1. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 1 Section 16 Acid-Base Equilibria

2. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 2 Solutions of a Weak Acid or Base The simplest acid-base equilibria are those in which a single acid or base solute reacts with water. –we will first look at solutions of pure weak acids or bases. –Next, we will also consider solutions of salts, which can have acidic or basic properties as a result of the reactions of their ions with water. Salts can be neutral, acidic, or basic.

3. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 3 Acid-Ionization Equilibria Acid ionization (or acid dissociation) is the reaction of an acid with water to produce hydronium ion (hydrogen ion) and the conjugate base anion (hydrolysis). –Because acetic acid is a weak electrolyte, it ionizes to a small extent in water, hence double arrow; not 100% like strong acid which wrote single arrow.

4. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 4 Acid-Ionization Equilibria For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid- ionization constant (also called the acid- dissociation constant, K a ). –Consider the generic monoprotic acid, HA.

5. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 5 Acid-Ionization Equilibria –The corresponding equilibrium expression is: –Since the concentration of water remains relatively constant (liquid = 1), we rearrange the equation to get:

6. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 6 Acid-Ionization Equilibria acid + water is K a –Thus, K a, the acid-ionization constant, equals the constant [H 2 O]K c.

7. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 7 Experimental Determination of Ka The degree of ionization of a weak electrolyte is the fraction of molecules that react with water to give ions. –Electrical conductivity or some other colligative property can be measured to determine the degree of ionization. –With weak acids, the pH can be used to determine the equilibrium composition of ions in the solution because it gives the concentration of hydronium ions at equil.

8. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 8 A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO 2. A 0.012 M solution of nicotinic acid has a pH of 3.39 at 25°C. Calculate the acid-ionization constant for this acid at 25°C. –It is important to realize that the solution was made 0.012 M in nicotinic acid, however, some molecules ionize making the equilibrium concentration of nicotinic acid less than 0.012 M, just a small amount less.

9. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 9 A Problem To Consider Starting 0.012 00 Change Equilibrium –Let x be the moles per liter of product formed.

10. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 10 A Problem To Consider –We can obtain the value of x from the given pH.

11. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 11 A Problem To Consider –Substitute this value of x in our equilibrium expression. –Note first, however, that the concentration of unionized acid remains virtually unchanged. Important point to remember for later. –Substitute this value of x in our equilibrium expression. HW 28

12. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 12 A Problem To Consider –To obtain the degree of ionization: –weak acids typically less than 5%. The lower the %ionized, less H + formed, the lower K a, weaker the acid. K a is measure of strength of acid or K b if base. Strength based on ionization not pH directly. Lower pH more acidic, does not mean stronger acid

13. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 13 K a is a measure of strength of acid. It indicates the amount ionized to form hydronium ions. The more hydronium ions formed, naturally the more acidic and lower pH. However, strength of acid is measure of amount ionized, not conc of acid. Conc can vary which ultimately will vary the pH of solution. To determine the strength of an acid you must compare K a values not pH of solution. pH is dependent on K a as well as conc. of solution. Must compare apples and apples. 10M HC 2 H 3 O 2 K a = 1.8 x 10 -5 pH = 1.87 0.1M HF K a = 6.9x10 -4 pH = 2.08 Which stronger acid?

14. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 14 Calculations With K a Once you know the value of K a, you can calculate the equilibrium concentrations of species HA, A-, and H 3 O + for solutions of different molarities. –The general method for doing this is same way we did K c or K p except we can simplify the math because of percent ionize is so small compared to initial conc of acid; remember 0.012-x=0.012. This wasn't the case for K c or K p problems so can't neglect in those type problems

15. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 15 Calculations With K a Note that in our previous example, the degree of dissociation was so small that “x” was negligible compared to the concentration of nicotinic acid. It is the small value of the degree of ionization that allowed us to ignore the subtracted x in the denominator of our equilibrium expression. The degree of ionization of a weak acid depends on both the K a and the concentration of the acid solution

16. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 16 Calculations With K a How do you know when you can use this simplifying assumption? – then this simplifying assumption of ignoring the subtracted x gives an acceptable error of less than 5% otherwise must do quadratic formula. Another way look at it factor of two - three difference in power of 10 is needed between K a and conc of acid. –It can be shown that if the acid concentration, C a, divided by the K a exceeds 100, that is,

17. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 17 ex. Calculate the pH in a 1.00 M solution of nitrous acid, HNO 2, for which K a = 6.0 x 10 -4. HNO 2 + H 2 O H 3 O + + NO 2 -

18. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 18 A Problem To Consider Really two equil: K w as well but neglect most of the time as long as [H + ] is greater than 10 -6 M from the acid. Really [H + ] tot = [H + ] acid +[H + ] H2O (water so small neglect)

19. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 19 ex. Calculate the pH of 10.00 mL of 1.00 M solution of nitrous acid, HNO 2, diluted to 100.0 mL with water for which K a = 6.0 x 10 -4. HNO 2 + H 2 O H 3 O + + NO 2 -

20. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 20 A Problem To Consider What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetyl salicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25°C. –The molar mass of HC 9 H 7 O 4 is 180.2 g.

21. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 21 A Problem To Consider What is the pH at 25°C of a solution obtained by dissolving 0.325 g of acetylsalicylic acid (aspirin), HC 9 H 7 O 4, in 0.500 L of water? The acid is monoprotic and K a =3.3 x 10 -4 at 25°C. Starting[] 0.003607 00 Change [] Equilibrium[] –These data are summarized below.

22. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 22 A Problem To Consider –Note that which is less than 100 or tell factor 10 -3 vs 10 -4 not factor 2-3; therefore, won't work, so we must solve the equilibrium equation exactly.

23. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 23 –If we substitute the equilibrium concentrations and the K a into the equilibrium constant expression, we get Starting[] 0.003607 00 Change [] -x +x Equilibrium[] 0.003607 –x xx

24. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 24 –You can solve this equation exactly by using the quadratic formula.

25. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 25 –Taking the upper sign, we get and neglect water OK A Problem To Consider –Now we can calculate the pH. HW 29

26. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 26 Calculate the pH of a solution that has a concentration of 1.0 x 10 -8 M NaOH.

27. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 27 Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. –Sulfuric acid, for example, can lose two protons in aqueous solution.

28. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 28 Polyprotic Acids –For a weak diprotic acid like carbonic acid, H 2 CO 3, two simultaneous equilibria must be considered.

29. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 29 –Each equilibrium has an associated acid-ionization constant. Polyprotic Acids –For the loss of the first proton –For the loss of the second proton

30. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 30 –In the case of a triprotic acid, such as H 3 PO 4, the third ionization constant, K a3, is smaller than the second one, K a2. Polyprotic Acids –In general, the second ionization constant, K a2, for a polyprotic acid is smaller than the first ionization constant, K a1. Harder to pull proton off charged species. Total H + is sum of all equil H + but usually neglect all but the principal rxn.

31. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 31 –However, reasonable assumptions can be made that simplify these calculations as we show in the next example. Polyprotic Acids –When several equilibria occur at once, it might appear complicated to calculate equilibrium compositions.

32. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 32 –For diprotic acids, K a2 is so much smaller than K a1 that the smaller amount of hydronium ion produced in the second reaction can be neglected. A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12.

33. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 33 –The pH can be determined by simply solving the equilibrium problem posed by the first ionization. A Problem To Consider –the first ionization is: Starting 0.10 00 Change Equilibrium

34. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 34 A Problem To Consider –Assuming that x is much smaller than 0.10 (1265), you get

35. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 35 –The ascorbate ion, C 6 H 6 O 6 2-, which we will call y, is produced only in the second ionization of H 2 C 6 H 6 O 6. A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6. What is the pH of a 0.10 M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 2- ? The acid ionization constants are K a1 = 7.9 x 10 -5 and K a2 = 1.6 x 10 -12.

36. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 36 –Assume the starting concentrations for H C 6 H 6 O 6 - and H 3 O + to be those from the first equilibrium. Starting [] 0.0028 0 Change [] Equilibrium []

37. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 37 –Substituting into the equilibrium expression –Assuming y is much smaller than 0.0028 (1.75 x 10 9 > 100), the equation simplifies to –Hence, –typically, the concentration of the anion ion equals K a2 if starting with reactants of first equil. HW 30

38. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 38 Base-Ionization Equilibria Equilibria involving weak bases are treated similarly to those for weak acids. –Ammonia, for example, ionizes in water as follows. –The corresponding equilibrium constant is:

39. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 39 Base-Ionization Equilibria Equilibria involving weak bases are treated similarly to those for weak acids. –Ammonia, for example, ionizes in water as follows. –The concentration of water is nearly constant.

40. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 40 Base-Ionization Equilibria Equilibria involving weak bases are treated similarly to those for weak acids. Higher K b stronger base, ionize more OH -, more basic. –In general, a weak base B with the base ionization has a base ionization constant equal to HW 31

41. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 41 A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. 1.Write the equation and make a table of concentrations. 2.Set up the equilibrium constant expression. 3.Solve for x = [OH - ]. –As before, we will follow the three steps in solving an equilibrium.

42. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 42 A Problem To Consider What is the pH of a 0.20 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x 10 -9. –Pyridine ionizes by picking up a proton from water (as ammonia does). Starting 0.20 00 Change Equilibrium

43. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 43 A Problem To Consider –Note that which is much greater than 100, so we may use the simplifying assumption that (0.20-x)  (0.20).

44. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 44 A Problem To Consider Starting 0.20 00 Change -x +x Equilibrium 0.20-x xx –The equilibrium expression is

45. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 45 A Problem To Consider –Using our simplifying assumption that the x in the denominator is negligible, we get

46. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 46 A Problem To Consider –Solving for pOH –Since pH + pOH = 14.00 HW 32

47. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 47 Acid-Base Properties of a Salt Solution One of the successes of the Brønsted- Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. –A 0.1 M solution has a pH of 11.1 and is therefore fairly basic. –Consider a solution of sodium cyanide, NaCN.

48. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 48 –Sodium ion, Na +, is unreactive with water, but the cyanide ion, CN -, reacts to produce HCN and OH - making it a basic solution Acid-Base Properties of a Salt Solution –From the Brønsted-Lowry point of view, the CN - ion acts as a base, because it accepts a proton from H 2 O.

49. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 49 –You can also see that OH - ion is a product, so you would expect the solution to have a basic pH. This explains why NaCN solutions are basic. –The reaction of the CN - ion with water is referred to as the hydrolysis of CN -. Acid-Base Properties of a Salt Solution

50. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 50 The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion. –The CN - ion hydrolyzes to give the conjugate acid and hydroxide. Acid-Base Properties of a Salt Solution

51. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 51 –The hydrolysis reaction for CN - has the form of a base ionization so you write the K b expression for it. Acid-Base Properties of a Salt Solution

52. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 52 –The NH 4 + ion hydrolyzes to the conjugate base (NH 3 ) and hydronium ion. Acid-Base Properties of a Salt Solution An example of an cation that undergoes hydrolysis is the ammonium ion

53. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 53 Predicting Whether a Salt is Acidic, Basic, or Neutral How can you predict whether a particular salt will be acidic, basic, or neutral and undergoes hydrolysis? –The Brønsted-Lowry concept illustrates the inverse relationship in the strengths of conjugate acid-base pairs. –Consequently, the anions of weak acids (poor proton donors) are good proton acceptors. –Anions of weak acids are basic.

54. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 54 Predicting Whether a Salt is Acidic, Basic, or Neutral –On the other hand, the anions of strong acids (good proton donors) have virtually no basic character, that is, they do not hydrolyze. – For example, the Cl - ion, which is conjugate to the strong acid HCl, shows no appreciable reaction with water. If did, strong acid breaks up 100% and hydroxide produced cancelled with hydronium of strong acid and make neutral solution

55. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 55 Predicting Whether a Salt is Acidic, Basic, or Neutral –Small highly charged (attract O more weakens OH bond) cations have an attraction for the O on water and cause the production of H + while large low charged cations do not. Low charged large ions are group I and II metals (cations of strong bases) and do not undergo hydrolysis. –For example, In summary, If cation is that of the strong base, it will have neutral hydrolysis; otherwise, cation will have acidic hydrolysis If anion of monoprotic strong acid, it will have neutral hydrolysis; otherwise, anion will have basic hydrolysis

56. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 56 Salts may be acidic/basic/neutral and are composed of cations (positive ions) and anions (negative ions) cation anion Acidic or neutral basic or neutral Cations of strong bases Anions of monoprotic strong are neutral; otherwise,acids are neutral; otherwise, cation contributes acidity anion contributes basicity Na +, K +, Li +, Ca 2+, Sr 2+, Ba 2+ Cl -, Br -, I -, NO 3 -, ClO 4 - add add neutral; rest of cations addneutral; rest of anions add acidity to salt basicity to salt Depending on the two components (cation/anion) the overall salt will be acidic/neutral/basic – nn  n; an  a; nb  b.

57. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 57 Predicting Whether a Salt is Acidic, Basic, or Neutral To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt.

58. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 58 Predicting Whether a Salt is Acidic, Basic, or Neutral

59. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 59 Predicting Whether a Salt is Acidic, Basic, or Neutral AlCl 3 Zn(NO 3 ) 2 KClO 4 Na 3 PO 4 LiF NH 4 FK a = 5.6x10 -10 > K b = 1.4x10 -11 NH 4 ClOK a = 5.6x10 -10 < K b = 3.6x10 -7 NH 4 C 2 H 3 O 2 K a = 5.6x10 -10 = K b = 5.6x10 -10 one with larger K will dictate acidity/basicity - stronger species HW 33

60. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 60 The pH of a Salt Solution To calculate the pH of a salt solution would require the K a of the acidic cation or the K b of the basic anion. –The ionization constants of ions are not listed directly in tables because the values are easily related to their conjugate species. –Thus the K b for CN - is related to the K a for HCN.

61. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 61 The pH of a Salt Solution To see the relationship between K a and K b for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN -. KaKa KbKb –When two reactions are added, their equilibrium constants are multiplied. KwKw

62. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 62 KaKa KbKb –Therefore, KwKw HW 34

63. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 63 The pH of a Salt Solution For a solution of a salt in which only one ion hydrolyzes, the calculation of equilibrium composition follows that of weak acids and bases. –The only difference is first must obtain the K a or K b for the ion that hydrolyzes.

64. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 64 Attack of hydrolysis problem When attacking pH (hydrolysis problems) always ask yourself if acid, base, or salt -- if acid or base, is it strong (straight forward) or weak (K a or K b problem) -- If salt, write dissociation eq and examine cation/anions involved to determine if acid or base hydrolysis is available then K a or K b problem.

65. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 65 A Problem To Consider What is the pH of a 0.10 M NaCN solution at 25 °C? The K a for HCN is 4.9 x 10 -10.

66. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 66 A Problem To Consider

67. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 67 example: What is the pH of 0.10 M NH 4 I solution? K b NH 3 = 1.8 x 10 -5 HW 35

68. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 68 The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.

69. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 69 The Common Ion Effect –The equilibrium composition would shift to the left and the degree of ionization of the acetic acid is decreased, meaning less hydronium formed and higher pH of solution as compared to pure acid. –This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect. Another source of ion coming from an additional solute added to the solution.

70. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 70 Buffers When common ion effect involves weak acid/conjugate base it is referred to as a buffer. Buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. –Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid. Typically conj is in form of salt. –Thus, a buffer contains both an acid species and a base species in equilibrium which allows it to absorb any added acid or base and keep pH of solution relatively constant

71. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 71 Buffers –Consider a buffer with equal molar amounts of HA and its conjugate base A -.

72. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 72 A Problem To Consider An aqueous solution is 0.025 M in formic acid, HCHO 2 and 0.018 M in sodium formate, NaCHO 2. What is the pH of the solution. The K a for formic acid is 1.7 x 10 -4.

73. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 73 A Problem To Consider

74. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 74 A solution is prepared to be 0.15 M HC 2 H 3 O 2 and 0.20 M NaC 2 H 3 O 2. What is the pH of this solution at 25 o C? K a HC 2 H 3 O 2 @ 25 o C = 1.8 x 10 -5

75. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 75 Calculate the pH of 75 mL of a buffer solution (0.15 M HC 2 H 3 O 2 / 0.20 M NaC 2 H 3 O 2 pH = 4.87) to which 9.5 mL of 0.10 M HCl is added. K a HC 2 H 3 O 2 @ 25 o C = 1.8 x 10 -5

76. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 76 HW 36

77. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 77 Buffers –Two important characteristics of a buffer are its buffer capacity and its pH. –Buffer capacity depends on the amount of acid and conjugate base present in the solution. –The next topic illustrates how to calculate the pH of a buffer by special eq but can do like we just did as well to keep with thought process and not memorize a equation.

78. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 78 The Henderson-Hasselbalch Equation How do you prepare a buffer of given pH? –A buffer must be prepared from selecting a conjugate acid-base pair in which the K a of the acid is approximately equal to the desired H 3 O + concentration. –Note: you can add extra conjugate base, higher pH, or less base, lower pH, to get to exact pH wanted. –To illustrate, consider a buffer of a weak acid HA and its conjugate base A -. The acid ionization equilibrium is:

79. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 79 The Henderson-Hasselbalch Equation –The acid ionization constant is: –By rearranging, you get an equation for the H 3 O + concentration.

80. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 80 The Henderson-Hasselbalch Equation –Taking the negative logarithm of both sides of the equation we obtain: –The previous equation can be rewritten

81. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 81 The Henderson-Hasselbalch Equation –More generally, you can write –This equation relates the pH of a buffer to the concentrations of the conjugate acid and base. It is known as the Henderson-Hasselbalch equation (for buffers only). Also way to work previous buffer problem. Note pH = pKa when conc base=acid

82. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 82 Let’s prepare a buffer with pH of 4.90 –So to prepare a buffer of a given pH (for example, pH 4.90) we need a conjugate acid-base pair with a pK a close to the desired pH or 10 -4.90 to find K a close to 1.25 x 10 -5 (desired hydronium conc. equivalent to pH desired) HC 2 H 3 O 2 K a = 1.8 x 10 -5 H 2 CO 3 K a = 4.3 x 10 -7 HNO 2 K a = 4.5 x 10 -4 –The K a for acetic acid is 1.8 x 10 -5, and its pK a is 4.74 or pH of same value if equal conc of acid/salt are used. –You could get a buffer of pH 4.90 by increasing the ratio of [base]/[acid], meaning more basic salt.

83. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 83 What if I wanted to use 0.100 M HC 2 H 3 O 2 (K a = 1.8 x 10 -5 ), what concentration of NaC 2 H 3 O 2 would I need to obtain a buffer with pH =4.90. HW 37

84. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 84 Acid-Ionization Titration Curves (or mixtures) An acid-base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid). –Such curves are used to gain insight into the titration process. –You can use titration curves to choose an appropriate indicator that will show when the titration is complete.

85. Figure: Curve for the titration of a strong acid by a strong base.

86. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 86 Figure shows a curve for the titration of HCl with NaOH. –Note that the pH changes slowly until the titration approaches the equivalence point. –The equivalence point is the point in a titration when a stoichiometric amount of reactant has been added; contrary to end point (observable point). –At this equivalence point of SA and SB, the pH of the solution is 7.0 because it contains a salt, NaCl, that does not hydrolyze. –However, the pH changes rapidly from a pH of about 3 to a pH of about 11. –To detect the equivalence point, you need an acid-base indicator that changes color beyond equivalence point. –Phenolphthalein can be used because it changes color in the pH range 8.2-10.

87. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 87 Previous example equivalent pt was at 7 but not all neutralization reactions eq pts are at 7. During a titration or just mixing an acid and base together to make a solution, the resulting solution will contain a salt and water (Acid + Base --> salt + water). One of the chemical properties of acids and bases is that they neutralize one another; however, that doesn’t mean that the product will be neutral. The misconception is that if the acid and base are in stoich proportions that the resulting solution is neutral. This is not true. The salt formed may be a acidic, basic, or neutral salt and will dictate the pH of the solution. Common sense can help you predict the pH of a mixture. SA + SB --> neutral salt only true neutralization pH =7 WA + SB --> basic salt original acid and base neutralize but product has basic properties and basic pH SA + WB --> acidic salt original acid and base neutralize but product has acidic properties and acidic pH

88. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 88 When working acid-base titration problems or mixtures treat them all the same but must be able to examine the components remaining after neutralization to determine what affects pH or not (strong, weak, salts). They are all done the same. The problem will dictate where you go. We can derive titration curve by doing a calculation at different points and draw curve or measure with pH meter. 1.) Calculate mmols initially 2.) write the neutralization reaction and find mmols of components left after neutralization 3.) find new conc. of species that affect pH 4.) do hydrolysis of main species that affect pH (like earlier in chapter)

89. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 89 A Problem To Consider Calculate the pH of a solution in which 50.0 mL of 1.000 M NaOH is added to 50.0 mL of 1.000 M HCl.

90. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 90 A Problem To Consider Calculate the pH of a solution in which 49.99 mL of 1.000 M NaOH is added to 50.00 mL of 1.000 M HCl.

91. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 91 A Problem To Consider Calculate the pH of a solution in which 50.01 mL of 1.000 M NaOH is added to 50.00 mL of 1.000 M HCl. HW 38

92. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 92 A Problem To Consider Calculate the pH of a solution in which 20.00 mL of 1.000 M NaOH is added to 50.00 mL of 1.000 M HC 2 H 3 O 2. K a HC 2 H 3 O 2 = 1.8x10 -5

93. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 93 A Problem To Consider continue

94. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 94 A Problem To Consider Calculate the pH of a solution in which 50.00 mL of 1.000 M NaOH is added to 50.00 mL of 1.000 M HC 2 H 3 O 2. K a HC 2 H 3 O 2 = 1.8x10 -5

95. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 95 A Problem To Consider continue

96. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 96 A Problem To Consider Calculate the pH of a solution in which 51.00 mL of 1.000 M NaOH is added to 50.00 mL of 1.000 M HC 2 H 3 O 2. K a HC 2 H 3 O 2 = 1.8x10 -5

97. Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 97 A Problem To Consider continue HW 39