1. Chapter 17 Acid-Base Equilibria

2. The simplest acid–base equilibria are those in which a weak acid or a weak base reacts with water. We can write an acid equilibrium reaction for the generic acid, HA. HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq)

3. Acetic acid is a weak acid. It reacts with water as follows: HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 - (aq) Here A from the previous slide = C 2 H 3 O 2 - (aq)

4. The equilibrium constant for the reaction of a weak acid with water is called the acid-ionization constant (or acid- dissociation constant), K a. Liquid water is not included in the equilibrium constant expression.

7. HCN Highest pH Lowest pH HF HNO 2 HC 2 H 3 O 2 HCN HF HNO 2 HC 2 H 3 O 2 1.7 × 10 -5 4.5 × 10 -4 K a 6.8 × 10 -4 4.9 × 10 -10 In order of decreasing K a :

8. Calculations with K a Given the value of K a and the initial concentration of HA, you can calculate the equilibrium concentration of all species. Given the value of K a and the initial concentration of HA, you can calculate the degree of ionization and the percent ionization. Given the pH of the final solution and the initial concentration of HA, you can find the value of K a and the percent ionization.

9. We can be given pH, percent or degree ionization, initial concentration, and K a. From pH, we can find [H 3 O + ]. From percent or degree ionization, we can find K a. Using what is given, we can find the other quantities.

10. Sore-throat medications sometimes contain the weak acid phenol, HC 6 H 5 O. A 0.10 M solution of phenol has a pH of 5.43 at 25°C. a.What is the acid-ionization constant, K a, for phenol at 25°C? b.What is the degree of ionization?

11. We were told that pH = 5.43. That allows us to find [H 3 O + ] = = x = [C 6 H 5 O - ]. Now we find [HC 6 H 5 O] = 0.10 – x = HC 6 H 5 O(aq) +H 2 O(l)H 3 O + (aq) +C 6 H 5 O - (aq) Initial Change Equilibrium

12. Finally, we write the expression for K a and substitute the concentrations we now know.

13. The degree of ionization is the ratio of ionized concentration to original concentration: Percent ionization is the degree of ionization × 100%: Percent ionization = 3.7 x 10 -3 % or 0.0037%

14. Simplifying Assumption for Acid and Base Ionizations The equilibrium concentration of the acid is most often ([HA] 0 – x). If x is much, much less than [HA] 0, we can assume that subtracting x makes no difference to [HA]: ([HA] 0 – x) = [HA] 0 This is a valid assumption when the ratio of [HA] 0 to K a is > 10 3. If it is not valid, you must use the quadratic equation to solve the problem.

15. Para-hydroxybenzoic acid is used to make certain dyes. What are the concentrations of this acid, of hydronium ion, and of para- hydroxybenzoate ion in a 0.200 M aqueous solution at 25°C? What is the pH of the solution and the degree of ionization of the acid? The K a of this acid is 2.6 × 10 -5. We will use the generic formula HA for para-hydroxybenzoic acid and the following equilibrium: HA + H 2 O H 3 O + + A -

16. HA(aq) +H 2 O(l)H 3 O + (aq) +A - (aq) Initial Change Equilibrium

18. Polyprotic Acids A polyprotic acid has more than one acidic proton—for example, H 2 SO 4, H 2 SO 3, H 2 CO 3, H 3 PO 4. These acids have successive ionization reactions with K a1, K a2,... The next example illustrates how to do calculations for a polyprotic acid.

19. Tartaric acid, H 2 C 4 H 4 O 6, is a diprotic acid used in food products. What is the pH of a 0.10 M solution? What is the concentration of the C 4 H 4 O 6 2  ion in the same solution? K a1 = 9.2  10  4 ; K a2 = 4.3  10  5. First, we will use the first acid-ionization equilibrium to find [H + ] and [HC 4 H 4 O 6 - ]. In these calculations, we will use the generic formula H 2 A for the acid. Next, we will use the second acid-ionization equilibrium to find [C 4 H 4 O 6 2- ].

20. H 2 A(aq) +H 2 O(l)H 3 O + (aq) +HA - (aq) Initial Change Equilibrium

23. HA - (aq) +H 2 O(l)H 3 O + (aq) +A 2- (aq) Initial Change Equilibrium

25. Base-Ionization Equilibrium The simplest acid–base equilibria are those in which a weak acid or a weak base reacts with water. We can write a base equilibrium reaction for the generic base, B. B(aq) + H 2 O(l) HB + (aq) + OH - (aq)

26. Ammonia is a weak base. It reacts with water as follows: NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) Here the generic B from the previous slide is NH 3 (aq).

27. The equilibrium constant for the reaction of a weak base with water is called the base-ionization constant, K b. Liquid water is not included in the equilibrium constant expression.

29. Writing K b Reactions The bases in Table 16.2 (previous slide) are nitrogen bases; that is, the proton they accept adds to a nitrogen atom. Next we’ll practice writing the K b reactions. Ammonium becomes ammonium ion: NH 3 + H 2 O  NH 4 + + OH - Ethylamine becomes ethyl ammonium ion: C 2 H 5 NH 2 + H 2 O  C 2 H 5 NH 3 + + OH -

30. Dimethylamine becomes dimethylammonium ion: (CH 3 ) 2 NH 2 + H 2 O  (CH 3 ) 2 NH 3 + + OH - Pyridine becomes pyridinium ion: C 5 H 5 N+ H 2 O  C 5 H 5 NH + + OH - Hydrazine becomes hydrazinium ion: N 2 H 4 + H 2 O  N 2 H 5 + + OH -

31. We can be given pH, initial concentration, and K b. From the pH, we can find first [H 3 O + ] and then [OH - ]. Using what is given, we can find the other quantities. We can also use a simplifying assumption: When [B] 0 / K b > 10 3, the expression ([B] 0 – x) = [B] 0.

32. Aniline, C 6 H 5 NH, is used in the manufacture of some perfumes. What is the pH of a 0.035 M solution of aniline at 25°C? K b = 4.2 × 10 -10 at 25°C. We will construct an ICE chart and solve for x.

33. We are told that K b = 4.2 × 10 -10. That allows us to substitute into the K b expression to solve for x. C 6 H 5 H(aq) +H 2 O(l)C 6 H 5 NH + (aq) +OH - (aq) Initial Change Equilibrium

35. The question asks for the pH: