1. Strength of Acids and Bases Do they ionize 100%?

2. Strong Acids :Give up H + easily Dissociate completely (100%) in water HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 4, HClO 3

3. Weak acids: (all others) Hold onto H + Few molecules dissociate Ex: HCH 3 O 2 Ex: HC 2 H 3 O 2 Strong/Weak Acid Animation http://educypedia.karadimov.info/library/acid13.swf http://educypedia.karadimov.info/library/acid13.swf

4. HA Let’s examine the behavior of an acid, HA, in aqueous solution. What happens to the HA molecules in solution?

5. HA H+H+ A-A- Strong Acid 100% dissociation of HA Would the solution be conductive? Oh yeah…

6. HA H+H+ A-A- Weak Acid Partial dissociation of HA Would the solution be conductive? Not really…

7. HA H+H+ A-A- Weak Acid HA  H + + A - At any one time, only a fraction of the molecules are dissociated.

8. Strong Bases: Dissociate completely (100%) in water - Group I metal hydroxides (NaOH, LiOH, etc.) - Group I metal hydroxides (NaOH, LiOH, etc.) - Some Group II metal hydroxides - Some Group II metal hydroxides Ca(OH) 2, Ba(OH) 2, Sr(OH) 2 Weak Bases Only a few ions dissociate Ex: NH 3 (ammonia)

9. Strength and Reactivity Acids/bases of the same initial molar concentration can react differently and conduct electricity differently if one is weak and the other strong. Acids/bases of the same initial molar concentration can react differently and conduct electricity differently if one is weak and the other strong. Ex: 2M HCl Strong Acid, Ex: 2M HCl Strong Acid, very conductive very reactive 2M HC 2 H 3 O 2 Weak Acid Weak Conduction Salad Dressing!!!

10. Conjugate Acid/Base Pairs Strong acid will have a weak conjugate base Strong acid will have a weak conjugate base Strong base will have a weak conjugate acid Strong base will have a weak conjugate acid

11. Hydrolysis Opposite reaction to neutralization Opposite reaction to neutralization Salt + Water Acid + Base

12. Parent Acid/Base If you know the salt involved you should be able to determine which acid and base it would form if water is added. If you know the salt involved you should be able to determine which acid and base it would form if water is added. Salt + Water Acid + Base Ex: NaCl with water (HOH) would form HCl and NaOH

13. You Try It Name the “parent” acid and base that would be produced from these salts. Name the “parent” acid and base that would be produced from these salts. Ex:Potassium chloride Ex:Potassium chloride Magnesium carbonate

14. pH and Hydrolysis Salts can yield neutral, acidic or basic solutions depending on what type of acid or base they produce. Salts can yield neutral, acidic or basic solutions depending on what type of acid or base they produce. SA/SB = Neutral SA/WB = Acidic WA/SB = Basic WA/WB = Undetermined

15. The Acid and Base Dissociation Constant, Ka & Kb

16. Setting up Ka and Kb Expressions weak acid:CH 3 COOH + H 2 O ↔ H 3 O + + CH 3 COO - Acid ionization constant: K a = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] [CH 3 COOH] weak base:NH 3 + H 2 O ↔ NH 4 + + OH - Base ionization constant: K b = [NH 4 + ][OH - ] Base ionization constant: K b = [NH 4 + ][OH - ] [NH 3 ] [NH 3 ] Acid and base ionization constants are the measure of the strengths of acids and bases. The larger the Ka/Kb value the stronger the acid or base

18. Ka Weak acids: Weak acids: only ionize to a small extent only ionize to a small extent come to a state of chemical equilibrium. come to a state of chemical equilibrium. Determine how much it ionizes by calculating the equilibrium constant (Ka) Determine how much it ionizes by calculating the equilibrium constant (Ka) Larger Ka: Larger Ka: stronger acid stronger acid more ions found in solution more ions found in solution more easily donate a proton more easily donate a proton

19. HCOOH (aq) + H 2 O (l) H 3 O + (aq) + HCOO - (aq) Ka = [H 3 O + ][HCOO - ] Ka = [H 3 O + ][HCOO - ][HCOOH] Notice how the Ka ignores the water since we are dealing with dilute solutions of acids, water is considered a constant and doesn’t have a concentration value Notice how the Ka ignores the water since we are dealing with dilute solutions of acids, water is considered a constant and doesn’t have a concentration value

20. Practice: 1. A solution of a weak acid, “HA”, is made up to be 0.15 M. Its pH was found to be 2.96. Calculate the value of Ka. Steps to follow: Steps to follow: 1. Write balanced equation 2. Calculate [H + ] using 10 -pH 3. Set up chart for equilibrium (ICE) 4. Solve using Ka expression

21. Answer 1. HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) 2. [H 3 O+]= 10 -2.96 = 0.0011 M HAH2OH2O H3O+H3O+ A-A- 0.15 -.0011 0.1390.0011 I C E

22. 4. Ka = [H 3 O + ][ A - ] [HA] = [0.0011][0.0011] = [0.0011][0.0011][0.139] = 8.7 x 10 -6 = 8.7 x 10 -6

23. Percent Ionization The fraction of acid molecules that dissociate compared with the initial concentration of the acid. Percent Ionization = [H 3 O + ] x 100% [HA i ] [HA i ] For the previous question: Percent Ionization = [0.0011] x 100% =0.73 % [0.15] [0.15]

24. Practice: Ka, for a hypothetical weak acid, HA, at 25°C is 2.2 x 10 -4. Ka, for a hypothetical weak acid, HA, at 25°C is 2.2 x 10 -4. a) Calculate [H 3 O + ] of a 0.20 M solution of HA. b) Calculate the percent ionization of HA.

25. Answer a) HA(aq) + H2O(l) H3O+(aq) + A-(aq) Ka = [H 3 O + ] [A-] = 2.2 x 10 -4 [HA] [HA] 2.2 x 10 -4 = (x)(x) (0.20M - x) (0.20M - x) x 2 = (2.2 x 10 -4 ) (0.20 M) x = 0.0066 M The [H 3 O + ] is 0.0066 M. Because it is a 1:1 ratio they are both the same concentration (x) Since Ka is rather small this number can be disregarded

26. b) % ionization = 0.0066 M x 100% = 3.3% 0.20 M 0.20 M

27. The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 C E The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 C E

28. The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 C E pH=1.28 The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 C E pH=1.28

29. The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 C E pH=1.28 [H + ]=10 -1.28 The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 C E pH=1.28 [H + ]=10 -1.28

30. The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 C E pH=1.28 [H + ]=10 -1.28 [H + ]=0.05248 M The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 C E pH=1.28 [H + ]=10 -1.28 [H + ]=0.05248 M

31. The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 C E0.052480.05248 pH=1.28 [H + ]=10 -1.28 [H + ]=0.05248 M The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 C E0.052480.05248 pH=1.28 [H + ]=10 -1.28 [H + ]=0.05248 M

32. The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 C - 0.05248 E0.047520.052480.05248 pH=1.28 [H + ]=10 -1.28 [H + ]=0.05248 M The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 C - 0.05248 E0.047520.052480.05248 pH=1.28 [H + ]=10 -1.28 [H + ]=0.05248 M

33. 1. The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 C - 0.05248 E0.047520.052480.05248 Ka=[H + ][HC 2 O 4 - ] =(0.05248) 2 [H 2 C 2 O 4 ]0.04752 Ka = 5.8 x 10 -2 1. The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I0.100 C - 0.05248 E0.047520.052480.05248 Ka=[H + ][HC 2 O 4 - ] =(0.05248) 2 [H 2 C 2 O 4 ]0.04752 Ka = 5.8 x 10 -2

34. Kb When using weak bases, the same rules apply as with weak acids, except you are solving for pOH and using [OH - ] When using weak bases, the same rules apply as with weak acids, except you are solving for pOH and using [OH - ]

35. If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 C E If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 C E

36. If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 C E pH=11.427 pOH=2.573 If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 C E pH=11.427 pOH=2.573

37. If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 C E pH=11.427 pOH=2.573 [OH - ]=10 -2.573 [OH - ]=0.002673 M If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 C E pH=11.427 pOH=2.573 [OH - ]=10 -2.573 [OH - ]=0.002673 M

38. If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 C - 0.002673 E0.39730.002673 0.002673 [OH - ]=0.002673 M If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 C - 0.002673 E0.39730.002673 0.002673 [OH - ]=0.002673 M

39. If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 C - 0.002673 E0.39730.002673 0.002673 [OH - ]=0.002673 M Kb=[NH 4 + ][OH - ]=(0.002673) 2 [NH 3 ] 0.3973 Kb = 1.8 x 10 -5 If the pH of 0.40 M NH 3 @ 25 o C is 11.427, calculate the Kb. NH 3 + H 2 O ⇄ NH 4 + +OH - I0.40 C - 0.002673 E0.39730.002673 0.002673 [OH - ]=0.002673 M Kb=[NH 4 + ][OH - ]=(0.002673) 2 [NH 3 ] 0.3973 Kb = 1.8 x 10 -5

40. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -.

41. CN - + H 2 O ⇄ HCN+OH - pH = 11.456 pOH = 2.55 [OH-] = 10 -2.55 [OH-] =.002858M The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -. CN - + H 2 O ⇄ HCN+OH - pH = 11.456 pOH = 2.55 [OH-] = 10 -2.55 [OH-] =.002858M

42. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -. CN - + H 2 O ⇄ HCN+OH - I0.20 C - 0.002858 E0.19710.002858 0.002858 The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -. CN - + H 2 O ⇄ HCN+OH - I0.20 C - 0.002858 E0.19710.002858 0.002858

43. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -. CN - + H 2 O ⇄ HCN+OH - I0.20 C- 0.002858 E0.19710.002858 0.002858 Kb= [HCN][OH - ] = (0.002858) 2 = 4.1 x 10 -5 [CN - ] 0.1971 The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN -. CN - + H 2 O ⇄ HCN+OH - I0.20 C- 0.002858 E0.19710.002858 0.002858 Kb= [HCN][OH - ] = (0.002858) 2 = 4.1 x 10 -5 [CN - ] 0.1971

44. Calculate the pH of 0.020 M H 3 BO 3 (Ka = 3.8 x 10 -10 ) (weak acid dissociates one H+ at a time) H 3 BO 3 ⇄ H + +H 2 BO 3 - I0.020 M C- x E0.020 - xxx Calculate the pH of 0.020 M H 3 BO 3 (Ka = 3.8 x 10 -10 ) (weak acid dissociates one H+ at a time) H 3 BO 3 ⇄ H + +H 2 BO 3 - I0.020 M C- x E0.020 - xxx disregard small Ka x 2 = 3.8 x 10 -10 0.020 x = [H + ] =2.76 x 10 -6 M pH =-Log[2.76 x 10 -6 ] pH =5.42 2 sig figs due to molarity and Ka