1. Acids and Bases 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!

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

3. Acids and Bases 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. Acids and Bases

5. Acids and Bases WARNING! Black text slides are embedded notes

6. Acids and Bases Strong Acids You will recall that the seven strong acids are HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 3, and HClO 4. These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. For the monoprotic strong acids, [H 3 O + ] = [acid].

7. Acids and Bases Strong Bases Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca 2+, Sr 2+, and Ba 2+ ). Again, these substances dissociate completely in aqueous solution.

8. Acids and Bases Weak Acids They do not completely ionize in aqueous solution. HA (aq) + H 2 O (l) A − (aq) + H 3 O + (aq) You know it’s a weak acid by the equilibrium arrows

9. Acids and Bases Dissociation Constants We can calculate to the extent of how MUCH will ionize by using the acid dissociation constant, K a [H 3 O + ] [A − ] [HA] K a = HA (aq) + H 2 O (l) A − (aq) + H 3 O + (aq) Look familiar Products on top and reactants on bottom.

10. Acids and Bases Dissociation Constants The greater the value of K a, the stronger the acid.

11. Acids and Bases Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCHO 2, at 25°C is 2.38. Calculate K a for formic acid at this temperature.

12. Acids and Bases Calculating K a from the pH First step- is to write out the equilibrium expression for K a [H + ] [CHO 2 - ] [HCHO 2 ] K a = HCHO 2 (aq) H + (aq) + CHO 2 - (aq) [x] [0.10-x] K a =

13. Acids and Bases Calculating K a from the pH Second step- is to find out ALL three concentrations at equilibrium. [H + ] [CHO 2 - ] [HCHO 2 ] K a = We can get information from the pH given. pH = -log[H+] = 2.38 [H+] = 4.2 x10 -3 M We know that [H+] = [CHO 2 -] so its 4.2 x10 -3 M We know that [HCHO 2 ] = 0.10M initially

14. Acids and Bases Calculating K a from pH Third step- find out what the concentration of HCHO 2 is at equilibrium [HCHO 2 ], M[H + ], M[CHO 2 - ], M Initially0.1000 Change −4.2  10 -3 +4.2  10 -3 +4.2  10 −3 At Equilibrium 0.10 − 4.2  10 −3 = 0.0958 = 0.10 4.2  10 −3

15. Acids and Bases Calculating K a from pH [4.2  10 −3 ] [0.10] K a = = 1.8  10 −4 Fourth step- Plug equilibrium concentrations in the K a expression

16. Acids and Bases Lactic Acid, HC 3 H 5 O 3, has one acidic hydrogen. A 0.10M solution of Lactic Acid has a pH of 2.44. Calculate K a. First step- is to write out the equilibrium expression for K a [H + ] [C 3 H 5 O 3 - ] [HC 3 H 5 O 3 ] K a = HC 3 H 5 O 3 (aq) H + (aq) + C 3 H 5 O 3 - (aq)

17. Acids and Bases Lactic Acid, HC 3 H 5 O 3, has one acidic hydrogen. A 0.10M solution of Lactic Acid has a pH of 2.44. Calculate K a. 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+] = [C 3 H 5 O 3 -] so its 3.6 x10 -3 M We know that [HC 3 H 5 O 3 ] = 0.10M initially [H + ] [C 3 H 5 O 3 - ] [HC 3 H 5 O 3 ] K a =

18. Acids and Bases Lactic Acid, HC 3 H 5 O 3, has one acidic hydrogen. A 0.10M solution of Lactic Acid has a pH of 2.44. Calculate K a. Third step- find out what the concentration of HCHO 2 is at equilibrium [HC 3 H 5 O 3 ], M[H + ], M[C 3 H 5 O 3 - ], M Initially0.1000 Change −3.6  10 -3 +3.6  10 -3 +3.6  10 −3 At Equilibrium 0.10 − 3.6  10 −3 = 0.0964 = 0.10 3.6  10 −3

19. Acids and Bases Lactic Acid, HC 3 H 5 O 3, has one acidic hydrogen. A 0.10M solution of Lactic Acid has a pH of 2.44. Calculate K a. [3.6  10 −3 ] [0.10] K a = = 1.3  10 −4 Fourth step- Plug equilibrium concentrations in the K a expression

20. Acids and Bases 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 [x] [0.10-x] K a = So 0.10 – x  0.10

21. Acids and Bases A phenylacetic acid, HC 8 H 7 O 2, solution containing a molarity of 0.85M. It has a pH of 2.68. Calculate the Ka. We can short cut it. [H + ] [HC 8 H 7 O 2 - ] [HC 8 H 7 O 2 ] K a = HC 8 H 7 O 2 (aq) H + (aq) + C 8 H 7 O 2 - (aq) pH = -log[H+] = 2.68 [H+] = 2.1 x10 -3 M We know that [H+] = [C 8 H 7 O 2 -] so its 2.1 x10 -3 M [2.1 x10 -3 ] [0.85 ] K a = Ka = 5.2 x 10 -6

22. Acids and Bases Calculating Percent Ionization Percent ionization =  100 In this example, [H + ] eq = 4.2  10  3 M [HCHO 2 ] initial = 0.10 M [H + ] eq [HA] initial Percent ionization =  100 4.2  10  3 0.10 = 4.2%

23. Acids and Bases 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 [HCHO 2 ] initial = 0.020 M [H + ] eq [HA] initial Percent ionization =  100 5.5  10  4 0.020 = 2.7%

24. Acids and Bases 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? [5.5 x10 -4 ] [0.020 ] K a = Ka = 1.5 x 10 -5

25. Acids and Bases Calculating pH from K a Calculate the pH of a 0.30 M solution of acetic acid, HC 2 H 3 O 2, at 25  C. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2  (aq) K a for acetic acid at 25  C is 1.8  10  5.

26. Acids and Bases Calculating pH from K a The equilibrium constant expression is [H 3 O + ] [C 2 H 3 O 2  ] [HC 2 H 3 O 2 ] K a = HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2  (aq)

27. Acids and Bases Calculating pH from K a We next set up a table… [HC 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2  ], M Initially0.3000 Change At equilibrium We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2  (aq)

28. Acids and Bases Calculating pH from K a [HC 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2  ], M Initially0.3000 Change xx +x+x+x+x At equilibrium HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2  (aq) We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

29. Acids and Bases Calculating pH from K a [HC 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2  ], M Initially0.3000 Change xx +x+x+x+x At equilibrium 0.30  x  0.30 xx We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2  (aq)

30. Acids and Bases Calculating pH from K a Now, (x) 2 (0.30) 1.8  10  5 = (1.8  10  5 ) (0.30) = x 2 5.4  10  6 = x 2 2.3  10  3 = x

31. Acids and Bases Calculating pH from K a pH =  log [H + ] pH =  log (2.3  10  3 ) pH = 2.64

32. Acids and Bases Calculating pH from K a Calculate the pH of a 0.20 M solution of HCN with a Ka value of 4.9 x 10 -10. HCN (aq) + H 2 O (l) H 3 O + (aq) + CN  (aq) K a is 4.9  10  10

33. Acids and Bases Calculating pH from K a The equilibrium constant expression is [H 3 O + ] [CN  ] [HCN] K a = HCN (aq) + H 2 O (l) H 3 O + (aq) + CN  (aq)

34. Acids and Bases Calculating pH from K a SHORT CUT, (x) 2 (0.20) 4.9  10  10 = (4.9  10  10 ) (0.20) = x 2 9.8  10  11 = x 2 9.9  10  6 = x

35. Acids and Bases Calculating pH from K a pH =  log [H + ] pH =  log (9.9  10  6 ) pH = 5.0

36. Acids and Bases Polyprotic Acids Polyprotic acids have more than one acidic proton. If the difference between the K a for the first dissociation and subsequent K a values is 10 3 or more, the pH generally depends only on the first dissociation.

37. Acids and Bases Weak Bases Bases react with water to produce hydroxide ion.

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

39. Acids and Bases Weak Bases K b can be used to find [OH  ] and, through it, pH.

40. Acids and Bases pH of Weak Basic Solutions What is the pH of a 0.15 M solution of NH 3 ? The Kb value is 1.8  10  5 [NH 4 + ] [OH  ] [NH 3 ] K b = = 1.8  10  5 NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH  (aq)

41. Acids and Bases pH of Weak Basic Solutions (1.8  10  5 ) (0.15) = x 2 2.7  10  6 = x 2 1.6  10  3 = x (x) 2 (0.15) 1.8  10  5 =

42. Acids and Bases 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

43. Acids and Bases Types of Weak Bases Amines- Neutral compounds containing nitrogen or non-bonded pairs of electrons serves as proton acceptors.  Example: Ammonia NH 3 Anions of weak acids (conjugate bases)  Example: NaClO ( Some salts can fall into this category)

44. Acids and Bases Pre-Lab Determination of Ka Labs