1. Acids and Bases

2. Acid/Base Definitions  Arrhenius Model  Acids produce hydrogen ions in aqueous solutions  Bases produce hydroxide ions in aqueous solutions  Bronsted-Lowry Model  Acids are proton donors  Bases are proton acceptors  Lewis Acid Model  Acids are electron pair acceptors  Bases are electron pair donors

3. Acid Dissociation HA  H + + A - Acid Proton Conjugate base Alternately, H + may be written in its hydrated form, H 3 O + (hydronium ion)

4. Dissociation of Strong Acids Strong acids are assumed to dissociate completely in solution. Large K a or small K a ? Reactant favored or product favored?

5. Dissociation Constants: Strong Acids AcidFormula Conjugate Base KaKa Perchloric HClO 4 ClO 4 - Very large Hydriodic HI I - Very large Hydrobromic HBr Br - Very large Hydrochloric HCl Cl - Very large Nitric HNO 3 NO 3 - Very large Sulfuric H 2 SO 4 HSO 4 - Very large Hydronium ion H 3 O + H 2 O 1.0

6. Dissociation of Weak Acids Weak acids are assumed to dissociate only slightly (less than 5%) in solution. Large K a or small K a ? Reactant favored or product favored?

7. Dissociation Constants: Weak Acids AcidFormula Conjugate Base KaKa Iodic HIO 3 IO 3 - 1.7 x 10 -1 Oxalic H 2 C 2 O 4 HC 2 O 4 - 5.9 x 10 -2 Sulfurous H 2 SO 3 HSO 3 - 1.5 x 10 -2 Phosphoric H 3 PO 4 H 2 PO 4 - 7.5 x 10 -3 Citric H 3 C 6 H 5 O 7 H 2 C 6 H 5 O 7 - 7.1 x 10 -4 Nitrous HNO 2 NO 2 - 4.6 x 10 -4 Hydrofluoric HF F - 3.5 x 10 -4 Formic HCOOH HCOO - 1.8 x 10 -4 Benzoic C 6 H 5 COOH C 6 H 5 COO - 6.5 x 10 -5 Acetic CH 3 COOH CH 3 COO - 1.8 x 10 -5 Carbonic H 2 CO 3 HCO 3 - 4.3 x 10 -7 Hypochlorous HClO ClO - 3.0 x 10 -8 Hydrocyanic HCN CN - 4.9 x 10 -10

8. PROBLEM: Write the simple dissociation (ionization) reaction (omitting water) for each of the following acids. a.Hydrochloric acid (HCl) b.Acetic acid (HC 2 H 3 O 2 ) c.The ammonium ion (NH 4 + ) d.The anilinium ion (C 6 H 5 NH 3 + ) e.The hydrated aluminum(III) ion [Al(H 2 O) 6 ] 3+

9. Self-Ionization of Water H 2 O + H 2 O  H 3 O + + OH - At 25 , [H 3 O + ] = [OH - ] = 1 x 10 -7 K w is a constant at 25  C: K w = [H 3 O + ][OH - ] K w = (1 x 10 -7 )(1 x 10 -7 ) = 1 x 10 -14

10. Calculating pH, pOH pH = -log 10 (H 3 O + ) pOH = -log 10 (OH - ) Relationship between pH and pOH pH + pOH = 14 Finding [H 3 O + ], [OH - ] from pH, pOH [H 3 O + ] = 10 -pH [OH - ] = 10 -pOH

11. pH and pOH Calculations

12. pH Scale

13. Problem: Calculate [H + ] or [OH - ] as required for each of the following solutions @ 25 o C, and state whether the solution is neutral, acidic, or basic. a.1.0 x 10 -5 M OH - b.1.0 x 10 -7 M OH - c.10.0M H +

14. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #1: Write the dissociation equation HC 2 H 3 O 2  C 2 H 3 O 2 - + H +

15. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #2: ICE it! HC 2 H 3 O 2  C 2 H 3 O 2 - + H + I C E 0.50 0 0 - x +x 0.50 - x xx

16. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #3: Set up the law of mass action HC 2 H 3 O 2  C 2 H 3 O 2 - + H + 0.50 - xxx E

17. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #4: Solve for x, which is also [H + ] HC 2 H 3 O 2  C 2 H 3 O 2 - + H + 0.50 - xxx E [H + ] = 3.0 x 10 -3 M

18. A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x 10 -5 ? Step #5: Convert [H + ] to pH HC 2 H 3 O 2  C 2 H 3 O 2 - + H + 0.50 - xxx E

19. Percent Dissociation = Amount dissociated (mol/L) Initial concentration (mol/L) X 100%

20. PRACTICE TIME! PLEASE BEGIN TO SOLVE THE FOLLOWING PROBLEMS RELATING TO: NATURE OF ACIDS-n-BASES, AUTOIONIZATION OF WATER, pH SCALE AND THE SOLUTIONS OF ACIDS. pp. 673-75: #28-30,33-36,38,40,44,50 52,55,58,59,62 and 64a-d Please complete problems #28-30,33-36,38 and 40 on a separate sheet of paper (Will be collected at end of block!)

21. Dissociation of Strong Bases  Strong bases are metallic hydroxides  Group I hydroxides (NaOH, KOH) are very soluble  Group II hydroxides (Ca, Ba, Mg, Sr) are less soluble  pH of strong bases is calculated directly from the concentration of the base in solution MOH(s)  M + (aq) + OH - (aq)

22. Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: CH 3 NH 2 + H 2 O  CH 3 NH 3 + + OH - K b = 4.38 x 10 -4

23. K b for Some Common Weak Bases BaseFormula Conjugate Acid KbKb Ammonia NH 3 NH 4 + 1.8 x 10 -5 Methylamine CH 3 NH 2 CH 3 NH 3 + 4.38 x 10 -4 Ethylamine C 2 H 5 NH 2 C 2 H 5 NH 3 + 5.6 x 10 -4 Diethylamine (C 2 H 5 ) 2 NH (C 2 H 5 ) 2 NH 2 + 1.3 x 10 -3 Triethylamine (C 2 H 5 ) 3 N (C 2 H 5 ) 3 NH + 4.0 x 10 -4 Hydroxylamine HONH 2 HONH 3 + 1.1 x 10 -8 HydrazineH 2 NNH 2 H 2 NNH 3 + 3.0 x 10 -6 Aniline C 6 H 5 NH 2 C 6 H 5 NH 3 + 3.8 x 10 -10 Pyridine C 5 H 5 N C 5 H 5 NH + 1.7 x 10 -9 Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?

24. Reaction of Weak Bases with Water The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion: B + H 2 O  BH + + OH - (Yes, all weak bases do this – DO NOT endeavor to make this complicated!)

25. A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? Step #1: Write the equation for the reaction NH 3 + H 2 O  NH 4 + + OH -

26. A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? Step #2: ICE it! I C E 0.50 0 0 - x +x 0.50 - x xx NH 3 + H 2 O  NH 4 + + OH -

27. A Weak Base Equilibrium Problem Step #3: Set up the law of mass action 0.50 - xxx E What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? NH 3 + H 2 O  NH 4 + + OH -

28. A Weak Base Equilibrium Problem Step #4: Solve for x, which is also [OH - ] 0.50 - xxx E [OH - ] = 3.0 x 10 -3 M NH 3 + H 2 O  NH 4 + + OH - What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ?

29. A Weak Base Equilibrium Problem Step #5: Convert [OH - ] to pH 0.50 - xxx E What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x 10 -5 ? NH 3 + H 2 O  NH 4 + + OH -

30. Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is from a strong base, anion from a strong acid KCl, KNO 3 NaCl NaNO 3 Both ions are neutral Neutral These salts simply dissociate in water: KCl(s)  K + (aq) + Cl - (aq)

31. Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is from a strong base, anion from a weak acid NaC 2 H 3 O 2 KCN, NaF Cation is neutral, Anion is basic Basic C 2 H 3 O 2 - + H 2 O  HC 2 H 3 O 2 + OH- base acid acid base The basic anion can accept a proton from water:

32. Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is the conjugate acid of a weak base, anion is from a strong acid NH 4 Cl, NH 4 NO 3 Cation is acidic, Anion is neutral Acidic NH 4 + (aq)  NH 3 (aq) + H + (aq) Acid Conjugate Proton base The acidic cation can act as a proton donor:

33. Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is the conjugate acid of a weak base, anion is conjugate base of a weak acid NH 4 C 2 H 3 O 2 NH 4 CN Cation is acidic, Anion is basic See below  IF K a for the acidic ion is greater than K b for the basic ion, the solution is acidic  IF K b for the basic ion is greater than K a for the acidic ion, the solution is basic  IF K b for the basic ion is equal to K a for the acidic ion, the solution is neutral

34. Acid-Base Properties of Salts Type of SaltExamplesCommentpH of solution Cation is a highly charged metal ion; Anion is from strong acid Al(NO 3 ) 2 FeCl 3 Hydrated cation acts as an acid; Anion is neutral Acidic Step #1: AlCl 3 (s) + 6H 2 O  Al(H 2 O) 6 3+ (aq) + Cl - (aq) Salt water Complex ion anion Step #2: Al(H 2 O) 6 3+ (aq)  Al(OH)(H 2 O) 5 2+ (aq) + H + (aq) Acid Conjugate base Proton

35. Effect of Structure on Acid-Base Properties