1. Obj 16.5, 16.6

2. A.) You will recall that the seven strong acids are HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 3, and HClO 4. B.) These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. C.) For the monoprotic strong acids, [H 3 O + ] = [acid].

3. Sample Exercise 16.8 Calculating the pH of a Strong Acid What is the pH of a 0.040 M solution of HClO 4 ? An aqueous solution of HNO 3 has a pH of 2.34. What is the concentration of the acid? Practice Exercise

4. D.) Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca 2+, Sr 2+, and Ba 2+ ). E.) Again, these substances dissociate completely in aqueous solution.

5. Sample Exercise 16.9 Calculating the pH of a Strong Base What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH) 2 ? What is the concentration of a solution of (a) KOH for which the pH is 11.89; (b) Ca(OH) 2 for which the pH is 11.68? Practice Exercise

6. 1.) For a generalized acid dissociation, the equilibrium expression would be 2.) 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)

7. B.) The greater the value of K a, the stronger the acid.

8. 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. We know that [H 3 O + ] [COO - ] [HCOOH] K a =

9. 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. We can find [H 3 O + ], which is the same as [HCOO - ], from the pH.

10. 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 - ]

11. [HCOOH], M[H 3 O + ], M[HCOO - ], M Initially0.1000 Change - 4.2  10 -3 + 4.2  10 -3 At Equilibrium 0.10 - 4.2  10 -3 = 0.0958 = 0.10 4.2  10 -3 © 2009, Prentice-Hall, Inc. Now we can set up a table…

12. [4.2  10 -3 ] [0.10]* K a = = 1.8  10 -4

13. Niacin, one of the B vitamins, has the following molecular structure: A 0.020 M solution of niacin has a pH of 3.26. What is the acid-dissociation constant, K a, for niacin? Practice Exercise

14. 1.) Percent Ionization =  100 2.) In this example [H 3 O + ] eq = 4.2  10 -3 M [HCOOH] initial = 0.10 M [H 3 O + ] eq [HA] initial Percent Ionization =  100 4.2  10 -3 0.10 = 4.2%

15. A 0.020 M solution of niacin has a pH of 3.26. Calculate the percent ionization of the niacin. Practice Exercise

16. 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.

17. The equilibrium constant expression is [H 3 O + ] [C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] K a =

18. [C 2 H 3 O 2 ], M[H 3 O + ], M[C 2 H 3 O 2 - ], M Initially0.3000 Change-x+x At Equilibrium 0.30 - x  0.30 xx We next set up a table… *We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

19. Now, (x) 2 (0.30-x)* 1.8  10 -5 = (1.8  10 -5 ) (0.30) = x 2 5.4  10 -6 = x 2 2.3  10 -3 = x

20. pH = -log [H 3 O + ] pH = -log (2.3  10 -3 ) pH = 2.64

21. Sample Exercise 16.12 Using K a to Calculate pH Calculate the pH of a 0.20 M solution of HCN. (K a = 4.9 X 10 -10 )

22. The Ka for niacin is 1.5 × 10 -5. What is the pH of a 0.010 M solution of niacin? Practice Exercise

23. Sample Exercise 16.13 Using K a to Calculate Percent Ionization Calculate the percentage of HF (K a of HF = 6.8 X 10 -4 ) molecules ionized in (a) a 0.10 M HF solution, (b) a 0.010 M HF solution.

24. In Practice Exercise 16.11, we found that the percent ionization of niacin (K a = 1.5 × 10 -5 ) in a 0.020 M solution is 2.7%. Calculate the percentage of niacin molecules ionized in a solution that is (a) 0.010 M, (b) 1.0 × 10 -3 M. Practice Exercise

25. …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.

26. Sample Exercise 16.14 Calculating the pH of a Polyprotic Acid Solution The solubility of CO 2 in pure water at 25 ºC and 0.1 atm pressure is 0.0037 M. The common practice is to assume that all of the dissolved CO 2 is in the form of carbonic acid (H 2 CO 3 ), which is produced by reaction between the CO 2 and H 2 O: What is the pH of a 0.0037 M solution of H 2 CO 3 ?

27. Sample Exercise 16.14 Calculating the pH of a Polyprotic Acid Solution (a) Calculate the pH of a 0.020 M solution of oxalic acid (H 2 C 2 O 4 ). (See Table 16.3 for K a1 and K a2.) (b) Calculate the concentration of oxalate ion [C 2 O 4 2– ], in this solution. Practice Exercise