1. Aqueous Ionic Equilibrium

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

3. Demo of a pH Buffer
Aqueous Solution pH
Δ pH
Bottled Water (30 mL)
5.6
0.7
Bottled Water + 1 drop of 0.1M HNO3
4.9
Deionized Water (30 mL)
6.0
2.2
Deionized Water + 1 drop of 0.1M HNO3
3.8
pH Buffer
7.8
0.0
pH Buffer + 1 drop of 0.1M HNO3
What should the pH be when we add 1 drop of 0.1M HNO3 to 30mL
of deionized water ?
Assume 1 drop = 0.05mL
When this 1 drop is dissolved in 30 mL of H2O

4. Why are pH buffers important?
Life on Earth is water-based
Human 48-75% water
Plants as high as 95% water
H+ and OH- are chemically and structurally reactive in cells
The functioning of the cell is very pH dependent
Blood plasma pH = 7.4
pH < 6.9 fatal (acidosis)
pH > 7.9 fatal (alkalosis)
Fish die if the pH of the water goes below 5 or above 9

5. Making an Acid Buffer

6. How Acid Buffers Work HA(aq) + H2O(l) A−(aq) + H3O+(aq)
buffers work by applying Le Châtelier’s Principle to weak acid equilibrium
Adding OH- to the buffer: the OH- reacts with the H3O+ to make 2H2O(l) so it removes H3O+ so drives the equilibrium to the right effectively doing the following
Effectively 𝐻𝐴 (𝑎𝑞) + 𝑂𝐻 (𝑎𝑞) − → 𝐴 𝑎𝑞 − + 𝐻 2 𝑂 (𝑙)
By Le Châtelier it drives the equilibrium to the right causing more HA to make A- and maintaining [H3O+]
Adding H3O+ to the buffer: By Le Châtelier it drives the equilibrium to the left (because we are adding more product) causing A- to react with the H3O+ to make more HA maintaining [H3O+]
Effectively 𝐴 𝑎𝑞 − + 𝐻 3 𝑂 (𝑎𝑞) + ⇢ 𝐻𝐴 (𝑎𝑞) + 𝐻 2 𝑂 (𝑙)

7. How Buffers Work
H2O
new HA HA HA A− A−
H3O+ +
Added
H3O+

8. How Buffers Work
H2O
new A− A− HA A− HA
H3O+ +
Added
HO−

9. 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

10. Common Ion Effect

11. What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5
HC2H3O2 + H2O C2H3O2- + H3O+
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
[HA]
[A-]
[H3O+]
initial
0.100
≈ 0
change
equilibrium

12. What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5
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 x x x

13. What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2 given 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 x x

14. the approximation is valid
What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2 given 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

15. What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5
substitute x into the equilibrium concentration definitions and solve
[HA]
[A-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium
1.8E-5 x x x
x = 1.8 x 10-5

16. What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5
substitute [H3O+] into the formula for pH and solve
[HA]
[A-]
[H3O+]
initial
0.100
≈ 0
change -x +x
equilibrium
1.8E-5

17. What is the pH of a buffer that is 0. 100 M HC2H3O2 and 0
What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2 given 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

18. What is the pH of a buffer that is 0. 14 M HF (pKa = 3. 15) and 0
What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
Your turn!

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?
HF + H2O F- + H3O+
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
[HA]
[A-]
[H3O+]
initial
0.14
0.071
≈ 0
change
equilibrium

20. What is the pH of a buffer that is 0. 14 M HF (pKa = 3. 15) and 0
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

21. What is the pH of a buffer that is 0. 14 M HF (pKa = 3. 15) and 0
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

22. What is the pH of a buffer that is 0. 14 M HF (pKa = 3. 15) and 0
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 x
Ka for HF = 7.0 x 10-4
check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init
x = 1.4 x 10-3
the approximation is valid

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?
[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

24. What is the pH of a buffer that is 0. 14 M HF (pKa = 3. 15) and 0
What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF?
substitute [H3O+] into the formula for pH and solve
[HA]
[A-]
[H3O+]
initial
0.14
0.071
≈ 0
change -x +x
equilibrium
0.072
1.4E-3

25. What is the pH of a buffer that is 0. 14 M HF (pKa = 3. 15) and 0
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

26. 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

27. Deriving the Henderson-Hasselbalch Equation

28. What is the pH of a buffer that is 0. 050 M HC7H5O2 and 0
What is the pH of a buffer that is M HC7H5O2 and M NaC7H5O2 where Ka=6.5x10-5 for HC7H5O2?

29. What is the pH of a buffer that is 0. 050 M HC7H5O2 and 0
What is the pH of a buffer that is M HC7H5O2 and M NaC7H5O2 where Ka=6.5x10-5 for HC7H5O2?
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

30. What is the pH of a buffer that is 0. 050 M HC7H5O2 and 0
What is the pH of a buffer that is M HC7H5O2 and M NaC7H5O2 where Ka=6.5x10-5 for HC7H5O2?
HC7H5O2 + H2O C7H5O2- + H3O+
[HA]
[A-]
[H3O+]
initial
0.050
0.150
≈ 0
change -x +x
equilibrium
0.050-x
0.150-x x

31. Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation?
It is not clear that the Henderson-Hasselbalch Equation is an approximation, but the way we use it, (by placing the initial concentration of [HA]o and [A-]o into the equation assuming they are equilibrium values), makes it an approximation
For this reason 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

32. 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
𝐴 𝑎𝑞 − + 𝐻 3 𝑂 (𝑎𝑞) + ⇢ 𝐻𝐴 (𝑎𝑞) + 𝐻 2 𝑂 (𝑙)
added base reacts with the HA to make more A−
𝐻𝐴 (𝑎𝑞) + 𝑂𝐻 (𝑎𝑞) − ⇢ 𝐴 𝑎𝑞 − + 𝐻 2 𝑂 (𝑙)
an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]

33. What is the pH of a buffer that has 0. 100 mol HC2H3O2 and 0
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
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 HA A-
OH−
mols Before
0.100
mols added -
0.010
mols After

34. What is the pH of a buffer that has 0. 100 mol HC2H3O2 and 0
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

35. What is the pH of a buffer that has 0. 100 mol HC2H3O2 and 0
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+
Substitute into the Henderson-Hasselbalch Equation
Check the “x is small” approximation
pKa for HC2H3O2 = 4.745

36. Compare the effect on pH of adding 0
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?
Pure Water
Buffer
HC2H3O2 + H2O C2H3O2- + H3O+
pKa for HC2H3O2 = 4.745

37. 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)

38. What is the pH of a buffer that is 0. 50 M NH3 (pKb = 4. 75) and 0
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

39. 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
(2) the absolute concentrations of acid and base

40. Buffering Effectiveness and Buffering Range
a buffer will be most effective when the [base]:[acid] = 1
equal concentrations of acid and base
a buffer will be most effective when the [acid] and the [base] are large
The Buffer is effective when 0.1 < [base]/[acid] < 10
𝑝𝐻 𝑚𝑖𝑛 = 𝑝𝐾 𝑎 +log⁡(0.1), 𝑝𝐻 𝑚𝑎𝑥 = 𝑝𝐾 𝑎 +log⁡(10),
The buffering range is therefore
𝑝𝐻 𝑟𝑎𝑛𝑔𝑒 = 𝑝𝐾 𝑎 ±1

41. Effect of Relative Amounts of Acid and Conjugate Base
a buffer is most effective with equal concentrations of acid and base
pH change after adding 0.010M NaOH
Buffer 2
0.18 mol HA & mol A-
Initial pH = 4.05
Buffer 1
0.100 mol HA & mol A-
Initial pH = 5.00
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

42. Effect of Absolute Concentrations of Acid and Conjugate Base
a buffer is most effective when the concentrations of acid and base are largest
pH change after adding 0.010M NaOH
Buffer 1
0.50 mol HA & 0.50 mol A-
Initial pH = 5.00
Buffer 2
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

43. Example from Homework
You have two acetic acid buffers (pKa = 4.475) and to these buffers you add mole of NaOH
Buffer 1
0.50 mol HC2H3O2 & 0.50 mol C2H3O2 -
Buffer 2
0.050 mol HC2H3O2 & mol C2H3O2 –
Calculate the pH before adding the NaOH to each buffer 
Calculate the pH after the NaOH to each buffer
Calculate the percentage change in the pH of each buffer
Which is the best buffer?

44. 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

45. 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.

46. 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

47. Example from Homework You have four weak acids HA
Chlorous Acid, HClO2 pKa = 1.95
Nitrous Acid, HNO2 pKa = 3.34
Formic Acid, HCHO2 pKa = 3.74
Hypochlorous Acid, HClO pKa = 7.54
Which would be the best acid for making a buffer at pH = 7.0 ?
What ratio will you need [A-]/[HA] to be to obtain the correct pH?

48. 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]

49. Buffering Capacity
a concentrated buffer can neutralize more added acid or base than a dilute buffer

50. 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

51. 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

52. Monitoring pH During a Titration

53. 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

54. Titration Curves for Strong and Weak Acids with a Strong Base

55. 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
pKa2
pKa1

56. Titration Curve of a Weak Base with a Strong Acid

57. 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
pKa = 4.2 Methyl Orange
pKa = 9.7 Phenolphthalein

58. 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 ≈ 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

59. Titration

60. Indicators: Phenolphthalein
Used to detect pH = 8 good for titrations of weak acids with a strong base

61. Indicators: Methyl Red
Used to detect pH = 5 good for titrations of weak bases with a strong acid

62. Acid-Base Indicators

63. Titrating Weak Acid with a Strong Base: Calculating the pH using Ka
the initial pH is that of the weak acid solution
calculate like a weak acid equilibrium problem
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

64. Titrating Weak Acid with a Strong Base: Calculating the pH using Ka
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
[A−]init = mol A−/total liters
calculate like a weak base equilibrium problem
beyond equivalence point, the OH is in excess
[OH−] = mol MOH xs/total liters
[H3O+][OH−]=1 x 10-14

65. Example from homework
A 10.0 mL sample of M HNO2 is titrated with M KOH.
Calculate the pH before any base is added
Calculate the volume of base needed to reach the equivalence point
Calculate the volume of base needed to reach the half equivalence point
Calculate the pH at the half equivalence point 
Calculate the pH at the equivalence point
Calculate the pH when 20 mL of base is added

66. 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

67. 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

69. 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)

70. 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

71. 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

72. Determine the Ksp of PbBr2 if its molar solubility in water at 25°C is 1.05 x 10-2 M

73. PbBr2(s) Pb2+(aq) + 2 Br−(aq)
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)

74. Ksp = [Pb2+][Br−]2 Ksp = (1.05 x 10-2)(2.10 x 10-2)2
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)

75. 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

76. 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

77. Calculate the molar solubility of CaF2 in 0.100 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

78. Calculate the molar solubility of CaF2 in 0.100 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

79. 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) Mn+(aq) + nCO32−(aq)
H3O+(aq) + CO32− (aq) HCO3− (aq) + H2O(l)

80. 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

81. precipitation occurs if Q > Ksp
a supersaturated solution will precipitate if a seed crystal is added

82. 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

83. What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059M]) from seawater assuming Ksp=2.06x10-13)?
precipitating may just occur when Q = Ksp

84. What is the [Mg2+] when Ca2+ (with [0
What is the [Mg2+] when Ca2+ (with [0.011M]) just begins to precipitate from seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M

85. precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
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

86. 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

87. Qualitative Analysis

89. 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

90. 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)

91. 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

92. Formation Constants

93. Cu2+(aq) + 4 NH3(aq) Cu(NH3)42+(aq)
200.0 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)42+(aq)

94. Cu2+(aq) + 4 NH3(aq) Cu(NH3)42+(aq)
200.0 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)42+(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

95. Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)
200.0 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

96. 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+

98. Solubility of Amphoteric Metal Hydroxides
many metal hydroxides are insoluble eg Fe(OH)3, Al(OH)3, Co(OH)2
all metal hydroxides become more soluble in acidic solution
shifting the equilibrium to the right by removing OH−
Fe(OH)3(s) [Fe(OH)2+](aq) + OH-(aq)
H3O+(aq)+OH-(aq)  2H2O(l)
Amphoteric metal hydroxides also become more soluble in basic solution
acting as a Lewis base forming a complex ion
some cations that form amphoteric hydroxides include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+

99. 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)