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Entry Task: Feb 20 th Thursday

Entry Task: Feb 20 th Thursday. AGENDA: Discuss titration lab Pick up titration calculation notes Titration ws. Titration calculations Break out calculators. Refresh on Strong Acid-Strong Base Titration curve. Strong Acid- Strong Base Titration 4 regions of a titration curve.

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Entry Task: Feb 20 th Thursday

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  1. Entry Task: Feb 20th Thursday AGENDA: Discuss titration lab Pick up titration calculation notes Titration ws

  2. Titration calculations Break out calculators

  3. Refresh on Strong Acid-Strong Base Titration curve

  4. Strong Acid- Strong Base Titration4 regions of a titration curve 1st- Initial pH- its really low- probably a strong acid. 2nd- Between initial and equivalence pt. rises slowly then rapidly around the ~SAME~ volume as the unknown.

  5. Strong Acid- Strong Base Titration4 regions of a titration curve 3rd Equivalence pt [H+] = [OH-] = pH 7 4th After equivalence pt. Has plateaued with excess base

  6. How many milliliters of 0.105 M of HCl are needed to titration each of the following solutions to the equivalence point: (a) 35.0 mls of 0.0950 M NaOH; (b) 4.07 ml of 0.117 M KOH; (c)50.0 ml of a solution that contains 1.35g of NaOH per liter?Solution for (a) (xml)(0.105 M HCl) = (35.0 ml)(0.0950M NaOH) xml = 31.7 ml of HCl

  7. How many milliliters of 0.105 M of HCl are needed to titration each of the following solutions to the equivalence point: (a) 35.0 mls of 0.0950 M NaOH; (b) 40.7 ml of 0.117 M KOH; (c)50.0 ml of a solution that contains 1.35g of NaOH per liter?Solution for (b) (xml)(0.105 M HCl) = (40.7 ml)(0.117M KOH) xml = 45.4 ml of HCl

  8. How many milliliters of 0.105 M of HCl are needed to titration each of the following solutions to the equivalence point: (a) 35.0 mls of 0.0950 M NaOH; (b) 40.7 ml of 0.117 M KOH; (c)50.0 ml of a solution that contains 1.35g of NaOH per liter?Solution for (c) Convert to moles then molarity 1.35/39.90 = 0.0338 mole/1 L = 0.0338 M of NaOH (xml)(0.105 M HCl) = (50.0 ml)(0.0338M NaOH) xml = 16.1 ml of HCl

  9. Calculate the pH when the following of quantities of 0.100 M NaOH solution have been added to 50.00 ml of 0.100 M HCl solution: What is the initial pH of each solution? NaOH has a pH of 13 HCl has a pH of 1

  10. Calculate the pH when the following of quantities of 0.100 M NaOH solution have been added to 50.00 ml of 0.100 M HCl solution: (a) 49.00 ml (b) 51.00 ml How much of each solution would produce a pH of 7? 50.0 mls of each solution would produce a pH of 7

  11. Calculate the pH when the following of quantities of 0.100 M NaOH solution have been added to 50.00 ml of 0.100 M HCl solution: (a) 49.00 ml (b) 51.00 ml Predict what would happen for each scenario: (a) would have more HCl than NaOH so the pH would be less than 7 (b) would have less HCl than NaOH so the pH would be more than 7

  12. Calculate the pH when the following of quantities of 0.100 M NaOH solution have been added to 50.00 ml of 0.100 M HCl solution: (a) 49.00 ml (b) 51.00 mlThe Solution for (a) (49.00 ml)(0.100M) = 4.9 OH- mmol (50.00 ml)(0.100M) = 5.0 H+ mmol More H+ Find the difference- 4.9 - 5.0 = 0.1/99.0 ml 1.0 x 10-3 [H+] = pH of 3

  13. Calculate the pH when the following of quantities of 0.100 M NaOH solution have been added to 50.00 ml of 0.100 M HCl solution: (a) 49.00 ml (b) 51.00 mlThe Solution for (b) (51.00 ml)(0.100M) = 5.1 OH- mmol More OH- (50.00 ml)(0.100M) = 5.0 H+ mmol Find the difference- 5.1 - 5.0 = 0.1/101 ml 9.9 x 10-4 [OH-] = pOH of 3 – 14 = pH of 11

  14. YOU TRY!!Calculate the pH when the following of quantities of 0.100 M HNO3 solution have been added to 25.00 ml of 0.100 M KOH solution: (a) 24.90 ml (b) 25.10ml Predict what would happen for each scenario: (a) would have more KOH than HNO3 so the pH would be MORE than 7 (b) would have less KOH than HNO3 so the pH would be LESS than 7

  15. YOU TRY!!Calculate the pH when the following of quantities of 0.100 M HNO3solution have been added to 25.00 ml of 0.100 M KOH solution: (a) 24.90 ml (b) 25.10mlThe Solution for (a) (24.90 ml)(0.100M) = 2.49 H+ mmol (25.00 ml)(0.100M) = 2.50 OH- mmol More OH- Find the difference- 2.49 – 2.50 = 0.01/49.90 ml 2.0 x 10-4 [OH-] = pOH of 3.7 – 14 = pH 10.30

  16. YOU TRY!!Calculate the pH when the following of quantities of 0.100 M HNO3solution have been added to 25.00 ml of 0.100 M KOH solution: (a) 24.90 ml (b) 25.10mlThe Solution for (b) (25.10 ml)(0.100M) = 2.51 H+ mmol More H+ (25.00 ml)(0.100M) = 2.50 OH- mmol Find the difference- 2.51 – 2.50 = 0.01/50.10 ml 1.99 x 10-4 [H+] = pH 3.70

  17. Refresh on Strong Base-Weak Acid Titration curve

  18. Titration of a Weak Acid with a Strong Base 1st- Initial pH- ~3 or 4 is a “stronger” weak acid 2nd- Between initial and equivalence pt. 2 things to consider 1. neutralization of weak acid by strong base 2. Strong base acts as a buffer so it resists the titration pH of your weak acid is ½ the amount for neutralization of base- ½ of 50 mls (25 mls) and pH of ~4.8

  19. Titration of a Weak Acid with a Strong Base 3rd At equivalence pt ~The pH here is above 7 which is what we expected 4th After equivalence pt The curve looks very similar to a strong acid/strong base curve.

  20. Calculate the pH of the solution formed when 45.0ml of 0.100M NaOH is added to 50.0 ml of 0.100M HC2H3O2 (Ka= 1.8 x10-5)TREAT AS NEUTRALIZATION!! (45.00 ml)(0.100M) = 4.5 NaOHmmol (50.00 ml)(0.100M) = 5.0 HC2H3O2mmol Plug back into Equilibrium- ICE Table HC2H3O2 + OH-  C2H3O2 - 5.0 mmol 4.5 mmol 0 mmol -4.5 mmol -4.5 mmol +4.5 mmol 0.5 mmol 0 mmol +4.5 mmol

  21. Calculate the pH of the solution formed when 45.0ml of 0.100M NaOH is added to 50.0 ml of 0.100M HC2H3O2 (Ka= 1.8 x10-5) Plug back into Equilibrium- ICE Table HC2H3O2 + OH-  C2H3O2 - 5.0 mmol 4.5 mmol 0 mmol -4.5 mmol -4.5 mmol +4.5 mmol 0.5 mmol 0 mmol +4.5 mmol 0.5 mmol / 95 mls = 5.26 x10-3 HC2H3O2 4.5 mmol / 95 mls = 4.74 x10-2 C2H3O2-

  22. Calculate the pH of the solution formed when 45.0ml of 0.100M NaOH is added to 50.0 ml of 0.100M HC2H3O2 (Ka= 1.8 x10-5) Plug back into Equilibrium expression to solve H+ HC2H3O2 + OH-  C2H3O2 - [H+][4.74 x10-2] [5.3 x 10-3] = 1.8 x10-5 (1.8 x10-5 )(5.3 x 10-3) = H+ 4.74 x10-2 H+ = 2.13 x 10-6 pH = 5.70

  23. Calculate the pH of the solution formed when 45.0ml of 0.100M NaOH is added to 50.0 ml of 0.100M HC2H3O2 (Ka= 1.8 x10-5)TREAT AS NEUTRALIZATION!! (45.00 ml)(0.100M) = 4.5 NaOHmmol (50.00 ml)(0.100M) = 5.0 HC2H3O2mmol Find the difference- 4.5 - 5.0 = 0.5mmol of HC2H3O2 in 95.0 ml 5.26 x 10-3 M [HC2H3O2] This means that there is 4.5mmol of C2H3O2- in 95.0 ml 4.74 x 10-2 M [C2H3O2-] Plug into Ka Expression to solve for H+

  24. YOU TRY!!!(a) Calculate the pH in the solution formed by adding 10.0 mls of 0.050M NaOH to 40.0 mls of 0.0250M benzoic acid (HC7H5O2, Ka= 6.3 x10-5)TREAT AS NEUTRALIZATION!! (10.00 ml)(0.050M) = 0.50 NaOHmmol (40.0 ml)(0.0250M) = 1.0 HC7H5O2mmol Find the difference: 0.50 - 1.0 = 0.5mmol of HC7H5O2 in 50.0 ml 1.0 x 10-2 M [HC7H5O2] This means that there is 0.50 mmol of C7H5O2- in 50.0 ml 1.0 x 10-2 M [C7H5O2-] Plug into Ka Expression to solve for H+

  25. (a) Calculate the pH in the solution formed by adding 10.0 mls of 0.050M NaOH to 40.0 mls of 0.0250M benzoic acid (HC7H5O2, Ka= 6.3 x10-5) HC7H5O2 + OH-  C7H5O2 - There is more weak acid than strong base so we use Ka. [H+][1.0 x 10-2 ] [1.0 x 10-2 ] = 6.3 x10-5 (6.3 x10-5 )(1.0 x 10-2 ) = H+ 1.0 x 10-2 H+ = 6.3 x 10-5 pH = 4.20

  26. YOU TRY!!!(b) Calculate the pH in the solution formed by adding 10.0 mls of 0.100M HCl to 20.0 mls of 0.100M NH3.Kb= 1.8 x10-5.TREAT AS NEUTRALIZATION!! (10.00 ml)(0.100 M) = 1.0 OH- mmol (20.0 ml)(0.100 M) = 2.0 NH4+ mmol Find the difference: 1.0 - 2.0 = 1.0 mmol of NH4+ in 30.0 ml 3.33 x 10-2 M [NH4+] This means that there is 1.0 mmol of OH- in 30.0 ml 3.33 x 10-2 [OH-] Plug into Ka Expression to solve for H+

  27. Calculate the pH in the solution formed by adding 10.0 mls of 0.100M HCl to 20.0 mls of 0.100M NH3 Kb= 1.8 x10-5. NH3 + HCl NH4+ + OH- Change Kb into Ka because most of the NH3 will be mostly neutralized. 1.0 x 10-14 / 1.8 x 10-5 = Kb= 5.56 x10-10 [H+][3.33 x 10-2 ] [3.33 x 10-2 ] = 5.56 x10-10 (5.56 x10-10)(3.33 x 10-2 ) = H+ 3.33 x 10-2 H+ = 5.56 x10-10 pH = 9.26

  28. Calculating equivalence pts Calculating equivalence points with strong acid and strong base is like the first problem in these notes. What about equivalence points of strong species with a weak one?

  29. Calculate the pH at the equivalence point in the titration of 50.0 ml of 0.100M HC2H3O2 with 0.100 M NaOH. 1st deal with them just like strong acid-strong base 5.0 mmol = (x mls)(0.100 M NaOH) (50.00 ml)(0.100M) = 5.0 mmol HC2H3O2 X = 50 mls of NaOH so that makes 100 mls of solution 5.00 mmol of HC2H3O2 in 100 mls of solution 5.00/100 mls = 0.0500 mmol of C2H3O2-

  30. Calculate the pH at the equivalence point in the titration of 50.0 ml of 0.100M HC2H3O2 with 0.100 M NaOH. Ka= 1.8 x 10-5 Now we plug into Ka or Kb expression to get to pH [HC2H3O2][OH-] [C2H3O2-] 5.00/100 mls = 0.0500 mmol of C2H3O2- Since we are using the conjugate base to find the pH we need the Kb value. 1.0 x10-14 / 1.8 x10-5 = Kb = 5.6 x10-10 [x][x] [0.0500] = 5.6 x10-10 = 5.6 x10-10 x2=( 5.6 x10-10)(0.0500) pH=8.72 x=5.29 x10-6 pOH=5.28 -14 x2=2.8 x10-11

  31. You Try!Calculate the pH at the equivalence point when (a) 40.0 mls of 0.025M benzoic acid (HC7H5O2Ka= 6.3 x10-5) is titrated with 0.050 M NaOH 1st deal with them just like strong acid-strong base 1.0 mmol = (x mls)(0.050 M NaOH) (40.00 ml)(0.025M) = 1.0 mmol HC7H5O2 X = 20 mls of NaOH so that makes 60 mls of solution 1.00 mmol of HC7H5O2 in 60 mls of solution 1.00/60 mls = 0.0167 mmol of C7H5O2-

  32. You Try!Calculate the pH at the equivalence point when (a) 40.0 mls of 0.025M benzoic acid (HC7H5O2Ka= 6.3 x10-5) is titrated with 0.050 M NaOH Now we plug into Ka or Kb expression to get to pH [HC7H5O2][OH-] [C7H5O2-] 1.00/60 mls = 0.0167 mmol of C7H5O2- Since we are using the conjugate base to find the pH we need the Kb value. 1.0 x10-14 / 6.3 x10-5 = Kb = 1.59 x10-10 [x][x] [0.0167] = 1.59 x10-10 = 1.59 x10-10 x2=( 1.59 x10-10)(0.0167) pH=8.21 x=1.63 x10-6 pOH=5.78 -14 x2=2.6 x10-12

  33. You Try!Calculate the pH at the equivalence point when (b) 40.0 mls of 0.100M NH3 is titrated with 0.100 M HCl.Kb= 1.8 x10-5 1st deal with them just like strong acid-strong base 4.0 mmol = (x mls)(0.100 M HCl) (40.00 ml)(0.100M) = 4.0 mmol NH3 X = 40 mls of HCl so that makes 80 mls of solution 4.00 mmol of NH3 in 80 mls of solution 4.00/80 mls = 0.05 mmol of NH4+

  34. You Try!Calculate the pH at the equivalence point when (b) 40.0 mls of 0.100M NH3 is titrated with 0.100 M HCl.Ka= 1.8 x10-5 Now we plug into Ka or Kb expression to get to pH [NH4+][OH-] [NH3] 4.00/80 mls = 0.05 mmol of NH4+ Since we are using the conjugate acid to find the pH we need the Ka value. 1.0 x10-14 / 1.8 x10-5 = Ka= 5.6 x10-10 [x][x] [0.050] = 5.6 x10-10 = 5.6 x10-10 x2=(5.6 x10-10)(0.050) x=5.29 x10-6 pH=5.28 x2=2.8 x10-11

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