# Entry Task: Feb 12 th Tuesday - PowerPoint PPT Presentation  Download Presentation Entry Task: Feb 12 th Tuesday

Entry Task: Feb 12 th Tuesday
Download Presentation ## Entry Task: Feb 12 th Tuesday

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript

1. Entry Task: Feb 12th Tuesday Define Buffer capacity You have 5 minutes

2. Agenda • Discuss Buffer ws 1 • In-class notes little more buffer info and practice on Titrations • HW: Buffers ws #2

3. Explain why a mixture of HCl and KCl does not function as a buffer, whereas a mixture of HC2H3O2 and NaC2H3O2 does? HCl and KCl are conjugate-pairs, problem is that potassium is an alkali metal and will stay dissociated and add more + to the system making it more acidic.

4. 2. What factors determine a) the pH, and b) the buffer capacity of a buffer solution? • The pH of a buffer is determined by Ka for the conjugate acid present and the ratio of the conjugate base concentration to the conjugate acid concentration. • The buffering capacity of buffer is determined by the absolute concentrations of the conjugate acid and conjugate base present. Higher the concentration the higher the capacity.

5. 3. In a solution, when the concentrations of a weak acid and its conjugate base are equal, A) the system is not at equilibrium. B) the buffering capacity is significantly decreased. C) the -log of the [H+] and the -log of the Ka are equal. D) all of the above are true.

6. 4. Of the following solutions, which has the greatest buffering capacity? A) 0.821 M HF and 0.217 M NaF B) 0.821 M HF and 0.909 M NaF C) 0.100 M HF and 0.217 M NaF D) 0.121 M HF and 0.667 M NaF E) They are all buffer solutions and would all have the same capacity.

7. 5. The addition of hydrofluoric acid and __________ to water produces a buffer solution. A) HCl B) NaNO3 C) NaCl D) NaOH E) NaBr

8. 6. Which of the following could be added to a solution of sodium acetate to produce a buffer? A) acetic acid only B) acetic acid or hydrochloric acid C) hydrochloric acid only D) potassium acetate only E) sodium chloride or potassium acetate

9. 7. A solution is prepared by dissolving 0.23 mol of hydrazoic acid and 0.27 mol of sodium azide in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of NaOH to this buffer solution causes the pH to increase slightly. The pH does not increase drastically because the NaOH reacts with the __________ present in the buffer solution. The Kaof hydrazoic acid is 1.9 × 10-5. A) H2O B) H3O+ C) azide D) hydrazoic acid E) This is a buffer solution: the pH does not change upon addition of acid or base.

10. 8. What is the pH of a buffer solution that is 0.211 M in lactic acid and 0.111 M in sodium lactate? The Ka of lactic acid is 1.4 × 10-4. Ka = [x][0.111] [0.211] Rearrange to get X by itself 1.4 x 10-4 = [x][0.111] [0.211] x= (1.4 x 10-4)(0.211) 0.111 x = [H+]= 2.66 x 10-4 pH = -log(2.66 x 10-4) = 3.57

11. Things we have not considered*volume changes*grams mole Molarity*effect on pH

12. Buffer Problems Hurdles- 1st -Which species in problem is “acid” or “base”? 2nd Kb value for OH- or Ka for H+ 3rd Setting up Ka or Kb expression correctly (1st) 4th Is there any changes in concentrations from given: *Addition of a new species (ICE table) *Mixing two different volumes (M x L)= mol

13. a) Calculate the pH of a buffer that is 0.100 M in NaHCO3 and 0.125M in Na2CO3. b) Calculate the pH of a solution formed by mixing 55mls of 0.20M NaHCO3 with 65mls of 0.15 M Na2CO3. Who is the conjugate in this reaction? HCO3- so we use H2CO3Ka value= 5.6x10-11 Provide the Kaexpress with # Provide the Ka expression 5.6 x 10-11= [x][0.125] [0.100] Ka = [H+][CO3-2] [HCO3-] Rearrange to get X by itself x= (5.6 x 10-11)(0.100) 0.125 x = [H+]= 4.48 x 10-11 pH = -log(4.48 x 10-11) = 10.35

14. [base] [acid] [0.125] [0.100] 0.0969 0.0969 pH = pKa + log pH = 10.25 + log pH = 10.25 + 10.35 = 10.25 + a) Calculate the pH of a buffer that is 0.100 M in NaHCO3 and 0.125M in Na2CO3. b) Calculate the pH of a solution formed by mixing 55mls of 0.20M NaHCO3 with 65mls of 0.15 M Na2CO3. 5.6 x 10-11= [x][0.125] [0.100] Ka = [H+][CO3-2] [HCO3-] OR use the H-H equation

15. a) Calculate the pH of a buffer that is 0.100 M in NaHCO3 and 0.125M in Na2CO3. b) Calculate the pH of a solution formed by mixing 55mls of 0.20M NaHCO3 with 65mls of 0.15 M Na2CO3. NOTICE we have different volumes!! 120 mls What will be the TOTAL VOLUME? Convert from Molarity BACK to mols* mol= M x L 0.20M x 0.055L = 0.011 mol of NaHCO3 / 120 = 9.17x10-5(new M) 0.15M x 0.065L = 0.00975 mol of Na2CO3 / 120 = 8.125x10-5(new M) Ka = [H+][CO3-2] [HCO3-] 5.6 x 10-11= [x][8.125x10-5] [9.17x10-5] Rearrange to get X by itself x= (5.6 x 10-11)(9.17x10-5) 8.125x10-5 x = [H+]= 6.32 x 10-11 pH = -log(6.32 x 10-11) = 10.20

16. [8.125 x10-5] [9.17x10-5] -0.0525 pH = 10.25 + log 10.20= 10.25 + a) Calculate the pH of a buffer that is 0.100 M in NaHCO3 and 0.125M in Na2CO3. b) Calculate the pH of a solution formed by mixing 55mls of 0.20M NaHCO3 with 65mls of 0.15 M Na2CO3. NOTICE we have different volumes!! 120 mls What will be the TOTAL VOLUME? Convert from Molarity BACK to mols* mol= M x L 0.20M x 0.055L = 0.011 mol of NaHCO3 / 120 = 9.17x10-5(new M) 0.15M x 0.065L = 0.00975 mol of Na2CO3 / 120 = 8.125x10-5(new M) Ka = [H+][CO3-2] [HCO3-] 5.6 x 10-11= [x][8.125x10-5] [9.17x10-5]

17. [base] [acid] [0.11] [0.12] -0.0378 pH = pKa + log pH = 3.85 + log 3.8= 3.85 + a) Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11M in sodium lactate. b) Calculate the pH of a solution formed by mixing 85mls of 0.13M lactic acid with 95mls of 0.15M sodium lactate. 1.4 x 10-4= [x][0.11] [0.12]

18. [7.92 x10-5] [6.14x10-5] 0.011 pH = 3.85 + log 3.96= 3.85 + a) Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11M in sodium lactate. b) Calculate the pH of a solution formed by mixing 85mls of 0.13M lactic acid with 95mls of 0.15M sodium lactate. 180 mls What will be the TOTAL VOLUME? Convert from Molarity BACK to mols* mol= M x L 0.13M x 0.085L = 0.0111 mol of lactic acid/ 180 = 6.14x10-5(new M) 0.15M x 0.095L = 0.0143 mol of sodium lactate/ 180 = 7.92x10-5(new M)

19. Titration: • A laboratory method for determining the concentration of an unknown acid or base using a neutralization reaction. • A standard solution,(a solution of known concentration-titrant), is used.

20. Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base). After titration- we have a known volume and concentration from titrant (M1)(V1). The unknown has a known volume(V2) so we can calculate (M2). (M1)(V1)= (M2)(V2) Titrant-known of concentration Unknown concentration with a known volume

21. Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

22. pH Titration Curve pH titration curve, a graph of pH as a function of volume of the added titrant. The pH curve can: *help determine equivalence point *determine the pH indicators needed for Ka or Kb determination.

23. pH Titration Curve Can you identify the titrant? Is it an acid or base. Base *pH is increasing *it levels off with lots of base.

24. 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.

25. 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

26. Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly.

27. Titration of a Strong Acid with a Strong Base Just before and after the equivalence point, the pH increases rapidly.

28. Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

29. Titration of a Strong Acid with a Strong Base As more base is added, the increase in pH again levels off.

30. We will learn how to CALCULATE the equivalence point later.

31. Titration of a Weak Acid with a Strong Base Which way will the equilibrium shift in the case of weak acid and strong base?

32. 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

33. 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.

34. Titration of a Weak Base with a Strong Acid Which way will the equilibrium shift in the case of weak base and strong acid?

35. Titration of a Weak base with a Strong acid 1st- Initial pH- its really high probably a strong base. 2nd- Between initial and equivalence pt. 2 things to consider 1. neutralization of weak base by strong acid 2. Strong acid acts as a buffer so it resists the titration

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

37. Titration of a Weak Acid with a Strong Base With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

38. Titrations of Polyprotic Acids In these cases there is an equivalence point for each dissociation.