1. The Danger of Antifreeze
each year, thousands of pets and wildlife die from consuming antifreeze
most brands of antifreeze contain ethylene glycol
sweet taste
initial effect drunkenness
metabolized in the liver to glycolic acid
HOCH2COOH
if present in high enough concentration in the bloodstream, it overwhelms the buffering ability of HCO3−, causing the blood pH to drop
when the blood pH is low, it ability to carry O2 is compromised
acidosis
the treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol
ethylene glycol
(aka 1,2–ethandiol)

2. Buffers
buffers are solutions that resist changes in pH when an acid or base is added
they act by neutralizing the added acid or base
but just like everything else, there is a limit to what they can do, eventually the pH changes
many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion
Tro, Chemistry: A Molecular Approach

3. Making an Acid Buffer
Tro, Chemistry: A Molecular Approach

4. How Acid Buffers Work HA(aq) + H2O(l)  A−(aq) + H3O+(aq)
buffers work by applying Le Châtelier’s Principle to weak acid equilibrium
buffer solutions contain significant amounts of the weak acid molecules, HA – these molecules react with added base to neutralize it
you can also think of the H3O+ combining with the OH− to make H2O; the H3O+ is then replaced by the shifting equilibrium
the buffer solutions also contain significant amounts of the conjugate base anion, A− - these ions combine with added acid to make more HA and keep the H3O+ constant
Tro, Chemistry: A Molecular Approach

5.  + How Buffers Work H2O HA HA A− A− new HA H3O+ Added H3O+
Tro, Chemistry: A Molecular Approach

6.  + How Buffers Work H2O A− HA A− HA new A− H3O+ Added HO−
Tro, Chemistry: A Molecular Approach

7. Common Ion Effect HA(aq) + H2O(l)  A−(aq) + H3O+(aq)
adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left
this causes the pH to be higher than the pH of the acid solution
lowering the H3O+ ion concentration
Tro, Chemistry: A Molecular Approach

8. Common Ion Effect
Tro, Chemistry: A Molecular Approach

9. Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
Ex What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2?
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0
HC2H3O2 + H2O  C2H3O2 + H3O+
[HA]
[A-]
[H3O+]
initial
0.100
≈ 0
change
equilibrium
Tro, Chemistry: A Molecular Approach

10. Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
Ex What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2?
HC2H3O2 + H2O  C2H3O2 + H3O+
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
[HA]
[A-]
[H3O+]
initial
0.100
change
equilibrium x +x +x
0.100 x x x
Tro, Chemistry: A Molecular Approach

11. Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
Ex What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2?
Ka for HC2H3O2 = 1.8 x 10-5
determine the value of Ka
since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x
[HA]
[A-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium x
0.100 x x
Tro, Chemistry: A Molecular Approach

12. Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
Ex What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2?
Ka for HC2H3O2 = 1.8 x 10-5
check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init
[HA]
[A-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium x
x = 1.8 x 10-5
the approximation is valid
Tro, Chemistry: A Molecular Approach

13. Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
Ex What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2?
[HA]
[A-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium
1.8E-5
substitute x into the equilibrium concentration definitions and solve
0.100 x x x
x = 1.8 x 10-5
Tro, Chemistry: A Molecular Approach

14. Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
Ex What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2?
[HA]
[A-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium
1.8E-5
substitute [H3O+] into the formula for pH and solve
Tro, Chemistry: A Molecular Approach

15. Ex 16. 1 - What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
Ex What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2?
Ka for HC2H3O2 = 1.8 x 10-5
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
[HA]
[A-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium
1.8E-5
the values match
Tro, Chemistry: A Molecular Approach

16. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
Tro, Chemistry: A Molecular Approach

17. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0
HF + H2O  F + H3O+
[HA]
[A-]
[H3O+]
initial
0.14
0.071
≈ 0
change
equilibrium
Tro, Chemistry: A Molecular Approach

18. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
HF + H2O  F + H3O+
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
[HA]
[A-]
[H3O+]
initial
0.14
0.071
change
equilibrium x +x +x
0.14 x x x
Tro, Chemistry: A Molecular Approach

19. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
Ka for HF = 7.0 x 10-4
determine the value of Ka
since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x
[HA]
[A-]
[H3O+]
initial
0.14
0.071
≈ 0
change -x +x
equilibrium
0.012
0.100 x
0.14 x x
Tro, Chemistry: A Molecular Approach

20. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
Ka for HF = 7.0 x 10-4
check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init
[HA]
[A2-]
[H3O+]
initial
0.14
0.071
≈ 0
change -x +x
equilibrium x
x = 1.4 x 10-3
the approximation is valid
Tro, Chemistry: A Molecular Approach

21. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
[HA]
[A2-]
[H3O+]
initial
0.14
0.071
≈ 0
change -x +x
equilibrium
0.072
1.4E-3
substitute x into the equilibrium concentration definitions and solve
0.14 x x x
x = 1.4 x 10-3
Tro, Chemistry: A Molecular Approach

22. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
[HA]
[A-]
[H3O+]
initial
0.14
0.071
≈ 0
change -x +x
equilibrium
0.072
1.4E-3
substitute [H3O+] into the formula for pH and solve
Tro, Chemistry: A Molecular Approach

23. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
Ka for HF = 7.0 x 10-4
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
[HA]
[A-]
[H3O+]
initial
0.14
0.071
≈ 0
change -x +x
equilibrium
0.072
1.4E-3
the values are close enough
Tro, Chemistry: A Molecular Approach

24. Henderson-Hasselbalch Equation
calculating the pH of a buffer solution can be simplified by using an equation derived from the Ka expression called the Henderson-Hasselbalch Equation
the equation calculates the pH of a buffer from the Ka and initial concentrations of the weak acid and salt of the conjugate base
as long as the “x is small” approximation is valid
Tro, Chemistry: A Molecular Approach

25. Deriving the Henderson-Hasselbalch Equation
Tro, Chemistry: A Molecular Approach

26. Ex 16. 2 - What is the pH of a buffer that is 0. 050 M HC7H5O2 and 0
Ex What is the pH of a buffer that is M HC7H5O2 and M NaC7H5O2?
HC7H5O2 + H2O  C7H5O2 + H3O+
Assume the [HA] and [A-] equilibrium concentrations are the same as the initial
Substitute into the Henderson-Hasselbalch Equation
Check the “x is small” approximation
Ka for HC7H5O2 = 6.5 x 10-5
Tro, Chemistry: A Molecular Approach

27. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
Tro, Chemistry: A Molecular Approach

28. Practice - What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
find the pKa from the given Ka
Assume the [HA] and [A-] equilibrium concentrations are the same as the initial
Substitute into the Henderson-Hasselbalch Equation
Check the “x is small” approximation
HF + H2O  F + H3O+
Tro, Chemistry: A Molecular Approach

29. Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation?
the Henderson-Hasselbalch equation is generally good enough when the “x is small” approximation is applicable
generally, the “x is small” approximation will work when both of the following are true:
the initial concentrations of acid and salt are not very dilute
the Ka is fairly small
for most problems, this means that the initial acid and salt concentrations should be over 1000x larger than the value of Ka
Tro, Chemistry: A Molecular Approach

30. How Much Does the pH of a Buffer Change When an Acid or Base Is Added?
though buffers do resist change in pH when acid or base are added to them, their pH does change
calculating the new pH after adding acid or base requires breaking the problem into 2 parts
a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other
added acid reacts with the A− to make more HA
added base reacts with the HA to make more A−
an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]
Tro, Chemistry: A Molecular Approach

31. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for it with A−.
Construct a stoichiometry table for the reaction
HC2H3O2 + OH−  C2H3O2 + H2O HA A-
OH−
mols Before
0.100
mols added -
0.010
mols After
Tro, Chemistry: A Molecular Approach

32. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
HC2H3O2 + OH−  C2H3O2 + H2O
Fill in the table – tracking the changes in the number of moles for each component HA A-
OH−
mols Before
0.100
≈ 0
mols added -
0.010
mols After
0.090
0.110
Tro, Chemistry: A Molecular Approach

33. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0, and using the new molarities of the [HA] and [A−]
HC2H3O2 + H2O  C2H3O2 + H3O+
[HA]
[A-]
[H3O+]
initial
0.090
0.110
≈ 0
change
equilibrium
Tro, Chemistry: A Molecular Approach

34. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
HC2H3O2 + H2O  C2H3O2 + H3O+
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
[HA]
[A-]
[H3O+]
initial
0.090
0.110
change
equilibrium x +x +x
0.090 x x x
Tro, Chemistry: A Molecular Approach

35. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
Ka for HC2H3O2 = 1.8 x 10-5
determine the value of Ka
since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x
[HA]
[A-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium
0.090
0.110 x
0.090 x x
Tro, Chemistry: A Molecular Approach

36. the approximation is valid
Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
Ka for HC2H3O2 = 1.8 x 10-5
check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init
[HA]
[A-]
[H3O+]
initial
0.090
0.110
≈ 0
change -x +x
equilibrium x
x = 1.47 x 10-5
the approximation is valid
Tro, Chemistry: A Molecular Approach

37. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
[HA]
[A-]
[H3O+]
initial
0.090
0.110
≈ 0
change -x +x
equilibrium
1.5E-5
substitute x into the equilibrium concentration definitions and solve
0.090 x x x
x = 1.47 x 10-5
Tro, Chemistry: A Molecular Approach

38. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
[HA]
[A-]
[H3O+]
initial
0.090
0.110
≈ 0
change -x +x
equilibrium
1.5E-5
substitute [H3O+] into the formula for pH and solve
Tro, Chemistry: A Molecular Approach

39. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
Ka for HC2H3O2 = 1.8 x 10-5
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
[HA]
[A-]
[H3O+]
initial
0.090
0.110
≈ 0
change -x +x
equilibrium
1.5E-5
the values match
Tro, Chemistry: A Molecular Approach

40. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
find the pKa from the given Ka
Assume the [HA] and [A-] equilibrium concentrations are the same as the initial
HC2H3O2 + H2O  C2H3O2 + H3O+
Ka for HC2H3O2 = 1.8 x 10-5
[HA]
[A-]
[H3O+]
initial
0.090
0.110
≈ 0
change -x +x
equilibrium x
Tro, Chemistry: A Molecular Approach

41. Ex What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it?
Substitute into the Henderson-Hasselbalch Equation
Check the “x is small” approximation
HC2H3O2 + H2O  C2H3O2 + H3O+
pKa for HC2H3O2 = 4.745
Tro, Chemistry: A Molecular Approach

42. Ex 16. 3 – Compare the effect on pH of adding 0
Ex 16.3 – Compare the effect on pH of adding mol NaOH to a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L to adding mol NaOH to 1.00 L of pure water?
HC2H3O2 + H2O  C2H3O2 + H3O+
pKa for HC2H3O2 = 4.745
Tro, Chemistry: A Molecular Approach

43. Basic Buffers B:(aq) + H2O(l)  H:B+(aq) + OH−(aq)
buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl−
H2O(l) + NH3 (aq)  NH4+(aq) + OH−(aq)
Tro, Chemistry: A Molecular Approach

44. Ex 16. 4 - What is the pH of a buffer that is 0. 50 M NH3 (pKb = 4
Ex What is the pH of a buffer that is 0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl?
NH3 + H2O  NH4+ + OH−
find the pKa of the conjugate acid (NH4+) from the given Kb
Assume the [B] and [HB+] equilibrium concentrations are the same as the initial
Substitute into the Henderson-Hasselbalch Equation
Check the “x is small” approximation
Tro, Chemistry: A Molecular Approach

45. Buffering Effectiveness
a good buffer should be able to neutralize moderate amounts of added acid or base
however, there is a limit to how much can be added before the pH changes significantly
the buffering capacity is the amount of acid or base a buffer can neutralize
the buffering range is the pH range the buffer can be effective
the effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base
Tro, Chemistry: A Molecular Approach

46. Effect of Relative Amounts of Acid and Conjugate Base
a buffer is most effective with equal concentrations of acid and base
Buffer 1
0.100 mol HA & mol A-
Initial pH = 5.00
Buffer 12
0.18 mol HA & mol A-
Initial pH = 4.05
pKa (HA) = 5.00
HA + OH−  A + H2O
after adding mol NaOH
pH = 5.09
after adding mol NaOH
pH = 4.25 HA A-
OH−
mols Before
0.100
mols added -
0.010
mols After
0.090
0.110
≈ 0 HA A-
OH−
mols Before
0.18
0.020
mols added -
0.010
mols After
0.17
0.030
≈ 0

47. Effect of Absolute Concentrations of Acid and Conjugate Base
a buffer is most effective when the concentrations of acid and base are largest
Buffer 1
0.50 mol HA & 0.50 mol A-
Initial pH = 5.00
Buffer 12
0.050 mol HA & mol A-
Initial pH = 5.00
pKa (HA) = 5.00
HA + OH−  A + H2O
after adding mol NaOH
pH = 5.02
after adding mol NaOH
pH = 5.18 HA A-
OH−
mols Before
0.50
0.500
mols added -
0.010
mols After
0.49
0.51
≈ 0 HA A-
OH−
mols Before
0.050
mols added -
0.010
mols After
0.040
0.060
≈ 0

48. Effectiveness of Buffers
a buffer will be most effective when the [base]:[acid] = 1
equal concentrations of acid and base
effective when 0.1 < [base]:[acid] < 10
a buffer will be most effective when the [acid] and the [base] are large
Tro, Chemistry: A Molecular Approach

49. 0.1 < [base]:[acid] < 10
Buffering Range
we have said that a buffer will be effective when
0.1 < [base]:[acid] < 10
substituting into the Henderson-Hasselbalch we can calculate the maximum and minimum pH at which the buffer will be effective
Lowest pH
Highest pH
therefore, the effective pH range of a buffer is pKa ± 1
when choosing an acid to make a buffer, choose one whose is pKa is closest to the pH of the buffer

50. Chlorous Acid, HClO2 pKa = 1.95
Ex. 16.5a – Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25?
Chlorous Acid, HClO2 pKa = 1.95
Nitrous Acid, HNO2 pKa = 3.34
Formic Acid, HCHO2 pKa = 3.74
Hypochlorous Acid, HClO pKa = 7.54
Tro, Chemistry: A Molecular Approach

51. Chlorous Acid, HClO2 pKa = 1.95
Ex. 16.5a – Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25?
Chlorous Acid, HClO2 pKa = 1.95
Nitrous Acid, HNO2 pKa = 3.34
Formic Acid, HCHO2 pKa = 3.74
Hypochlorous Acid, HClO pKa = 7.54
The pKa of HCHO2 is closest to the desired pH of the buffer, so it would give the most effective buffering range.
Tro, Chemistry: A Molecular Approach

52. Ex. 16.5b – What ratio of NaCHO2 : HCHO2 would be required to make a buffer with pH 4.25?
Formic Acid, HCHO2, pKa = 3.74
to make the buffer with pH 4.25, you would use 3.24 times as much NaCHO2 as HCHO2
Tro, Chemistry: A Molecular Approach

53. Buffering Capacity
buffering capacity is the amount of acid or base that can be added to a buffer without destroying its effectiveness
the buffering capacity increases with increasing absolute concentration of the buffer components
as the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves
buffers that need to work mainly with added acid generally have [base] > [acid]
buffers that need to work mainly with added base generally have [acid] > [base]
Tro, Chemistry: A Molecular Approach

54. Buffering Capacity
a concentrated buffer can neutralize more added acid or base than a dilute buffer
Tro, Chemistry: A Molecular Approach

55. Titration
in an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete
when the reaction is complete we have reached the endpoint of the titration
an indicator may be added to determine the endpoint
an indicator is a chemical that changes color when the pH changes
when the moles of H3O+ = moles of OH−, the titration has reached its equivalence point
Tro, Chemistry: A Molecular Approach

56. Titration
Tro, Chemistry: A Molecular Approach

57. Titration Curve a plot of pH vs. amount of added titrant
the inflection point of the curve is the equivalence point of the titration
prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH
the pH of the equivalence point depends on the pH of the salt solution
equivalence point of neutral salt, pH = 7
equivalence point of acidic salt, pH < 7
equivalence point of basic salt, pH > 7
beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH
Tro, Chemistry: A Molecular Approach

58. Titration Curve: Unknown Strong Base Added to Strong Acid
Tro, Chemistry: A Molecular Approach

59. Tro, Chemistry: A Molecular Approach

60. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(aq)
initial pH = -log(0.100) = 1.00
initial mol of HCl = L x mol/L = 2.50 x 10-3
before equivalence point
added 5.0 mL NaOH
5.0 x 10-4 mol NaOH
2.00 x 10-3 mol HCl
Tro, Chemistry: A Molecular Approach

61. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(aq)
at equivalence, 0.00 mol HCl and 0.00 mol NaOH
pH at equivalence = 7.00
after equivalence point
added 30.0 mL NaOH
5.0 x 10-4 mol NaOH xs
Tro, Chemistry: A Molecular Approach

62. Adding NaOH to HCl added 10.0 mL NaOH 0.00150 mol HCl pH = 1.37
equivalence point
pH = 7.00
25.0 mL M HCl
mol HCl
pH = 1.00
added 5.0 mL NaOH
mol HCl
pH = 1.18
added 35.0 mL NaOH
mol NaOH
pH = 12.22
added 30.0 mL NaOH
mol NaOH
pH = 11.96
added 15.0 mL NaOH
mol HCl
pH = 1.60
added 40.0 mL NaOH
mol NaOH
pH = 12.36
added 50.0 mL NaOH
mol NaOH
pH = 12.52
added 20.0 mL NaOH
mol HCl
pH = 1.95
Tro, Chemistry: A Molecular Approach

63. Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH
HCHO2(aq) + NaOH(aq)  NaCHO2(aq) + H2O(aq)
Initial pH:
Ka = 1.8 x 10-4
[HCHO2]
[CHO2-]
[H3O+]
initial
0.100
0.000
≈ 0
change -x +x
equilibrium x x
Tro, Chemistry: A Molecular Approach

64. Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH
HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq)
initial mol of HCHO2 = L x mol/L = 2.50 x 10-3
before equivalence
added 5.0 mL NaOH HA A-
OH−
mols Before
2.50E-3
mols added -
5.0E-4
mols After
2.00E-3
≈ 0
Tro, Chemistry: A Molecular Approach

65. Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH
HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq)
initial mol of HCHO2 = L x mol/L = 2.50 x 10-3
at equivalence
CHO2−(aq) + H2O(l)  HCHO2(aq) + OH−(aq)
added 25.0 mL NaOH
Kb = 5.6 x 10-11
[HCHO2]
[CHO2-]
[OH−]
initial
0.0500
≈ 0
change +x -x
equilibrium x
5.00E-2-x HA A-
OH−
mols Before
2.50E-3
mols added -
mols After
≈ 0
[OH-] = 1.7 x 10-6 M

66. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq)
after equivalence point
added 30.0 mL NaOH
5.0 x 10-4 mol NaOH xs
Tro, Chemistry: A Molecular Approach

67. Adding NaOH to HCHO2 added 10.0 mL NaOH 0.00150 mol HCHO2 pH = 3.56
mol NaOH xs
pH = 12.22
added 30.0 mL NaOH
mol NaOH xs
pH = 11.96
added 25.0 mL NaOH
equivalence point
mol CHO2−
[CHO2−]init = M
[OH−]eq = 1.7 x 10-6
pH = 8.23
added 5.0 mL NaOH
mol HCHO2
pH = 3.14
initial HCHO2 solution
mol HCHO2
pH = 2.37
added 12.5 mL NaOH
mol HCHO2
pH = 3.74 = pKa
half-neutralization
added 40.0 mL NaOH
mol NaOH xs
pH = 12.36
added 15.0 mL NaOH
mol HCHO2
pH = 3.92
added 50.0 mL NaOH
mol NaOH xs
pH = 12.52
added 20.0 mL NaOH
mol HCHO2
pH = 4.34
Tro, Chemistry: A Molecular Approach

68. Titrating Weak Acid with a Strong Base
the initial pH is that of the weak acid solution
calculate like a weak acid equilibrium problem
e.g., 15.5 and 15.6
before the equivalence point, the solution becomes a buffer
calculate mol HAinit and mol A−init using reaction stoichiometry
calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−init
half-neutralization pH = pKa
Tro, Chemistry: A Molecular Approach

69. Titrating Weak Acid with a Strong Base
at the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established
mol A− = original mole HA
calculate the volume of added base like Ex 4.8
[A−]init = mol A−/total liters
calculate like a weak base equilibrium problem
e.g., 15.14
beyond equivalence point, the OH is in excess
[OH−] = mol MOH xs/total liters
[H3O+][OH−]=1 x 10-14
Tro, Chemistry: A Molecular Approach

70. Ex 16. 7a – A 40. 0 mL sample of 0. 100 M HNO2 is titrated with 0
Ex 16.7a – A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the volume of KOH at the equivalence point
Write an equation for the reaction for B with HA.
Use Stoichiometry to determine the volume of added B
HNO2 + KOH  NO2 + H2O
Tro, Chemistry: A Molecular Approach

71. Ex 16. 7b – A 40. 0 mL sample of 0. 100 M HNO2 is titrated with 0
Ex 16.7b – A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the pH after adding 5.00 mL KOH
Write an equation for the reaction for B with HA.
Determine the moles of HAbefore & moles of added B
Make a stoichiometry table and determine the moles of HA in excess and moles A made
HNO2 + KOH  NO2 + H2O
HNO2
NO2-
OH−
mols Before
≈ 0
mols added -
mols After
Tro, Chemistry: A Molecular Approach

72. Ex 16. 7b – A 40. 0 mL sample of 0. 100 M HNO2 is titrated with 0
Ex 16.7b – A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the pH after adding 5.00 mL KOH.
HNO2 + H2O  NO2 + H3O+
Write an equation for the reaction of HA with H2O
Determine Ka and pKa for HA
Use the Henderson-Hasselbalch Equation to determine the pH
Table Ka = 4.6 x 10-4
HNO2
NO2-
OH−
mols Before
≈ 0
mols added -
mols After
Tro, Chemistry: A Molecular Approach

73. Ex 16. 7b – A 40. 0 mL sample of 0. 100 M HNO2 is titrated with 0
Ex 16.7b – A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the pH at the half-equivalence point
Write an equation for the reaction for B with HA.
Determine the moles of HAbefore & moles of added B
Make a stoichiometry table and determine the moles of HA in excess and moles A made
HNO2 + KOH  NO2 + H2O
at half-equivalence, moles KOH = ½ mole HNO2
HNO2
NO2-
OH−
mols Before
≈ 0
mols added -
mols After
Tro, Chemistry: A Molecular Approach

74. Ex 16. 7b – A 40. 0 mL sample of 0. 100 M HNO2 is titrated with 0
Ex 16.7b – A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the pH at the half-equivalence point.
HNO2 + H2O  NO2 + H3O+
Write an equation for the reaction of HA with H2O
Determine Ka and pKa for HA
Use the Henderson-Hasselbalch Equation to determine the pH
Table Ka = 4.6 x 10-4
HNO2
NO2-
OH−
mols Before
≈ 0
mols added -
mols After
Tro, Chemistry: A Molecular Approach

75. Titration Curve of a Weak Base with a Strong Acid
Tro, Chemistry: A Molecular Approach

76. Titration of a Polyprotic Acid
if Ka1 >> Ka2, there will be two equivalence points in the titration
the closer the Ka’s are to each other, the less distinguishable the equivalence points are
titration of 25.0 mL of M H2SO3 with M NaOH
Tro, Chemistry: A Molecular Approach

77. Monitoring pH During a Titration
the general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H3O+]
using a probe that specifically measures just H3O+
the endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve
if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator
Tro, Chemistry: A Molecular Approach

78. Monitoring pH During a Titration
Tro, Chemistry: A Molecular Approach

79. HInd(aq) + H2O(l)  Ind(aq) + H3O+(aq)
Indicators
many dyes change color depending on the pH of the solution
these dyes are weak acids, establishing an equilibrium with the H2O and H3O+ in the solution
HInd(aq) + H2O(l)  Ind(aq) + H3O+(aq)
the color of the solution depends on the relative concentrations of Ind:HInd
when Ind:HInd ≈ 1, the color will be mix of the colors of Ind and HInd
when Ind:HInd > 10, the color will be mix of the colors of Ind
when Ind:HInd < 0.1, the color will be mix of the colors of HInd
Tro, Chemistry: A Molecular Approach

80. Phenolphthalein

81. Methyl Red
Tro, Chemistry: A Molecular Approach

82. Monitoring a Titration with an Indicator
for most titrations, the titration curve shows a very large change in pH for very small additions of base near the equivalence point
an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH
pKa of HInd ≈ pH at equivalence point
Tro, Chemistry: A Molecular Approach

83. Acid-Base Indicators

84. Solubility Equilibria
all ionic compounds dissolve in water to some degree
however, many compounds have such low solubility in water that we classify them as insoluble
we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water
Tro, Chemistry: A Molecular Approach

85. Solubility Product
the equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, Ksp
for an ionic solid MnXm, the dissociation reaction is:
MnXm(s)  nMm+(aq) + mXn−(aq)
the solubility product would be
Ksp = [Mm+]n[Xn−]m
for example, the dissociation reaction for PbCl2 is
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
and its equilibrium constant is
Ksp = [Pb2+][Cl−]2
Tro, Chemistry: A Molecular Approach

86. Tro, Chemistry: A Molecular Approach

87. Molar Solubility
solubility is the amount of solute that will dissolve in a given amount of solution
at a particular temperature
the molar solubility is the number of moles of solute that will dissolve in a liter of solution
the molarity of the dissolved solute in a saturated solution
for the general reaction MnXm(s)  nMm+(aq) + mXn−(aq)
Tro, Chemistry: A Molecular Approach

88. PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C
Write the dissociation reaction and Ksp expression
Create an ICE table defining the change in terms of the solubility of the solid
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
Ksp = [Pb2+][Cl−]2
[Pb2+]
[Cl−]
Initial
Change +S
+2S
Equilibrium S 2S
Tro, Chemistry: A Molecular Approach

89. Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C
Substitute into the Ksp expression
Find the value of Ksp from Table 16.2, plug into the equation and solve for S
Ksp = [Pb2+][Cl−]2
Ksp = (S)(2S)2
[Pb2+]
[Cl−]
Initial
Change +S
+2S
Equilibrium S 2S
Tro, Chemistry: A Molecular Approach

90. Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M
Tro, Chemistry: A Molecular Approach

91. PbBr2(s)  Pb2+(aq) + 2 Br−(aq)
Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M
Write the dissociation reaction and Ksp expression
Create an ICE table defining the change in terms of the solubility of the solid
PbBr2(s)  Pb2+(aq) + 2 Br−(aq)
Ksp = [Pb2+][Br−]2
[Pb2+]
[Br−]
Initial
Change
+(1.05 x 10-2)
+2(1.05 x 10-2)
Equilibrium
(1.05 x 10-2)
(2.10 x 10-2)
Tro, Chemistry: A Molecular Approach

92. Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M
Substitute into the Ksp expression
plug into the equation and solve
Ksp = [Pb2+][Br−]2
Ksp = (1.05 x 10-2)(2.10 x 10-2)2
[Pb2+]
[Br−]
Initial
Change
+(1.05 x 10-2)
+2(1.05 x 10-2)
Equilibrium
(1.05 x 10-2)
(2.10 x 10-2)
Tro, Chemistry: A Molecular Approach

93. Ksp and Relative Solubility
molar solubility is related to Ksp
but you cannot always compare solubilities of compounds by comparing their Ksps
in order to compare Ksps, the compounds must have the same dissociation stoichiometry
Tro, Chemistry: A Molecular Approach

94. The Effect of Common Ion on Solubility
addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt
for example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2
PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
addition of Cl− shifts the equilibrium to the left
Tro, Chemistry: A Molecular Approach

95. CaF2(s)  Ca2+(aq) + 2 F−(aq)
Ex – Calculate the molar solubility of CaF2 in M NaF at 25C
Write the dissociation reaction and Ksp expression
Create an ICE table defining the change in terms of the solubility of the solid
CaF2(s)  Ca2+(aq) + 2 F−(aq)
Ksp = [Ca2+][F−]2
[Ca2+]
[F−]
Initial
0.100
Change +S
+2S
Equilibrium S S
Tro, Chemistry: A Molecular Approach

96. Ex 16. 10 – Calculate the molar solubility of CaF2 in 0
Ex – Calculate the molar solubility of CaF2 in M NaF at 25C
Substitute into the Ksp expression
assume S is small
Find the value of Ksp from Table 16.2, plug into the equation and solve for S
Ksp = [Ca2+][F−]2
Ksp = (S)( S)2
Ksp = (S)(0.100)2
[Ca2+]
[F−]
Initial
0.100
Change +S
+2S
Equilibrium S S
Tro, Chemistry: A Molecular Approach

97. The Effect of pH on Solubility
for insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide
and the lower the pH, the higher the solubility
higher pH = increased [OH−]
M(OH)n(s)  Mn+(aq) + nOH−(aq)
for insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility
M2(CO3)n(s)  2 Mn+(aq) + nCO32−(aq)
H3O+(aq) + CO32− (aq)  HCO3− (aq) + H2O(l)
Tro, Chemistry: A Molecular Approach

98. Precipitation
precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound
if we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur
Q = Ksp, the solution is saturated, no precipitation
Q < Ksp, the solution is unsaturated, no precipitation
Q > Ksp, the solution would be above saturation, the salt above saturation will precipitate
some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions
Tro, Chemistry: A Molecular Approach

99. a supersaturated solution will precipitate if a seed crystal is added
precipitation occurs if Q > Ksp
Tro, Chemistry: A Molecular Approach

100. Selective Precipitation
a solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others
a successful reagent can precipitate with more than one of the cations, as long as their Ksp values are significantly different
Tro, Chemistry: A Molecular Approach

101. precipitating may just occur when Q = Ksp
Ex What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059]) from seawater?
precipitating may just occur when Q = Ksp
Tro, Chemistry: A Molecular Approach

102. precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
Ex What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
Tro, Chemistry: A Molecular Approach

103. precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
Ex What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M
when Ca2+ just begins to precipitate out, the [Mg2+] has dropped from M to 4.8 x M
Tro, Chemistry: A Molecular Approach

104. Qualitative Analysis
an analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme
wet chemistry
a sample containing several ions is subjected to the addition of several precipitating agents
addition of each reagent causes one of the ions present to precipitate out
Tro, Chemistry: A Molecular Approach

105. Qualitative Analysis
Tro, Chemistry: A Molecular Approach

107. Group 1 group one cations are Ag+, Pb2+, and Hg22+
all these cations form compounds with Cl− that are insoluble in water
as long as the concentration is large enough
PbCl2 may be borderline
molar solubility of PbCl2 = 1.43 x 10-2 M
precipitated by the addition of HCl
Tro, Chemistry: A Molecular Approach

108. Group 2
group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+, Pb2+, Sb3+, and Hg2+
all these cations form compounds with HS− and S2− that are insoluble in water at low pH
precipitated by the addition of H2S in HCl

109. Group 3
group three cations are Fe2+, Co2+, Zn2+, Mn2+, Ni2+ precipitated as sulfides; as well as Cr3+, Fe3+, and Al3+ precipitated as hydroxides
all these cations form compounds with S2− that are insoluble in water at high pH
precipitated by the addition of H2S in NaOH

110. Group 4 group four cations are Mg2+, Ca2+, Ba2+
all these cations form compounds with PO43− that are insoluble in water at high pH
precipitated by the addition of (NH4)2HPO4
Tro, Chemistry: A Molecular Approach

111. Group 5 group five cations are Na+, K+, NH4+
all these cations form compounds that are soluble in water – they do not precipitate
identified by the color of their flame

112. Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq)
Complex Ion Formation
transition metals tend to be good Lewis acids
they often bond to one or more H2O molecules to form a hydrated ion
H2O is the Lewis base, donating electron pairs to form coordinate covalent bonds
Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq)
ions that form by combining a cation with several anions or neutral molecules are called complex ions
e.g., Ag(H2O)2+
the attached ions or molecules are called ligands
e.g., H2O
Tro, Chemistry: A Molecular Approach

113. Complex Ion Equilibria
if a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand
Ag(H2O)2+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) + 2 H2O(l)
generally H2O is not included, since its complex ion is always present in aqueous solution
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
Tro, Chemistry: A Molecular Approach

114. Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
Formation Constant
the reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
the equilibrium constant for the formation reaction is called the formation constant, Kf
Tro, Chemistry: A Molecular Approach

115. Formation Constants
Tro, Chemistry: A Molecular Approach

116. Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Ex – mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with mL of 0.20 M NH3. What is the [Cu2+] at equilibrium?
Write the formation reaction and Kf expression.
Look up Kf value
Determine the concentration of ions in the diluted solutions
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Tro, Chemistry: A Molecular Approach

117. Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Ex – mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with mL of 0.20 M NH3. What is the [Cu2+] at equilibrium?
Create an ICE table. Since Kf is large, assume all the Cu2+ is converted into complex ion, then the system returns to equilibrium
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
[Cu2+]
[NH3]
[Cu(NH3)22+]
Initial
6.7E-4
0.11
Change
-≈6.7E-4
-4(6.7E-4)
+ 6.7E-4
Equilibrium x

118. Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Ex – mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with mL of 0.20 M NH3. What is the [Cu2+] at equilibrium?
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Substitute in and solve for x
confirm the “x is small” approximation
[Cu2+]
[NH3]
[Cu(NH3)22+]
Initial
6.7E-4
0.11
Change
-≈6.7E-4
-4(6.7E-4)
+ 6.7E-4
Equilibrium x
since 2.7 x << 6.7 x 10-4, the approximation is valid
Tro, Chemistry: A Molecular Approach

119. The Effect of Complex Ion Formation on Solubility
the solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands
AgCl(s)  Ag+(aq) + Cl−(aq) Ksp = 1.77 x 10-10
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) Kf = 1.7 x 107
adding NH3 to a solution in equilibrium with AgCl(s) increases the solubility of Ag+
Tro, Chemistry: A Molecular Approach

121. Solubility of Amphoteric Metal Hydroxides
many metal hydroxides are insoluble
all metal hydroxides become more soluble in acidic solution
shifting the equilibrium to the right by removing OH−
some metal hydroxides also become more soluble in basic solution
acting as a Lewis base forming a complex ion
substances that behave as both an acid and base are said to be amphoteric
some cations that form amphoteric hydroxides include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+
Tro, Chemistry: A Molecular Approach

122. Al3+ Al3+ is hydrated in water to form an acidic solution
Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+(aq) + H3O+(aq)
addition of OH− drives the equilibrium to the right and continues to remove H from the molecules
Al(H2O)5(OH)2+(aq) + OH−(aq)  Al(H2O)4(OH)2+(aq) + H2O (l)
Al(H2O)4(OH)2+(aq) + OH−(aq)  Al(H2O)3(OH)3(s) + H2O (l)
Tro, Chemistry: A Molecular Approach

123. Tro, Chemistry: A Molecular Approach