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Entry Task: Jan 29 th Tuesday

Entry Task: Jan 29 th Tuesday. What is the [H+] and [OH-] of a solution with a pH of 4.67? You have 5 minutes!. Agenda. Discuss Ch. 16 sec. 5-7 In-class practice on weak acid/base. I can…. Explain how strong acid and strong base ionize in water and how weak acids and weak based dissociate.

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Entry Task: Jan 29 th Tuesday

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  1. Entry Task: Jan 29th Tuesday What is the [H+] and [OH-] of a solution with a pH of 4.67? You have 5 minutes!

  2. Agenda • Discuss Ch. 16 sec. 5-7 • In-class practice on weak acid/base

  3. I can… • Explain how strong acid and strong base ionize in water and how weak acids and weak based dissociate. • Use Ka and concentrations weak acids to find pH and likewise with bases (Kb). • Explain how the weak acid/base concentrations affect the magnitude of Ka or Kb.

  4. WARNING!Black text slides are embedded notes

  5. Strong Acids • You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4. • These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. • For the monoprotic strong acids, [H3O+] = [acid].

  6. Strong Bases • Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+). • Again, these substances dissociate completely in aqueous solution.

  7. HA(aq) + H2O(l) A−(aq) + H3O+(aq) Weak Acids They do not completely ionize in aqueous solution. You know it’s a weak acid by the equilibrium arrows

  8. Ka= [H3O+] [A−] [HA] HA(aq) + H2O(l) A−(aq) + H3O+(aq) Dissociation Constants • We can calculate to the extent of how MUCH will ionize by using the acid dissociation constant, Ka Look familiar  Products on top and reactants on bottom.

  9. Dissociation Constants The greater the value of Ka, the stronger the acid.

  10. Calculating Ka from the pH • The pH of a 0.10 M solution of formic acid, HCHO2, at 25°C is 2.38. Calculate Ka for formic acid at this temperature.

  11. HCHO2(aq) H+(aq) + CHO2-(aq) [H+] [CHO2- ] [HCHO2] [x] [x] [0.10-x] Ka = Ka = Calculating Ka from the pH • First step- is to write out the equilibrium expression for Ka

  12. [H+] [CHO2- ] [HCHO2] Ka = Calculating Ka from the pH • Second step- is to find out ALL three concentrations at equilibrium. We can get information from the pH given. pH = -log[H+] = 2.38 [H+] = 4.2 x10-3 M We know that [H+] = [CHO2-] so its 4.2 x10-3 M We know that [HCHO2] = 0.10M initially

  13. Calculating Ka from pH Third step- find out what the concentration of HCHO2 is at equilibrium

  14. [4.2  10−3] [4.2  10−3] [0.10] Ka = Calculating Ka from pH Fourth step- Plug equilibrium concentrations in the Ka expression = 1.8  10−4

  15. HC3H5O3(aq) H+(aq) + C3H5O3-(aq) [H+] [C3H5O3- ] [HC3H5O3] Ka = Lactic Acid, HC3H5O3, has one acidic hydrogen. A 0.10M solution of Lactic Acid has a pH of 2.44. Calculate Ka. • First step- is to write out the equilibrium expression for Ka

  16. [H+] [C3H5O3- ] [HC3H5O3] Ka = Lactic Acid, HC3H5O3, has one acidic hydrogen. A 0.10M solution of Lactic Acid has a pH of 2.44. Calculate Ka. • Second step- is to find out ALL three concentrations at equilibrium. We can get information from the pH given. pH = -log[H+] = 2.44 [H+] = 3.6 x10-3 M We know that [H+] = [C3H5O3-] so its 3.6 x10-3 M We know that [HC3H5O3] = 0.10M initially

  17. Lactic Acid, HC3H5O3, has one acidic hydrogen. A 0.10M solution of Lactic Acid has a pH of 2.44. Calculate Ka. Third step- find out what the concentration of HCHO2 is at equilibrium

  18. [3.6  10−3] [3.6  10−3] [0.10] Ka = Lactic Acid, HC3H5O3, has one acidic hydrogen. A 0.10M solution of Lactic Acid has a pH of 2.44. Calculate Ka. Fourth step- Plug equilibrium concentrations in the Ka expression = 1.3  10−4

  19. [x] [x] [0.10-x] Ka = When do we use the quadratic equation? • The values for Ka are so small that the equilibrium will lie far to the left (reactants) so the x values will be very very small in comparison to its initial concentration So 0.10 – x  0.10

  20. HC8H7O2(aq) H+(aq) + C8H7O2 -(aq) [2.1 x10-3 ] [2.1 x10-3 ] [0.85] [H+] [HC8H7O2- ] [HC8H7O2] Ka = Ka = A phenylaceticacid, HC8H7O2, solution containing a molarity of 0.85M. It has a pH of 2.68. Calculate the Ka. We can short cut it. pH = -log[H+] = 2.68 Ka = 5.2 x 10-6 [H+] = 2.1 x10-3 M We know that [H+] = [C8H7O2-] so its 2.1 x10-3 M

  21. [H+]eq [HA]initial 4.2  103 0.10 Calculating Percent Ionization • Percent ionization =  100 • In this example, [H+]eq = 4.2  103 M [HCHO2]initial= 0.10 M Percent ionization =  100 = 4.2%

  22. [H+]eq [HA]initial 5.5  104 0.020 A 0.020M solution of niacin has a pH of 3.26. a) What is the percentage of acid ionized in this solution? b) What is the acid dissociation constant, Ka, for niacin? • Percent ionization =  100 • In this example, [H+]eq = 5.5  104 M [HCHO2]initial = 0.020 M Percent ionization =  100 = 2.7%

  23. [5.5 x10-4 ] [5.5 x10-4 ] [0.020] Ka = A 0.020M solution of niacin has a pH of 3.26. a) What is the percentage of acid ionized in this solution? b) What is the acid dissociation constant, Ka, for niacin? Ka = 1.5 x 10-5

  24. Calculating pH from Ka Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25C. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2(aq) Ka for acetic acid at 25C is 1.8  105.

  25. [H3O+] [C2H3O2] [HC2H3O2] Ka = Calculating pH from Ka The equilibrium constant expression is HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2(aq)

  26. Calculating pH from Ka HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2(aq) We next set up a table… We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

  27. Calculating pH from Ka HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2(aq) We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

  28. Calculating pH from Ka HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2(aq) We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

  29. (x)2 (0.30) 1.8  105 = Calculating pH from Ka Now, (1.8  105) (0.30) = x2 5.4  106 = x2 2.3  103 = x

  30. Calculating pH from Ka pH = log [H+] pH = log (2.3  103) pH = 2.64

  31. Calculating pH from Ka Calculate the pH of a 0.20 M solution of HCN with a Ka value of 4.9 x 10-10. HCN(aq) + H2O(l) H3O+(aq) + CN(aq) Kais 4.9  1010

  32. [H3O+] [CN] [HCN] Ka = Calculating pH from Ka The equilibrium constant expression is HCN(aq) + H2O(l) H3O+(aq) + CN(aq)

  33. (x)2 (0.20) 4.9  1010= Calculating pH from Ka SHORT CUT, (4.9  1010) (0.20) = x2 9.8  1011= x2 9.9  106= x

  34. Calculating pH from Ka pH = log [H+] pH = log (9.9  106) pH = 5.0

  35. Polyprotic Acids Polyprotic acids have more than one acidic proton. If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

  36. Weak Bases Bases react with water to produce hydroxide ion.

  37. [HB] [OH] [B-] Kb = Weak Bases • We can calculate to the extent of how MUCH will ionize by using the base dissociation constant, Kb Just like acids but with bases Products on top and reactants on bottom.

  38. Weak Bases Kb can be used to find [OH] and, through it, pH.

  39. NH3(aq) + H2O(l) NH4+(aq) + OH(aq) [NH4+] [OH] [NH3] Kb = = 1.8  105 pH of Weak Basic Solutions What is the pH of a 0.15 M solution of NH3? The Kb value is 1.8 105

  40. pH of Weak Basic Solutions (x)2 (0.15) (1.8  105) (0.15) = x2 2.7  106 = x2 1.6  103 = x 1.8  105 =

  41. pH of Weak Basic Solutions Therefore, [OH] = 1.6  103 M pOH = log (1.6  103) pOH = 2.80 pH = 14.00  2.80 pH = 11.20

  42. Types of Weak Bases • Amines- Neutral compounds containing nitrogen or non-bonded pairs of electrons serves as proton acceptors. • Example: Ammonia NH3 • Anions of weak acids (conjugate bases) • Example: NaClO( Some salts can fall into this category)

  43. Pre-Lab Determination of Ka Labs

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