1. CHAPTER 15 REACTIONS AND EQUILIBRIA INVOLVING ACIDS, BASES, AND SALTS

2. Common ion effect- The addition of an ion already present(common) in a system causes equilibrium to shift away from the common ion.

3. For example, the addition of concentrated HCl to a saturated solution of NaCl will cause some solid NaCl to precipitate out of solution. The NaCl has become less soluble because of the addition of additional chloride ion. This can be explained by the use of LeChatelier's Principle. NaCl(s)  Na + + Cl -

4. The addition of a common ion to a solution of a weak acid makes the solution less acidic. HC 2 H 3 O 2  H + + C 2 H 3 O 2 - If we add NaC 2 H 3 O 2, equilibrium shifts to undissociated HC 2 H 3 O 2, raising pH. The new pH can be calculated by putting the concentration of the anion into the K a equation and solving for the new [H + ].

5. Buffered solution- A solution that resists changes in pH when hydroxide ions or protons are added. A buffer solution usually consists of a solution of a weak acid and its salt or a weak base and its salt.

6. Ex. HC 2 H 3 O 2 /C 2 H 3 O 2 - buffer system Addition of strong acid: H + + C 2 H 3 O 2 -  HC 2 H 3 O 2 Addition of strong base: OH - + HC 2 H 3 O 2  H 2 O + C 2 H 3 O 2 -

7. NH 3 /NH 4 + buffer system Addition of strong acid: H + + NH 3  NH 4 + Addition of strong base: OH - + NH 4 +  NH 3 + H 2 O

8. Buffer capacity- The amount of acid or base that can be absorbed by a buffer system without a significant change in pH. In order to have a large buffer capacity, a solution should have large concentrations of both buffer components.

9. One way to calculate the pH of a buffer system is with the Henderson-Hasselbach equation. pH = pK a + log [base] [acid] pH = pK a + log [A - ] [HA]

10. For a particular buffering system all solutions that have the same ratio of [A - ]/[HA] have the same pH. Optimum buffering occurs when [HA] = [A - ] and the pK a of the weak acid used should be as close as possible to the desired pH of the buffer system.

11. The Henderson-Hasselbach (HH) equation needs to be used cautiously. It is sometimes used as a quick, easy equation to plug numbers into. A K a or K b problem requires a greater understanding of the factors involved and can always be used instead of the HH equation.

12. Hints for Solving Buffer Problems: 1. Determine major species involved initially. 2. If chemical reaction occurs, write equation and solve stoichiometry in moles, then change to molarity. 3. Write equilibrium equation. 4. Set up equilibrium expression (K a or K b ) or HH equation. 5. Solve. 6. Check logic of answer.

13. Ex. A solution is 0.120 M in acetic acid and 0.0900 M in sodium acetate. Calculate the [H + ] at equilibrium. The K a of acetic acid is 1.8 x 10 -5. Rxn HC 2 H 3 O 2  H + + C 2 H 3 O 2 - Initial 0.120 0 0.0900 Change -x +x +x Equil. 0.120-x x 0.0900 + x K a = x (0.0900 + x)  x (0.0900) = 1.8 x 10 -5 0.120-x 0.120 x = 2.4 x 10 -5 M [H + ] = 2.4 x 10 -5 WA + CB = K a !

14. Using the Henderson-Hasselbach equation: pK a = -log 1.8 x 10 -5 = 4.74 pH = 4.74 + log (0.0900/0.120) = 4.62 [H + ] = antilog (-4.62) = 2.4 x 10 -5

15. Ex. Calculate the pH of the above buffer system when 100.0 mL of 0.100 M HCl is added to 455 mL of solution. 0.100 L HCl x 0.100 M = 0.0100 mol H + 0.455 L C 2 H 3 O 2 - x 0.0900 M = 0.0410 mol C 2 H 3 O 2 - 0.455 L HC 2 H 3 O 2 x 0.120 M = 0.0546 mol HC 2 H 3 O 2 H + + C 2 H 3 O 2 -  HC 2 H 3 O 2 Before 0.0100 mol 0.0410 mol 0.0546 mol Change -0.0100 mol -0.0100 mol +0.0100 mol After 0 0.0310 mol 0.0646 mol 0.0310 mol acetate / 0.555 L solution = 0.0559 M acetate 0.0646 mol acetic acid/0.555 L solution = 0.116 M acetic acid WA + CB = K a !

16. Rxn HC 2 H 3 O 2  H + + C 2 H 3 O 2 Initial 0.116 M 0 0.0559 M Change -x +x +x Equil. 0.116-x x 0.0559 + x K a = 1.8 x 10 -5 = x(0.0559+x)  x(0.0559) 0.116-x 0.116 x = 3.74 x 10 -5 M = [H + ] pH = 4.43

17. Acid-Base Titrations

18. titrant-solution of known concentration (in buret) The titrant is added to a solution of unknown concentration until the substance being analyzed is just consumed (stoichiometric point or equivalence point).

19. pH or titration curve -plot of pH as a function of the amount of titrant added.

20. Types of Acid-Base Titrations:

21. 1. Strong acid-strong base Simple reaction H + + OH -  H 2 O The pH is easy to calculate because all reactions go to completion. At the equivalence point, the solution is neutral.

22. The pH Curve for the Titration of 50.0 mL of 0.200 M HNO 3 with 0.100 M NaOH Only strong acid left Only neutral salt Strong base left

23. Ex. 100.0 mL of 1.00 M HCl is titrated with 0.500 M NaOH. Calculate the [H + ] after 50.0 mL of base has been added. 0.1000L x 1.00 M = 0.100 mol H + 0.0500L x 0.500 M = 0.0250 mol OH H + + OH -  H 2 O Before 0.100 mol 0.0250 mol 0 Change -0.0250 -0.0250 +0.0250 After 0.0750 mol 0 0.0250 mol 0.0750 mol H + / (0.100L + 0.0500 L) = 0.500 M H +

24. Calculate the [H + ] after 200 mL of base has been added. 0.200L x 0.500 M = 0.100 mol OH - H + + OH -  H 2 O Before 0.100 mol 0.100 mol 0 Change -0.100 -0.100 +0.100 After 0 0 0.100 mol [H + ] is not zero. The [H + ] of pure water is 1.0 x 10 -7, therefore pH = 7 [H + ] = 1.0 x 10 -7

25. Calculate the pH after 300 mL of base has been added. 0.300L x 0.500 M = 0.150 mol OH - H + + OH -  H 2 O Before 0.100 mol 0.150 mol --- Change-0.100 -0.100 --- After 0 0.050 mol --- [OH - ]= 0.050 mol/0.400L = 0.125 M OH - pOH = 0.913 pH = 13.097

26. 2. Weak acid - strong base The reaction of a strong base with a weak acid is assumed to go to completion. Before the equivalence point, the concentration of weak acid remaining and the conjugate base formed are determined. At halfway to the equivalence point, pH = pK a.

27. The pH Curve for the Titration of 50.0 mL of 0.100 M HC 2 H 3 O 2 with 0.100 M NaOH WA only K a WA > SB Stoich + K a WA >SB Stoich + K a WA>SB [HA]=[A - ] ½ eq pt pH = pK a SB>WA Stoich pOH = -log [SB] SB = WA Basic salt K b

28. At the equivalence point, a basic salt is present and the pH will be greater than 7. After the equivalence point, the strong base will be the dominant species and a simple pH calculation can be made after the stoichiometry is done.

30. Ex. 30.0 mL of 0.10 M NaOH is added to 50.0 mL of 0.10 M HF. (K a of HF = 7.2 x 10 -4 ) Determine the pH of the final solution. 0.0300L x 0.10 M = 0.00300 mol OH - 0.0500L x 0.10 M = 0.00500 mol HF Stoichiometry OH - + HF  H 2 O + F - Before 0.00300 mol 0.00500 mol --- 0 Change -0.003 -0.003 +0.003 After 0 0.00200 mol --- 0.00300 mol 0.00200 mol/(0.030L + 0.050L) = 0.0250M HF 0.00300 mol/(0.030L + 0.050L) = 0.0375M F - WA + CB = K a !

31. Equilibrium Rxn HF  H + + F - Initial 0.025 0 0.0375 Change -x +x +x Equil. 0.025-x x 0.0375 +x K a = 7.2 x 10 -4 = x (0.0375 +x)  x(0.0375) 0.0250-x 0.0250 x = 4.8 x 10 -4 [H + ] = 4.8 x 10 -4 pH = 3.32

32. Ex. 50.0 mL of 0.10 M NaOH is added to 50.0 mL of 0.10 M HF. (K a of HF = 7.2 x 10 -4 ) Determine the pH of the final solution. 0.050L x 0.10 M = 0.0050 mol OH - 0.050L x 0.10 M = 0.0050 mol HF Stoichiometry OH - + HF  H 2 O + F - Before 0.0050 mol 0.0050 mol ---- 0 After 0 0 ----- 0.0050 mol 0.0050 mol/(0.050L + 0.050L) = 0.050M F - CB only= K b !

33. Equilibrium Rxn F - + H 2 O  HF + OH - Initial 0.050 ---- 0 0 Change -x ----- +x +x Equil. 0.050-x ---- x x

34. K b for F - = 1.0 x 10 -14 /K a for HF K b = 1.4 x 10 -11 = [HF][OH - ] = x 2  x 2 [F - ] 0.050-x 0.050 x = 8.4 x 10 -7 M [OH - ] =8.4 x 10 -7 pOH = 6.08 pH = 7.92

35. Ex. 60.0 mL of 0.10 M NaOH is added to 50.0 mL of 0.10 M HF. (K a of HF = 7.2 x 10 -4 ) Determine the pH of the final solution. 0.0600L x 0.10 M OH - = 0.0060 mol OH - 0.0500L x 0.10 M HF = 0.0050 mol HF Stoichiometry OH - + HF  H 2 O + F - Before 0.0060 mol 0.0050 mol ------ 0 After 0.0010 mol 0 -------- 0.0050 mol [OH - ] = 0.0010 mol/0.110 L = 9.09 x 10 -3 M Strong base + weak base (ignore weak base!) pOH = 2.04 pH = 11.96

36. Weak base - Strong acid Before the equivalence point, a weak base equilibria exists. Calculate the stoichiometry and then the weak base equilibria. At the equivalence point, an acidic salt is present and the pH is below 7. After the equivalence point, the strong acid is the dominant species. Use the [H + ] to find the pH.

37. The pH Curve for the Titration of 100.0 mL of 0.050 M NH 3 with 0.10 M HCI Weak base only K b SA < WB Stoich + K b SA<WB [B] = [HB + ] ½ eq pt pOH = pK b SA < WB Stoich + K b WB = SA Acidic salt K a SA> WB Stoich, pH = -log [SA]

38. Ex. Calculate the pH when 100.0 mL of 0.050 M NH 3 is titrated with 10.0 mL of 0.10 M HCl. K b of NH 3 = 1.8 x 10 -5 0.100L x 0.050 M = 0.0050 mol NH 3 0.010L x 0.10 M = 0.0010 mol H + NH 3 + H +  NH 4 + Before 0.0050 mol 0.0010 mol 0 Change -0.0010 -0.0010 +0.0010 After 0.0040 mol 0 0.0010 mol 0.0010 mol/0.110 L = 9.09 x 10 -3 M NH 4 + 0.0040 mol/0.110 L = 3.64 x 10 -2 M NH 3 WB+ CA = K b !

39. Equilibrium Rxn NH 3 + H 2 O  NH 4 + + OH - Initial 0.0364 ----- 0.00909 0 Change -x +x +x Equil 0.0364-x 0.00909+x x K b = 1.8 x 10 -5 = (0.00909+x)x  (0.00909)x 0.0364-x 0.0364 x = [OH - ]= 7.21 x 10 -5 pOH = 4.14 pH = 9.86

40. Ex. Calculate the pH when 100.0 mL of 0.050 M NH 3 is titrated with 50.0 mL of 0.10 M HCl. K b of NH 3 = 1.8 x 10 -5 0.100L x 0.050 M = 0.0050 mol NH 3 0.050L x 0.10 M = 0.0050 mol H + NH 3 + H +  NH 4 + Before 0.0050 mol 0.0050 mol 0 Change -0.0050 -0.0050 +0.0050 After 0 0 0.0050 mol 0.0050 mol/ 0.150L = 0.0333M NH 4 + CA only = K a !

41. Equilibrium Rxn NH 4 + + H 2 O  NH 3 + H 3 O + Initial 0.0333 ----- 0 0 Change -x +x +x Equil. 0.0333-x x x K a for NH 4 + = 1.0 x 10 -14 /K b for NH 3 = 5.56 x 10 -10 5.56 x 10 -10 = [NH 3 ][H 3 O + ] = x 2  x 2 [NH 4 + ] 0.0333-x 0.0333 x = 4.30 x 10 -6 = [H + ] pH = 5.37

42. Ex. Calculate the pH when 100.0 mL of 0.050 M NH 3 is titrated with 60.0 mL of 0.10 M HCl. 0.100L x 0.050 M = 0.0050 mol NH 3 0.060L x 0.10 M = 0.0060 mol H + NH 3 + H +  NH 4 + Before 0.0050 mol 0.0060 mol 0 Change -0.0050 -0.0050 +0.0050 After 0 0.0010 mol 0.0050 mol 0.0010 mol/0.160L = 6.25 x 10 -3 M H 3 O + pH = 2.20 Strong acid and CA Ignore CA(weak)!

43. Acid-Base Indicators end point- point in titration where indicator changes color

44. When choosing an indicator, we want the indicator end point and the titration equivalence point to be as close as possible. Since strong acid-strong base titrations have a large vertical area, color changes will be sharp and a wide range of indicators can be used. For titrations involving weak acids or weak bases, we must be more careful in our choice of indicator.

45. Indicators are usually weak acids, HIn. They have one color in their acidic (HIn) form and another color in their basic (In - ) form. A very common indicator, phenolphthalein, is colorless in its HIn form and pink in its In - form. It changes color in the range of pH 8-10.

47. Usually 1/10 of the initial form of the indicator must be changed to the other form before a new color is apparent.

48. The following equations can be used to determine the pH at which an indicator will change color: For titration of an acid: pH = pK a + log 1/10 = pK a -1 For titration of a base: pH = pK a + log 10/1 = pK a +1

49. The useful range of an indicator is usually its pK a ±1. When choosing an indicator, determine the pH at the equivalence point of the titration and then choose an indicator with a pK a close to that.

51. The pH Curve for the Titration of 100.0 mL of 0.10 M HCI with 0.10 M NaOH

52. The pH Curve for the Titration of 50 mL of 0.1 M HC 2 H 3 O 2 with 0.1 M NaOH

53. SOLUBILITY EQUILIBRIA

54. Saturated solutions of salts are another type of chemical equilibria. For a saturated solution of AgCl, the equation would be: AgCl(s)  Ag + (aq) + Cl - (aq) The solubility product expression would be: K sp = [Ag + ][Cl - ] The AgCl(s) is left out since solids are left out of equilibrium expressions (constant concentrations). For Ag 2 CO 3, Ag 2 CO 3  2Ag + + CO 3 2- K sp = [Ag + ] 2 [CO 3 2- ]

55. The K sp of AgCl is 1.6 x 10 -10. This means that if the product of [Ag + ][Cl - ] 1.6 x 10 -10, the solution is saturated and a solid (precipitate) would form. The product of the ions (raised to the power of their coefficients) is called the ion product constant or Q. If Q > K sp, ppt forms. If Q < K sp, no ppt forms.

56. Ex. The molar solubility of silver sulfate is 1.5 x 10 -2 mol/L. Calculate the solubility product of the salt. Reaction Ag 2 SO 4 (s)  2Ag + + SO 4 2- Initial ---- 0 0 Change -x +2x +x Equil. ---- 3.0 x 10 -2 1.5 x 10 -2 x = 1.5 x 10 -2 Since 1.5 x 10 -2 mol/L of Ag 2 SO 4 dissolve, 1.5 x 10 -2 mol/L of SO 4 2- form and 2(1.5 x 10 -2 mol/L) of Ag + form. K sp = [Ag + ] 2 [SO 4 2- ] = (3.0 x 10 -2 ) 2 (1.5 x 10 -2 )= 1.4 x 10 -5 Remember that molar solubility is “x”!

57. Ex. Calculate the molar solubility of calcium phosphate. The K sp of calcium phosphate is 1.2 x 10 -26. Reaction Ca 3 (PO 4 ) 2  3Ca 2+ + 2PO 4 3- Initial --- 0 0 Change -x +3x +2x Equilibrium --- 3x 2x K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 1.2 x 10 -26 = (3x) 3 (2x) 2 = 108x 5 x 5 = 1.1 x 10 -28 x = 2.6 x 10 -6 M

58. Ex. What is the molar solubility of lead(II) iodide in a 0.050 M solution of sodium iodide? (Common ion effect problem) Reaction PbI 2  Pb 2+ + 2I - Initial ---- 0 0.050M Change -x +x +2x Equil. ---- x 0.050 + 2x K sp = [Pb 2+ ][I - ] 2 K sp = 1.4 x 10 -8 = (x)(0.050+2x) 2  x(0.050) 2 x = 5.6 x 10 -6 M The molar solubility of PbI 2 in pure water is 1.5 x 10 -3 M. This shows the decreased solubility of a salt in the presence of a common ion. Don’t forget to put in the initial concentration of the common ion!

59. Ex. Exactly 200 mL of 0.040 M BaCl 2 are added to exactly 600 mL of 0.080 M K 2 SO 4. Will a precipitate form? BaCl 2 + K 2 SO 4  BaSO 4 + 2KCl Barium sulfate is the likely precipitate. 0.200 L x 0.040 M BaCl 2 = 8.0 x 10 -3 mol Ba 2+ 8.0 x 10 -3 mol/0.800L total volume = 1.0 x 10 -2 M Ba 2+ 0.600 L x 0.080 M K 2 SO 4 = 4.8 x 10 -2 mol SO 4 2- 4.8 x 10 -2 mol/0.800L = 6.0 x 10 -2 M SO 4 2- K sp = [Ba 2+ ][SO 4 2- ] = 1.1 x 10 -10 (look this up in table) Q = (1.0 x 10 -2 )(6.0 x 10 -2 ) = 6.0 x 10 -4 Q > K sp 6.0 x 10 -4 > 1.1 x 10 -10 A precipitate of BaSO 4 forms.