O 2 bases, but C2H3O2- is weak [OH-] = 9.1x10-3 M pOH = 2.04 so pH = 11.96

  18. Calculate pH when 50.0 mL of 0.100 M HC2H3O2 is titrated with ? mL 0.10 M NaOH. 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 HC2H3O2 =5.00 mmole HC2H3O2 1 mL …all 5.0 mmole HC2H3O2 reacts, leaving 5.0 mmole C2H3O2- in 125 mL and 2.5 mmol OH- from NaOH Again the major species are: Na+, C2H3O2-, OH-, H2O using the stronger of the 2 bases… pH is determined by the OH- excess [OH-] = 2.0x10-2 M pOH = 1.07 so pH = 12.30

  19. 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. Equivalence point 8.72 

  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 (Ka=6.2x10-10) is titrated with 0.10 M NaOH, calculate the pH of the solution.  HCN H+ + CN-  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 .0724 – x x .0138+x x = [H+] = 3.25x10-9 M 6.2x10-10= x(.0138+x) .0724-x pH = 8.49

  23. 15.9 If a 50.0 mL sample of 0.100 M HCN (Ka=6.2x10-10) is titrated with 0.10 M NaOH, calculate the pH of the solution.  HCN H+ + CN-  B. At the ½ point of the titration 50.0 mL  0.100 mmole HCN =5.0 mmole HCN 1 mL Half of the 5.0 mmole HCN is 2.5 mmole HCN (also 2.5 mmole CN-) in what volume? 2.5 mmoleNaOH  1 mL = 25 mL NaOH .100 mmole NaOH .0333 – x x .0333+x 6.2x10-10= x(.0333+x) .0333-x x = [H+] = 6.2x10-10 M pH = 9.21

  24. 15.9 If a 50.0 mL sample of 0.100 M HCN (Ka=6.2x10-10) is titrated with 0.10 M NaOH, calculate the pH of the solution.  CN- + H2O OH- + HCN  C. At the equivalence point 50.0 mL  0.100 mmole HCN =5.0 mmole HCN 1 mL 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) 5.0 mmoleNaOH  1 mL = 50 mL NaOH .100 mmole NaOH Remember Kb=Kw Ka .05 – x x x 1.6x10-5= x2 .05-x x = [OH-] = 8.9x10-4 M pH = 10.96

  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. Ka = (1.0x10-6)(8.33x10-3 + 1.0x10-6) (8.33x10-3 – 1.0x10-6) HA  H+ + A- 8.33x10-3 -x x 8.33x10-3+x 1.00mmol1.00mmol 120 mL 120 mL Since pH = 6 …..[H+] = 1.0x10-6 M =1.0x10-6 20.0 mL  0.050 mmol NaOH = 1.00 mmol NaOH 1 mL

  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. HIn H+ + In-  Hypothetical indicator, HIn with a Ka = 1.8x10-8 redblue Ka = [H+][In-] [HIn] Ka = [In-] [H+] [HIn] = 1.0x10-8 .01 = 1 . 10 000 000 If we add indicator to [H+] = .01 M acidic solution… 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- H2O + In-  yellowblue Ka = 1.0x10-7 = [H+][In-] [HIn] [H+] 1 10 [H+] = 1.0x10-6 pH = -log(1.0x10-6) = 6 As ratio gets closer to 1/10, color change can be detected by the eye. 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-  10 1 pH = pKa + log [base] [acid] yellowblue 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 [In-] [HIn] 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. Choose an indicator for the titration of 100.00 mL of 0.100 M HCl with 0.100 M NaOH. We then choose an indicator so the indicator endpoint and titration equivalence point are close. 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 + -1 pKa = 8 or Ka = 1.0x10-8

  35. Now continue chapter 15 with show for Solubility Equilibria

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