1. Weak Acids & Bases Chapter 16

2. Dissociation Constants Since weak acids do not dissociate completely, [H 3 O + ] ≠ [acid] For a generalized acid dissociation, the equilibrium expression would be This equilibrium constant is called the acid-dissociation constant, K a. [H 3 O + ] [A − ] [HA] K c = HA (aq) + H 2 O (l) A − (aq) + H 3 O + (aq)

3. Dissociation Constants The greater the value of K a, the stronger the acid.

4. Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38 (this solution is at equilibrium). Calculate K a for formic acid at this temperature. HCOOH (aq) ↔H + (aq) + HCOO - (aq) We know that [H 3 O + ] [COO − ] [HCOOH] K a =

5. Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate K a for formic acid at this temperature. –To calculate K a, we need the equilibrium concentrations of all three things. Use an ICE Table –We can find [H 3 O + ], which is the same as [HCOO − ], from the pH based on the stoichiometry of the reaction.

6. Calculating K a from the pH pH = −log [H 3 O + ] 2.38 = −log [H 3 O + ] −2.38 = log [H 3 O + ] 10 −2.38 = 10 log [H 3 O + ] = [H 3 O + ] 4.2  10 −3 = [H 3 O + ] = [HCOO − ] (Concentration at equilibrium)

7. Calculating K a from pH Now we can set up a table… [HCOOH], M[H 3 O + ], M[HCOO − ], 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 Since weak acids do not dissociate much, you can neglect the value that is changed for the acid, it’s concentration at equilibrium is the same as the concentration at equilibrium

8. Calculating K a from pH [4.2  10 −3 ] [0.10] K a = = 1.8  10 −4

9. Practice problem A 0.020 M solution of niacin has a pH of 3.26. What is the acid-dissociation constant, K a, for niacin? [Niacin] M[H 3 O + ], M[niacin − ], M Initially0.02000 Changenegligible+ x At Equilibrium 0.020xx

10. 3.26 = -log [H 3 O + ] at equilibrium [H 3 O + ] = 5.50 x 10 -4 M = x K a = [5.50 x 10 -4 M ][5.50 x 10 -4 M ]/0.02 = K a = 1.5 x 10 -5 [Niacin] M[H 3 O + ], M[niacin − ], M Initially0.02000 Changenegligible+ x At Equilibrium 0.020xx

11. Calculating Percent Ionization Another way to measure an acid’s strength if calculate the percent ionization –The higher the percent, the stronger the acid Percent Ionization =  100 In this example [H 3 O + ] eq = 4.2  10 −3 M [HCOOH] initial = 0.10 M [H 3 O + ] eq [HA] initial

12. Calculating Percent Ionization Percent Ionization =  100 4.2  10 −3 0.10 = 4.2%

13. 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.

14. 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)

15. Calculating pH from K a We next set up an ICE table… [C 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. Since [H 3 O + ] <<<< [C 2 H 3 O 2 ]

16. 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

17. Calculating pH from K a pH = −log [H 3 O + ] pH = −log (2.3  10 −3 ) pH = 2.64

18. 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.

19. Weak Bases Bases react with water to produce hydroxide ion.

20. Weak Bases The equilibrium constant expression for this reaction is [HB] [OH − ] [B − ] K b = where K b is the base-dissociation constant.

21. Weak Bases K b can be used to find [OH − ] and, through it, pH.

22. pH of Basic Solutions What is the pH of a 0.15 M solution of NH 3 ? [NH 4 + ] [OH − ] [NH 3 ] K b = = 1.8  10 −5 NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH − (aq)

23. pH of Basic Solutions Tabulate the data. [NH 3 ], M[NH 4 + ], M[OH − ], M Initially0.1500 At Equilibrium 0.15 - x  0.15 xx

24. pH of Basic Solutions (1.8  10 −5 ) (0.15) = x 2 2.7  10 −6 = x 2 1.6  10 −3 = x 2 (x) 2 (0.15) 1.8  10 −5 =

25. pH of 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

26. K a and K b K a and K b are related in this way: K a  K b = K w Therefore, if you know one of them, you can calculate the other.

27. pH vs pK a pH = -log ([H 3 O+]) pK a = -log ([K a ]) pH measures the acidity of a given solution. pK a measures the strength of an acid –You can create a solution of a weak acid and a solution of a strong acid that have the same pH, as long as you have a high enough concentration of the weak acid

28. Factors Affecting Acid Strength The more polar the H-X bond in a period and/or the weaker the H-X bond, the more acidic the compound. Acidity increases from left to right across a row and from top to bottom down a group.

29. Factors Affecting Acid Strength Strong bonds (bonds that are less polar) in a molecule cause the acidity to decrease because the hydrogen does not transfer easily