1. EQUILIBRIA OF ACIDS, BASES, AND SALTS 18.3

2. The equilibrium constant can be used for weak acids and bases. By calculating the K a value, these weak acids can be compared quantitatively. K b is used for bases.

3. 18.3 The equilibrium constant can be used for weak acids and bases. H 2 CO 3 (aq) + H 2 O(l)  H 3 O + (aq) + HCO 3 1- (aq)

4. 18.3 The equilibrium constant can be used for weak acids and bases. H 2 CO 3 (aq) + H 2 O(l)  H 3 O + (aq) + HCO 3 1- (aq) The higher the K a value, the stronger the acid. The smaller the K a value, the weaker the acid.

5. 18.3 The equilibrium constant can be used for weak acids and bases. H 2 CO 3 (aq) + H 2 O(l)  H 3 O + (aq) + HCO 3 1- (aq) At equilibrium, the concentration of the acid is the initial concentration minus the amount that ionizes.

6. 18.3 ICE CHART: INITIAL CONCENTRATION CHANGE CONCENTRATION EQUILIBRIUM CONCENTRATION

7. 18.3 The equilibrium constant can be used for weak acids and bases. H 2 CO 3 (aq) + H 2 O(l)  H 3 O + (aq) + HCO 3 1- (aq) If the initial concentration of the acid was 0.0030M, and the concentration of H 3 O + is 0.00025M, then the concentration of the acid at equilibrium is INITIAL CONCENTRATION0.0030 CHANGE CONCENTRATION0.00025 EQUILIBRIUM CONCENTRATION 0.00275

8. 18.3 Why can’t K a be used for strong acids?

9. 18.3 What does the chart on p. 606 tell you about K a values for weak acids?

10. 18.3 Example 1: A student prepared a 0.10M solution of formic acid, HCOOH, and measured its pH to be 2.38. Calculate K a of formic acid.

11. 18.3 Example 2: Niacin is one of the B vitamins. A 0.020M solution of niacin has a pH of 3.26. What is the K a of niacin?

12. 18.3 Example 3: Calculate the pH of a 0.20M solution of HCN. The K a of HCN is 4.9 x 10 -10.

13. 18.3 Example 4: Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. A 0.035M solution of ephedrine has a pH of 11.33. What is the K b of ephedrine?

14. 18.3 Example 4: Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. At equilibrium, what are the concentrations of C 10 H 15 ON, C 10 H 15 ONH +, and OH - ?

15. 18.3 What is a buffer?

16. 18.3 Buffer – substance that resists changes in pH because it contains either:

17. 18.3 Buffer – substance that resists changes in pH because it contains either: a weak acid and the salt of the acid (conjugate base) (the acid can donate an H + and the conjugate base can accept an H + )

18. 18.3 Buffer – substance that resists changes in pH because it contains either: a weak acid and the salt of the acid (conjugate base) (the acid can donate an H + and the conjugate base can accept an H + ) or a weak base and the salt of the base (conjugate acid) (the base can accept an H + and the conjugate acid can donate an H + )

19. 18.3 A common buffer is below. How would it react if acid were added? Base added? CH 3 COOH + NaCH 3 COO CH 3 COOH (a weak acid can donate an H + ) Na 1+ + CH 3 COO 1- ( the salt of a weak acid can accept an H + )

20. 18.3 Example 7: Explain how each of the following solutions could function as a buffer. Use page 485 to justify your answer. H 2 CO 3 (aq) + NaHCO3(aq)

21. 18.3 Example 7: Explain how each of the following solutions could function as a buffer. Use page 485 to justify your answer. H 3 PO 4 (aq) + Na 3 PO 4 (aq)

22. 18.3 Example 7: Explain how each of the following solutions could function as a buffer. Use page 485 to justify your answer. NH 3 (aq) + NH 4 Cl(aq)

23. 18.3 Hydrolysis of salts Some salts when they dissolve in water can function as either a weak base or a weak acid.

24. 18.3 Hydrolysis of salts A neutral salt: NaCl(s) NaCl(s) + H 2 O(l)  Na + (aq) + Cl - (aq) + H 2 O(l) (There is nothing here to attract H 1+ or donate H 1+ )

25. 18.3 Hydrolysis of salts A basic salt: NaClO(s) NaClO(s) + H 2 O(l)  Na + (aq) + HClO(aq) + OH - (The ClO 1- readily pulls an H 1+ away from water to create the weak acid, HClO)

26. 18.3 Hydrolysis of salts An acidic salt: NH 4 Cl(s) NH 4 Cl(s) + H 2 O(l)  NH 3 (aq) + H 3 O + (aq) + Cl - (aq) (The NH 4 1+ readily gives up an H 1+ to water to form H 3 O 1+ )