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Entry Task: Feb 8 th Friday

Entry Task: Feb 8 th Friday. Question: Describe the common ion effect? You have 5 minutes. Agenda. Discuss Ch. 17. sec. 1-2 notes Get through as much of the notes/ in-class practice- it might spill into next time. HW: Common ion Effect ws. Equation Sheet. I can….

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Entry Task: Feb 8 th Friday

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  1. Entry Task: Feb 8th Friday Question: Describe the common ion effect? You have 5 minutes

  2. Agenda • Discuss Ch. 17. sec. 1-2 notes • Get through as much of the notes/ in-class practice- it might spill into next time. • HW: Common ion Effect ws

  3. Equation Sheet

  4. I can… • Explain and calculate the common ion effect. • Describe how a buffer system works and calculate the pH changes.

  5. AP Exam • The equilibrium concepts in this chapter are a big part of the AP Exam. Question 1 on the F.R. is always an equilibrium problem and is 20% of the F.R. score. There are also many questions regarding equilibrium on the M.C. section

  6. Control of solution composition Chap 16: What is the pH after adding X to water? Chap 17:How can we control the pH of a solution?How can we determine acid/base concentration? Chap 4:Solubility of ionic compounds. Chap 17:How soluble are they? Buffers Titrations Ksp=solubility product

  7. WARNINGBlack text slides are embedded notes

  8. HC2H3O2(aq) H+(aq) + C2H3O2−(aq) The Common-Ion Effect • Consider a solution of acetic acid in water: • What would happen if we introduce a substance with a common ion like NaC2H3O2?

  9. HC2H3O2(aq) H+(aq) + C2H3O2−(aq) The Common-Ion Effect • Consider a solution of acetic acid in water: • The Na+ + C2H3O2 – forming more (-)ions causing a decrease in H+ ions. • How would this affect the pH?

  10. Common ion effect The dissociation of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.

  11. HC2H3O2(aq) H+(aq) + C2H3O2−(aq) The Common-Ion Effect 17.1 Sample: What is the pH of a solution made with 0.30 mols of acetic acid, HC2H3O2, and 0.30 mols of sodium acetate, NaC2H3O2, to enough water to make 1.0 L solution? Ka for acetic acid at 25°C is 1.8  10-5. NaC2H3O2 Na+ + C2H3O2-

  12. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) The Common-Ion Effect Use the same method as in Chapter 16: ICE + equilibrium expression But, adjust initial concentrations to account for the salt. NaC2H3O2 Na+ + C2H3O2-

  13. Ka = Ka = [H+] [C2H3O2−] [HC2H3O2] [x] [0.30 + x] [0.30 - x] H+(aq) + C2H3O2−(aq) HC2H3O2(aq) = 1.8 x 10-5 = 1.8 x 10-5 The Common-Ion Effect Use the ICE table: Water was omitted for convenience

  14. Ka = [x] [0.30 + x] [0.30 - x] The Common-Ion Effect Suppose x is small relative to 0.30: ≈ X 0.30 0.30 Ka= x 1.8  10-5 = x = [H+] pH=–log(1.8  10-5)=4.74

  15. HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq) The Common-Ion Effect 17.1 Practice Example: Calculate the pH of a solution containing 0.085 M of nitrous acid, HNO2 (Ka= 4.5 x 10-4), and 0.10M potassium nitrite, KNO2. KNO2 K+ + NO2-

  16. Ka = Ka = [H3O+] [NO2−] [HNO2] [x] [0.10 + x] [0.085 - x] HNO2(aq) + H2O(l) H3O+(aq) + NO2−(aq) = 4.5 x 10-4 = 4.5 x 10-4 The Common-Ion Effect Use the ICE table: Water was omitted for convenience

  17. Ka = [x] [0.10 + x] [0.085 - x] The Common-Ion Effect Suppose x is small relative to 0.085: ≈ = 4.5 x 10-4 X 0.10 0.085 (4.5 x 10-4)(0.085) 0.10 3.825 x 10-5 0.10 X = 3.83 x10-4 x = [H+] pH=–log(3.83 10-4) = 3.42

  18. Ka = [H3O+] [F−] [HF] = 6.8  10-4 The Common-Ion Effect Sample 17.2 Calculate the fluoride ion concentration and pH of a solution that is 0.10 M in HCl and 0.20 M in HF in a 1 L solution. Ka for HF is 6.8  10−4.

  19. HF(aq) + H2O(l) H3O+(aq) + F−(aq) The Common-Ion Effect Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.

  20. (0.10) (x) (0.20) 6.8  10−4 = (0.20) (6.8  10−4) (0.10) The Common-Ion Effect = x 1.4  10−3 = x 1.4  10−3 = concentration of F- ions

  21. The Common-Ion Effect • Therefore, [F−] = x = 1.4  10−3 [H3O+] = 0.10 + x = 0.10 + 1.4  10−3 = 0.1014 M • So, pH = −log (0.1014) pH = 1.00

  22. Ka = [H3O+] [CHO2−] [HCHO2] = 1.8  10-4 The Common-Ion Effect 17.2 Practice Calculate the formate-ion concentration and pH of a solution that is 0.050 M in formic acid, HCHO2 and 0.10 M in HNO3 in a 1 L solution. Ka for HCHO2 is 1.8  10−4.

  23. HCHO2(aq) + H2O(l) H3O+(aq) + CHO2−(aq) The Common-Ion Effect Because HNO3, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.

  24. (0.10) (x) (0.050) 1.8  10−4 = (0.050) (1.8  10−4) (0.10) The Common-Ion Effect = x 9.0  10−5 = x 9.0  10−5= concentration of CHO2- ions

  25. The Common-Ion Effect • Therefore, [CHO2−] = x = 9.0  10−5 [H3O+] = 0.10 + x = 0.10 + 9.0  10−5 = 0.10 M • So, pH = −log (0.10) pH = (0.99) ~1.00

  26. Buffers: • Defined as a solutions of a weak conjugate acid-base pair. • They are particularly resistant to pH changes, even when strong acid or base is added.

  27. Buffers: • Buffers resist changes in pH because they contain BOTH an acidic species to neutralize OH- ions and a basic species to neutralize H+ ions. • Its important that this “combo” team does not consume itself. • Hence- weak acid-base conjugate pairing

  28. Buffers: • The combo team would be: • HC2H3O2 -- C2H3O2- • NH4+ -- NH3 • They are made by mixing a weak acid or weak base with a salt of that acid or base. • HC2H3O2 -- C2H3O2- • We add NaC2H3O2 • NH4+ -- NH3 • We add NH4Cl

  29. Buffers

  30. Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

  31. Buffers If acid is added, the F− reacts to form HF and water.

  32. H+ = Ka Ka = [H3O+] [A−] [HA] [HA] [A-] Buffers: • We rearrange the acid dissociation constant to provide a ratio between conjugate [A-] and its acid [HA]. • As long as the ratio is small so will the change in pH.

  33. Buffers: • Buffer capacity is the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree. • It depends on the amount of acid or base from which the buffer is made. • Example- • 1.0 M HC2H3O2 and 1.0 M NaC2H3O2 • 0.1 M HC2H3O2 and 0.1 M NaC2H3O2 • BOTH would have the same [H+]ratio • Greater amounts of conjugate a/b the more resistant the ratio of concentrations –hence pH is to change • This guy would have a greater buffering capacity

  34. Ka = [H3O+] [A−] [HA] HA + H2O H3O+ + A− Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA:

  35. [A−] [HA] [H3O+] Ka = base [A−] [HA] −log [H3O+] + −log −log Ka = pKa acid pH Buffer Calculations Rearranging slightly, this becomes Taking the negative log of both side, we get

  36. [base] [acid] pKa = pH − log [base] [acid] pH = pKa + log Buffer Calculations • So • Rearranging, this becomes • This is the Henderson–Hasselbalchequation.

  37. [base] [acid] (0.10) (0.12) pH = pKa + log pH = −log (1.4  10−4) + log Henderson–Hasselbalch Equation Sample 17.3 What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4  10−4. pH = 3.85 + (−0.08) pH = 3.77

  38. [base] [acid] (0.20) (0.12) pH = pKa + log pH = −log (6.3  10−5)+ log Henderson–Hasselbalch Equation Practice 17.3 What is the pH of a buffer that is composed of 0.12 M in benzoic acid and 0.20 M in sodium benzoate? Ka for benzoic acid is 6.3  10−5. pH = 4.20 + (0.22) pH = 4.42

  39. pH Range • The pH range is the range of pH values over which a buffer system works effectively. • It is best to choose an acid with a pKa close to the desired pH.

  40. When Strong Acids or Bases Are Added to a Buffer… …it is safe to assume that all of the strong acid or base is consumed in the reaction.

  41. Addition of Strong Acid or Base to a Buffer • Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. • Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

  42. Addition of Strong Acid or Base to a Buffer • Consider the acid-base neutralization reaction, and determine its effects on [HX] and [X-]. This stage of the procedure is a stoichiometry calculation • Use Ka and NEW concentrations of [HX] and [X-] to calculate [H+]. This stage of the procedure is equilibrium calculations.

  43. Calculating pH Changes in Buffers Sample 17.5 A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added. Ka = 1.8 x 10-5 Stoichiometry relationship between [HX] and [X-]. We set up – like an ICE table but it will provide the stoich changes that will occur BEFORE the equilibrium change

  44. Calculating pH Changes in Buffers The addition of 0.020 molNaOH HC2H3O2(aq) + OH−(aq) C2H3O2−(aq) + H2O(l)

  45. Calculating pH Changes in Buffers We have NEW concentrations to plug into a equilibrium ICE table HC2H3O2(aq) + OH−(aq) C2H3O2−(aq) + H2O(l)

  46. Calculating pH Changes in Buffers NOW THE EQUILIBRIUM!!! HC2H3O2(aq) H++ C2H3O2−(aq)

  47. Calculating pH Changes in Buffers Plug NEW concentrations into equilibrium expression Ka = [H+][C2H3O2-] [HC2H3O2] Ka = [x][0.320-x] [0.280-x]  Ka = [x][0.320] [0.280] Rearrange to get X by itself 1.8 x 10-5 = [x][0.320] [0.280] x= (1.8 x 10-5)(0.280) 0.320 x = [H+]= 1.6 x 10-5 pH = -log(1.6 x 10-5) = 4.80

  48. (0.320) (0. 280) pH = 4.74 + log Calculating pH Changes in Buffers OR we can use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + (0.06) pH = 4.80

  49. Calculating pH Changes in Buffers Now we are doing the same buffer but with the addition on HCl which will take on the weak base component of the buffer C2H3O2-. The addition of 0.020 molHCl C2H3O2-(aq) + H3O+(aq) HC2H3O2(aq) + H2O(l)

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