1. Chemistry 232 Applications of Aqueous Equilbria

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 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 a(H 2 O) is 1.00!

9. The Defination of K w K w = a(H + ) a(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 = a(H + ) a(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 activity in solution pH  -log a(H + ) u Single ion activities and activity coefficients can’t be measured

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

12. u For the dissociation of HCl in water HCl (aq)  Cl - (aq) + H + (aq) u We measure the mean activity of the acid a(HCl) = a(H + ) a(Cl - ) a(H + ) a(Cl - ) = (a  (HCl)) 2

13. u Under the assumption a(H + ) = a(Cl - ) u We obtain a´(H + ) = (a(HCl)) 1/2 = a  (HCl)

14. 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.

15. 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)

16. 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.

17. The Nonelectrolyte Activity HF (aq)  H + (aq) + F - (aq) u The undissociated HF is a nonelectrolyte  a(HF) =  (HF) m[HF]  m[HF]  (HF)  1

18. Equilibria of Weak Bases in Water u To calculate the percentage 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.

19. 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) u For the above system

20. 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).

21. 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 molality and the mean activity coefficient of the dissolved acid pH  -log (   (acid) m[H + ])

22. 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

23. u The pH of these solutions is obtained by first estimating the pOH from the molality and mean activity coefficient of the dissolved base pOH  -log (   (NaOH) m[OH - ]) pOH  -log{   (Ba(OH) 2 ) 2 m[Ba(OH) 2 ]} pH = 14.00 - pOH

24. Calculating the pH of a Weak Acid Solution u The pH of a weak acid solution is obtained via an iterative procedure. u We begin by making the assumption that the mean activity coefficient of the dissociated acid is 1.00. u We ‘correct’ the value of  (H + ) by calculating the mean activity coefficient of the dissociated acid. u Repeat the procedure until  (H + ) converges.

25. Measuring the pH of Solutions u Because the activity of a single ion cannot be measured, we can only measure our ‘best approximation’ to the hydrogen ion activity. u Let’s assume that we are going to couple a hydrogen electrode with another reference electrode, e.g., a calomel reference electrode.

26. A Cell for ‘Measuring’ the pH u Half-cell reactions. HgCl 2 (s) + 2e -  Hg (l) + 2 Cl - (aq) E  (SCE) = 0.2415 V 2 H + (aq) + 2e -  H 2 (g) E  (H + /H 2 ) = 0.000 V u Cell Reaction HgCl 2 (s) + H 2 (g)  Hg (l) + 2 H + (aq) + 2Cl - (aq) E  cell = (0.2415 - 0.0000 V) = 0.2415 V Pt H 2 (g), f=1 H + (aq) HgCl 2 Hg Cl - (aq), 3.5 M Pt

27. The Nernst Equation u For the above cell u Note since the concentration of the KCl on one side of the liquid junction is so large, the magnitude of the junction potential should be small!

28. The Practical Problem u The activity of the Cl - ion in the cell is not accurately known. u We try to place the cell in a reference solution with an accurately known pH (solution I). u Next place the solution whose pH we are attempting to measure into the cell (solution II).

29. u Assuming that the E LJ and the a(Cl - ) are the same in both cases, u Substituting the definition of the pH into the above expression,

30. Standard Solutions u Generally, two solutions are used as references. u Saturated aqueous solution of sodium hydrogen tartarate, pH = 3.557 at 25  C. u 0.0100 mol/kg disodium tetraborate, pH = 9.180 at 25  C.

31. The Glass Electrode u The glass electrode has replaced the hydrogen electrode in the operational definition of the pH.

32. Glass Electrodes u Measuring the pH – the glass electrode is immersed in the solution of interest. u Inner solution – solution is generally a phosphate buffer with a sufficient quantity of Cl - (aq). u Silver-silver chloride electrode is sealed within the cell and a calomel electrode is used as the reference electrode.

33. Glass Electrodes and pH u The potential difference across the special glass membrane arises to equilibrate the hydronium ions inside the membrane with those outside the membrane.

34. 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.

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

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

37. The Buffer Equation u Substituting and rearranging

38. 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!

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

40. 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 - ) StartA0B Change -  eq +  eq  m m(A-  eq )(  eq )(B+  eq )

41. u The pH of the solution will be almost entirely due to the original molalities 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)

42. 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?

43. 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.

44. 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.

45. Solubility Equilibria u Examine the following systems AgCl (s) ⇌ Ag + (aq) + Cl - (aq) BaF 2 (s) ⇌ Ba 2+ (aq) + 2 F - (aq) u Using the principles of chemical equilibrium, we write the equilibrium constant expressions as follows

47. u Calculate the solubility of a solid in the presence of a common ion. Examples of K sp Calculations u Calculate the solubility of a sparingly soluble solid in water. u Calculate the solubility of a solid in the presence of an inert electrolyte.

48. Solubility of Sparingly Soluble Solids in Water AgCl (s) ⇌ Ag + (aq) + Cl - (aq) u We approach this using the principles of chemical equilibrium. We set up the equilibrium data table, and calculate the numerical value of the activity of the dissolved ions in solution.

49. The Common Ion Effect u What about the solubility of AgCl in solution containing NaCl (aq)? AgCl (s) ⇌ Ag + (aq) + Cl - (aq) NaCl (aq)  Na + (aq) + Cl - (aq) AgCl (s) ⇌ Ag + (aq) + Cl - (aq) Equilibrium is displaced to the left by LeChatelier’s principle (an example of the common ion effect).

50. Solubility in the Presence of an Inert Electrolyte u What happens when we try to dissolve a solid like AgCl in solutions of an inert electrolyte (e.g., KNO 3 (aq))? u We must now take into account of the effect of the ionic strength on the mean activity coefficient!

51. The Salting-In Effect AgCl (s) ⇌ Ag + (aq) + Cl - (aq). u We designate the solubility of the salt in the absence of the inert electrolyte as s o = m(Ag + ) = m(Cl - ) at equilibrium.

52. u For a dilute solution u Designate s as the solubility of the salt in the presence of varying concentrations of inert electrolyte.