1. Acid/Base Chemical Equilibria

2. The Brønsted Definitions  Brønsted Acid  proton donor  Brønsted Base  proton acceptor  Conjugate acid - base pair  an acid and its conjugate base or a base and its conjugate acid

3. Example Acid-Base Reactions u Look at acetic acid dissociating CH 3 COOH(aq)  CH 3 COO - (aq) + H + (aq)  Brønsted acid Conjugate base u Look at NH 3 (aq) in water NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq)   Brønsted baseconjugate acid

4. Representing Protons in Aqueous Solution CH 3 COOH(aq)  CH 3 COO - (aq) + H + (aq) CH 3 COOH(aq) + H 2 O(l)  CH 3 COO - (aq) + H 3 O + (aq) HCl (aq)  Cl - (aq) + H + (aq) HCl(aq) + H 2 O(l)  Cl - (aq) + H 3 O + (aq)

5. What is H + (aq)? H3O+H3O+ H5O2+H5O2+ H+H+ H9O4+H9O4+

6. Representing Protons u Both representations of the proton are equivalent. u H 5 O 2 + (aq), H 7 O 3 + (aq), H 9 O 4 + (aq) have been observed. u We will use H + (aq)!

7. The Autoionization of Water u Water autoionizes (self-dissociates) to a small extent 2H 2 O(l)  H 3 O + (aq) + OH - (aq) H 2 O(l)  H + (aq) + OH - (aq) u These are both equivalent definitions of the autoionization (self ionization) reaction. Water is acting as a base and an acid in the above reaction  water is amphoteric.

8. The Autoionization Equilibrium u From the equilibrium chapter u But we know (H 2 O) is 1.00 (pure liquids are not included in an equilibrium expression)!

9. The Defination of K w K w = (H + ) (OH - ) Ion product constant for water, K w, is the product of the activities of the H + and OH - ions in pure water at a temperature of 298.15 K K w = (H + ) (OH - ) = 1.0x10 -14 at 298.2 K

10. The pH scale u Attributed to Sørenson in 1909 u We should define the pH of the solution in terms of the hydrogen ion concentration in solution pH  -log (H + )

11. Determination of pH u What are we really measuring when we measure the pH? pH  -log (H + ) u (H + ) is the best approximation to the hydrogen ion activity (physical chemistry term) in solution. u How do we measure (H + )?

12. Equilibria in Aqueous Solutions of Weak Acids/ Weak Bases u By definition, a weak acid or a weak base does not ionize completely in water (  <<100%). How would we calculate the pH of a solution of a weak acid or a weak base in water? u To obtain the pH of a weak acid solution, we must apply the principles of chemical equilibrium.

13. Equilibria of Weak Acids in Water : The K a Value u Define the acid dissociation constant K a u For a general weak acid reaction HA (aq)  H + (aq) + A - (aq)

14. Equilibria of Weak Acids in Water u For the dissolution of HF(aq) in water. HF (aq)  H + (aq) + F - (aq) u The small value of K a indicates that this acid is only ionized to a small extent at equilibrium.

15. Equilibria of Weak Bases in Water u To calculate the percent dissociation of a weak base in water (and the pH of the solutions) CH 3 NH 2 (aq) + H 2 O  CH 3 NH 3 + (aq) + OH - (aq) u We approach the problem as in the case of the weak acid above, i.e., from the chemical equilibrium viewpoint.

16. The K b Value u Define the base dissociation constant K b u For a general weak base reaction with water B (aq) + H 2 O (aq)  B + (aq) + OH - (aq)

17. Examples of Acid-Base Calculations u Determining the pH of a strong acid (or base solution). u Determining the pH of a weak acid (or base solution).

18. Calculating the pH of Solutions of Strong Acids For the dissolution of HCl, HI, or any of the other seven strong acids in water HCl (aq)  H + (aq) + Cl - (aq) HI (aq)  H + (aq) + I - (aq) %  eq = 100% The pH of these solutions can be estimated from the molarity of the dissolved acid pH  -log [H + ]

19. u For the dissolution of NaOH, Ba(OH) 2, or any of the other strong bases in water NaOH (aq)  Na + (aq) + OH - (aq) Ba(OH) 2 (aq)  Ba 2+ (aq) + 2OH - (aq) %  eq = 100% Calculating the pH of Solution of Strong Bases

20. u The pH of these solutions is obtained by first estimating the pOH from the molarity of the dissolved base pOH  -log [OH - ] pOH  -log[Ba(OH) 2 ] pH = 14.00 - pOH

21. The Definition of a Buffer u Buffer  a reasonably concentrated solution of a weak acid and its conjugate base that resists changes in the pH when an additional amount of strong acid or strong base is added to the solutions.

22. u How would we calculate the pH of a buffer solution? pH of a Buffer

23. note pH = -log (H + ) Define pK a = -log (K a )

24. The Buffer Equation u Substituting and rearranging

25. The Generalized Buffer Equation u The pH of the solution determined by the ratio of the weak acid to the conjugate base. This equation (the Henderson-Hasselbalch equation) is often used by chemists, biochemists, and biologists for calculating the pH of a solution of a weak acid and its conjugate base!

26. The Generalized Buffer Equation u Note: The Henderson-Hasselbalch equation is really only valid for pH ranges near the pK a of the weak acid!

27. The Generalized Buffer Equation u Buffer  CH 3 COONa (aq) and CH 3 COOH (aq)) CH 3 COOH (aq) ⇄ CH 3 COO - (aq) + H + (aq) The Equilibrium Data Table n(CH 3 COOH)n(H + )n(CH 3 COO - ) IA0B C-x-x+ x+x+x E(A-x)(x)(B+x)

28. u The pH of the solution will be almost entirely due to the original molarities of acid and base!! u This ratio will be practically unchanged in the presence of a small amount of added strong acid or base u The pH of the solution changes very little after adding strong acid or base (i.e., it is buffered)

29. u How does the pH change after the addition of strong acid or base? Example of Buffer Calculations u How do we calculate the pH of a buffer solution?

30. The pH of a Buffer Solution u The major task in almost all buffer calculations is to obtain the ratio of the concentrations of conjugate base to weak acid! u Using the K a of the appropriate acid, the pH of the solution is obtained from the Henderson-Hasselbalch equation.

31. Adding Strong Acid or Base to Buffer Solutions u To obtain the pH after the addition of a strong acid or base, we must calculate the new amount of weak acid and conjugate base from the reaction of the strong acid (or base) to the buffer system. u The pH of the solution may again be calculated with the Henderson-Hasselbalch equation.

32. Buffer Capacity u Buffer solutions do not have unlimited capacity to keep the pH relatively constant. When the weak acid of a buffer is used up, no more buffering action toward additional base is possible. u The buffer capacity is the amount of acid or base that can be added to a certain volume of a buffer solution before the Ph begins to change significantly.

33. Buffer Capacity u If you need to make a buffer solution up to a certain pH, use the Henderson-Hasselbalch Equation. u pH = pK a + log u A - = conjugate base (salt) u HA = acid u Generally easier to use than solving quadratic.

34. Calculating the pH of a Buffer Using the H-H Eq. u What is the pH of a buffer solution that is 0.10 M chloroacetic acid and 0.15 M sodium chloroacetate? K a = 1.4 X10 -3 u pH = pK a + log

35. Example u pH = -log(1.4 X10 -3 ) + log[0.15M]/[0.10M] u pH = 3.03

36. Calculating the pH of a Buffer from Given Volumes of Solution u What is the pH of a buffer that contains 60 mL of 0.100 M NH 3 with 40 mL of 0.100 M NH 4 Cl? u NH 3 + H 2 O  NH 4 + + OH - u The problem is to obtain the concentration of OH - in this mixture. We must, then, calculate the new concentrations of the NH 3 and NH 4 +.

37. Example u NH 3 M = 0.001 moles/L = X/0.060L u X = 0.0060 mols u NH 4 + M= 0.100 moles/L = X/0.40L u X = 0.0040 mols u NH 3 molarity = 0.0060 mols/0.100L u = 0.060M u NH 4 + M = 0.0040 mols/0.100L = 0.040M

38. Example u Must use K b since a weak base. Create table. u NH 3 + H 2 0  NH 4 + + OH - u

39. First Way to Complete u K b = u 1.8 X 10 -5 = (0.040 + X) (X)/(0.060 – X) u Assume X is small and delete u 1.8 X 10 -5 = 0.040 X/0.060 u X = 2.7 X 10 -5 = [OH - ] u pOH = -log[2.7 X 10 -5 ] = 4.56 u pH = 14 - 4.56 = 9.43

40. Second Way to Complete Using the Henderson-Hasslebach u pOH = pK b + log[salt]/[acid] u pOH = -log[1.8 X 10 -5 ] + log 0.040/0.060 u pOH = 4.74 + -0.176 u pOH = 4.56 u pH = 9.43