2. Properties of Acids  Taste sour  Contain H + ion  pH less than 7  React with bases to form a salt and water  React with some metals to produce hydrogen gas

3. Properties of Acids  Turn litmus paper red  Phenolphthalein is colorless in the presence of an acid  Bromothymol blue is yellow in the presence of an acid  Found in citrus fruits in the form of citric acid

4. Properties of Acids  Found in soured milk and in sore muscles in the form of lactic acid  Found in vitamin C in the form of ascorbic acid  Found in carbonated beverages in the form of carbonic acid…that’s also what you exhale

5. Properties of Bases  Taste bitter  Contain OH - ion  pH greater than 7  React with acids to form a salt and water  React with organic material

6. Properties of Bases  Feel slippery because they immediately begin to dissolve the outer layer of skin tissue  Turn litmus paper blue  Phenolphthalein is fuchsia in the presence of a base  Bromothymol blue is blue in the presence of a base

7. Properties of Bases  Found in drain cleaners usually in the form of sodium hydroxide  Found in ammonia-based cleaners like Windex  Lye (NaOH) is used to make soaps

8. Classifying Acids and Bases  Svante Arrhenius—Swedish guy who put forth his definitions of acids and bases in 1884 at the age of 25 Worked with our buddy van’t Hoff Received 1903 Nobel Chemistry Prize for electrolytic dissociation discoveries

9. Arrhenius Acids  are substances that will dissociate in water to yield hydrogen ion (H + )

10. Arrhenius Bases  are substances that will dissociate in water to yield hydroxide ion (OH - )

11. Classifying Acids and Bases  Johannes Brønsted (Danish) and Thomas Lowry (English) came up with a new way to classify acids and bases and their conjugates (pairs that have features in common but are opposites) in the 1920’s never received a Nobel for furthering these acid/base concepts, and Arrhenius never accepted them!

12. Brønsted-Lowry Acid  A reactant that donates a proton in a chemical reaction The proton is actually the hydrogen ion…since a hydrogen atom has 1 proton and 1 electron and the ion with a 1+ charge indicates that it has lost an electron

13. Brønsted-Lowry Base  A reactant that accepts a proton in a chemical reaction

14. Brønsted-Lowry Conjugate Acid  A product that is formed when a base accepts a proton in the reverse reaction, will donate a proton

15. Brønsted-Lowry Conjugate Base  A product that is formed when an acid donates a proton (what’s left after the donation occurs) in the reverse reaction, will accept a proton

16. Classifying Acids and Bases  Around the same time that Brønsted and Lowry were devising their acid/base scheme, our buddy Gilbert Lewis (yep, the same guy who did the dots) came up with yet another method of classifying them…it’s a broader method than Arrhenius, Brønsted, or Lowry ever postulated

17. Lewis Acid  A reactant that accepts an electron pair

18. Lewis Base  A reactant that donates an electron pair

19. Example #1 HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl - (aq) Or HCl (aq)  H + (aq) + Cl - (aq)

20. Example #1  HCl is an acid It dissociates to yield H 3 O + (hydronium ion), which is really water with an extra H +. (Arrhenius) It donates a proton (H + ) to water in the first reaction written. (Brønsted-Lowry)

21. Example #1  H 2 O is an base It accepts a proton (H + ) from HCl in the first reaction written. (Brønsted-Lowry)

22. Example #1  H 3 O + is a conjugate acid It is produced when the water accepts a proton (H + ) from HCl in the first reaction written. (Brønsted-Lowry) In the reverse reaction, it will donate a proton (H + ) to Cl - in the first reaction written. (Brønsted- Lowry)

23. Example #1  Cl - is a conjugate base It is produced when the HCl donates a proton (H + ) to water in the first reaction written. (Brønsted-Lowry) In the reverse reaction, it will accept a proton (H + ) from H 3 O + in the first reaction written. (Brønsted-Lowry)

24. Example #2 NH 3 (g) + H 2 O (l)  NH 4 + (aq) + OH - (aq)

25. Example #2  NH 3 is a base It accepts a proton (H + ) from water. (Brønsted-Lowry)

26. Example #2  H 2 O is an acid It donates a proton (H + ) to ammonia. (Brønsted-Lowry)

27. Example #2  NH 4 + is a conjugate acid It is produced when the ammonia accepts a proton (H + ) from water. (Brønsted-Lowry) In the reverse reaction, it will donate a proton (H + ) to OH -. (Brønsted-Lowry)

28. Example #2  OH - is a conjugate base It is produced when the water donates a proton (H + ) to ammonia. (Brønsted-Lowry) In the reverse reaction, it will accept a proton (H + ) from NH 4 +. (Brønsted-Lowry)

29. Water—our special friend  Did you notice that it behaved as a base in the first example and as an acid in the second example? n A substance that can behave as either an acid or a base is called, amphoteric or amphiprotic.

30. Example #3 H H—N—H + H +  H—N—H H H

31. Example #3  NH 3 is a base It donates a pair of electrons to H +. (Lewis)

32. Example #3  H + is an acid It accepts a pair of electrons from NH 3. (Lewis)

33. Autoionization of Water  Every acid and base will dissociate in water…even water (since it’s amphoteric)! 2H 2 O (l)  H 3 O + (aq) + OH - (aq) or H 2 O (l)  H + (aq) + OH - (aq)

34. Autoionization of Water  Usually, the 2 nd reaction is the one we will use since H 3 O + is just water with an extra H +.  Write the K expression for the 2 nd reaction, keeping in mind that we only include gaseous and aqueous phases.

35. Autoionization of Water  K = [H + ][OH - ] note that water is not included because it is a liquid n This expression is known as the K w, or equilibrium constant for water, expression n K w = [H + ][OH - ]

36. Autoionization of Water  The value of K w is 1 x 10 -14 M 2 at 25°C.  This is a small K value. n If the temperature changes, so does the value of K w

37. Autoionization of Water  Make an equilibrium chart for the dissociation, or autoionization, of water.

38. Autoionization of Water [H 2 O][H + ][OH - ] Initial --00 Change Eq.

39. Autoionization of Water [H 2 O][H + ][OH - ] Initial --00 Change --+x Eq.

40. Autoionization of Water [H 2 O][H + ][OH - ] Initial --00 Change --+x Eq. --xx

41. Autoionization of Water  Determine the equlibrium concentrations of both the hydrogen ion and the hydroxide ion by plugging into the K w expression. 1 x 10 -14 M 2 = [x][x]

42. Autoionization of Water 1 x 10 -14 M 2 = x 2 1 x 10 -7 M = x [H + ] = 1 x 10 -7 M [OH - ] = 1 x 10 -7 M

43. Autoionization of Water  Because the H + and OH - concentrations are equal, the solution is neutral.  If [H + ] > [OH - ], then the solution is an acid.  If [H + ] < [OH - ], then the solution is a base.

44. Autoionization of Water  Since every acid or base dissociation we will entertain occurs in water, then the K w expression is applicable to any of these dissociations.

45. Autoionization of Water  Thus, if you know the [H + ] concentration of a solution, you can determine the [OH - ] concentration.  And, if you know the [OH - ] concentration of a solution, you can determine the [H + ] concentration.

46. pH  Represents the “power of hydrogen”  Calculated by taking the opposite of the logarithm of the hydrogen ion concentration… pH = -log [H + ]

47. pH  Calculate the pH of water at 25°C knowing that the [H + ] is 1 x 10 -7 M. pH = -log [1 x 10 -7 ] pH = 7

48. pH  If you already know the pH of a solution, then you can find the [H + ] using: [H + ] = 10 -pH So, [H + ] = 10 -7 [H + ] = 1 x 10 -7 M

49. pOH  Represents the “power of hydroxide”  Calculated by taking the opposite of the logarithm of the hydroxide ion concentration… pOH = -log [OH - ]

50. pOH  Calculate the pOH of water at 25°C knowing that the [OH - ] is 1 x 10 -7 M. pOH = -log [1 x 10 -7 ] pOH = 7

51. Ooh, ah!  So, the sum of pH and pOH for all aqueous solutions will be 14. pH + pOH = 14

52. Example #4—show all work  Find the pH, pOH, and [OH - ] and state whether the solution is acidic, neutral or basic if the hydrogen ion concentration is 3.48 x 10 -4 M. pH = 3.46 pOH = 10.54 [OH - ] = 2.87 x 10 -11 M acidic

53. Example #5—show all work  Find the pOH, [H + ], and [OH - ] and state whether the solution is acidic, neutral or basic if the pH is 9.84. pOH = 4.16 [H + ] = 1.45 x 10 -10 M [OH - ] = 6.91 x 10 -5 M basic

54. Example #6—show all work  Find the pH, [H + ], and [OH - ] and state whether the solution is acidic, neutral or basic if the pOH is 12.7. pH = 1.30 [H + ] = 5.01 x 10 -2 M [OH - ] = 1.99 x 10 -13 M acidic

55. Example #7—show all work  Find the pH, pOH, and [H + ], and state whether the solution is acidic, neutral or basic if the [OH - ] is 5.26 x 10 -2 M. pH = 12.7 pOH = 1.28 [H + ] = 1.90 x 10 -13 M basic