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 ions. 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 HC2H3O2  H+ + C2H3O2 If we add NaC2H3O2, equilibrium shifts to undissociated HC2H3O2, raising pH. The new pH can be calculated by putting the concentration of the anion into the Ka 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. HC2H3O2/C2H3O2 buffer system Addition of strong acid: H+ + C2H3O2  HC2H3O2 Addition of strong base: OH  + HC2H3O2 H2O + C2H3O2 
(Easiest way to show a strong acid is to use H+)
(Easiest way to show a strong base is to use OH)
Notice the one-way arrows. When a strong acid or base is added,
the reaction is driven to completion.

7. NH3/NH4+ buffer system Addition of strong acid: H+ + NH3  NH4+ Addition of strong base: OH  + NH4+  NH3 + H2O

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 = pKa + log [base] [acid] pH = pKa + 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 pKa 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
The Henderson-Hasselbach (HH) equation needs to be used cautiously. It is sometimes used as a quick, easy equation to plug numbers into. A Ka or Kb 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
Hints for Solving Buffer Problems: 1. Determine major species involved initially. 2. If chemical reaction occurs(Is there a strong acid or strong base?), write equation and solve stoichiometry in moles, then change to molarity. 3. Write equilibrium equation. 4. Set up equilibrium expression (Ka or Kb). 5. Solve. 6. Check logic of answer.

13. Rxn HC2H3O2  H+ + C2H3O2 Initial 0.120 0 0.0900 Change x +x +x
Ex. A solution is M in acetic acid and M in sodium acetate. Calculate the [H+] at equilibrium. The Ka of acetic acid is 1.8×10-5.
WA + CB = Ka!
No strong acid or base. No stoich!
Rxn HC2H3O2  H C2H3O2
Initial
Change x x x
Equil x x x
Ka = x ( x)  x (0.0900) = 1.8 × 10 x
x = 2.4×105 M
[H+] = 2.4×105

14. Using the Henderson-Hasselbach equation: pKa = log 1. 8×105 = 4
Using the Henderson-Hasselbach equation: pKa = log 1.8×105 = pH = log (0.0900/0.120) = [H+] = 10(4.62) = 2.4×105

15. Ex. Calculate the pH of the above buffer system when 100. 0 mL of 0
Ex. Calculate the pH of the above buffer system when mL of M HCl is added to 455 mL of solution.
0.100 L HCl × M = mol H+
0.455 L C2H3O2 × M = mol C2H3O2
0.455 L HC2H3O2 × M = mol HC2H3O H C2H3O2  HC2H3O2
Before mol mol mol
Change  mol  mol mol
After mol mol
mol acetate / L solution = M acetate
mol acetic acid/0.555 L solution = M acetic acid
WA + CB = Ka!

16. Rxn HC2H3O2  H C2H3O2
Initial M M
Change x x x
Equil x x x Ka = 1.8×105 = x( x)  x(0.0559)
0.116x
x = 3.74×10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 (analyte) 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  H2O 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 HNO3 with 0
The pH Curve for the Titration of 50.0 mL of M HNO3 with M NaOH
Strong base left
Only neutral salt
Only strong acid left

23. Ex. 100. 0 mL of 1. 00 M HCl is titrated with 0. 500 M NaOH
Ex mL of 1.00 M HCl is titrated with M NaOH. Calculate the [H+] after 50.0 mL of base has been added.
0.1000L × 1.00 M = mol H+
0.0500L × M = mol OH
H OH  H2O
Before mol mol
Change
After mol mol
mol H+/ (0.100L L)
= M H+

24. Calculate the [H+] after 200 mL of base has been added.
0.200L x M = mol OH
H OH  H2O
Before mol mol
Change  
After 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 × M = mol OH
H OH  H2O
Before mol mol
Change  
After mol
[OH]= mol/0.400L = M OH
pOH = pH =

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 = pKa.

27. The pH Curve for the Titration of 50. 0 mL of 0. 100 M HC2H3O2 with 0
The pH Curve for the Titration of 50.0 mL of M HC2H3O2 with M NaOH
SB>WA
Stoich
pOH = -log [SB]
SB = WA
Basic salt Kb
WA>SB
[HA]=[A-] ½ eq pt
pH = pKa
WA >SB
Stoich + Ka
WA only Ka
WA > SB
Stoich + Ka

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. Titration Curve for a Diprotic Acid

31. Ex. 30. 0 mL of 0. 10 M NaOH is added to 50. 0 mL of 0. 10 M HF
Ex mL of 0.10 M NaOH is added to 50.0 mL of 0.10 M HF. (Ka of HF = 7.2 × 104) Determine the pH of the final solution.
0.0300L × 0.10 M = mol OH
0.0500L × 0.10 M = mol HF
Stoichiometry OH HF  H2O F
Before mol mol
Change  
After mol mol
mol/(0.030L L) = M HF
mol/(0.030L L) = M F
WA + CB = Ka!

32. Rxn HF  H+ + F Initial 0.025 0 0.0375 Change x +x +x
Equil x x x
Ka = 7.2×104 = x ( x)  x(0.0375) x
x = 4.8 × 104
[H+] = 4.8×104
pH = 3.32
Equilibrium

33. 0.050L×0.10 M = 0.0050 mol HF Stoichiometry
Ex mL of 0.10 M NaOH is added to 50.0 mL of 0.10 M HF. (Ka of HF = 7.2×104) Determine the pH of the final solution.
0.050L×0.10 M = mol OH
0.050L×0.10 M = mol HF Stoichiometry
OH HF  H2O F
Before mol mol
After mol
mol/(0.050L L) = 0.050M F
CB only= Kb!

34. Equilibrium Rxn F + H2O  HF + OH Initial 0.050 ---- 0 0
Change x x x
Equil x x x

35. Kb for F = 1.0×1014/Ka for HF x = 8.4×107 M [OH] =8.4×107
Kb = 1.4×1011 = [HF][OH] = x  x [F] x
x = 8.4×107 M
[OH] =8.4×107
pOH = 6.08
pH = 7.92

36. Ex. 60. 0 mL of 0. 10 M NaOH is added to 50. 0 mL of 0. 10 M HF
Ex mL of 0.10 M NaOH is added to 50.0 mL of 0.10 M HF. (Ka of HF = 7.2×104) Determine the pH of the final solution.
0.0600L×0.10 M OH = mol OH
0.0500L×0.10 M HF = mol HF
Stoichiometry
OH HF  H2O F
Before mol mol
After mol mol
[OH] = mol/0.110 L = 9.09×103 M
Strong base + weak base (ignore weak base!)
pOH = pH = 11.96

37. 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.

38. The pH Curve for the Titration of 100. 0 mL of 0. 050 M NH3 with 0
The pH Curve for the Titration of mL of M NH3 with 0.10 M HCl
SA<WB
[B] = [HB+] ½ eq pt
pOH = pKb
Weak base only Kb
SA < WB
Stoich + Kb
SA < WB
Stoich + Kb
SA> WB
Stoich,
pH = log [SA]
WB = SA
Acidic salt Ka

39. Ex. Calculate the pH when mL of M NH3 is titrated with 10.0 mL of 0.10 M HCl. Kb of NH3 = 1.8 × 105
0.100L×0.050 M = mol NH3
0.010L×0.10 M = mol H+
NH H+  NH4+
Before mol mol 0
Change  
After mol mol
mol/0.110 L = 9.09×103M NH4+
mol/0.110 L = 3.64×102M NH3
WB+ CA = Kb!

40. Equilibrium Rxn NH3 + H2O  NH4+ + OH Initial 0.0364 ----- 0.00909 0
Change x x x
Equil x x x
Kb = 1.8×105 = ( x)x  ( )x x
x = [OH]= 7.21×105 pOH = 4.14
pH = 9.86

41. Ex. Calculate the pH when mL of M NH3 is titrated with 50.0 mL of 0.10 M HCl. Kb of NH3 = 1.8×105
0.100L×0.050 M = mol NH3
0.050L×0.10 M = mol H+
NH H  NH4+
Before mol mol
Change  
After mol mol/ 0.150L = M NH4+
CA only = Ka!

42. Equilibrium Rxn NH4+ + H2O  NH3 + H3O+ Initial 0.0333 ----- 0 0
Change x x x
Equil x x x
Ka for NH4+ = 1.0×1014/Kb for NH3 = 5.56×1010
5.56×1010 = [NH3][H3O+] = x  x [NH4+] x
x = 4.30×106= [H+] pH = 5.37

43. Ex. Calculate the pH when 100.0 mL of 0.050 M NH3 is titrated with 60.0 mL of 0.10 M HCl.
0.100L×0.050 M = mol NH3
0.060L×0.10 M = mol H NH H  NH4+
Before mol mol
Change  
After mol mol
mol/0.160L = 6.25×103 M H3O+
pH = 2.20
Strong acid and CA
Ignore CA(weak)!

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

45. 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.

46. Indicators are usually weak acids, HIn
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.

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

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

51. The pH Curve for the Titration of 100. 0 mL of 0. 10 M HCI with 0
The pH Curve for the Titration of 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 HC2H3O2 with 0
The pH Curve for the Titration of 50 mL of 0.1 M HC2H3O2 with 0.1 M NaOH