1. Chapter 17
Applying equilibrium

2. 17.1 The Common Ion Effect
When the salt with the anion of a weak acid is added to that acid, it reverses the dissociation of the acid.
Lowers the percent dissociation of the acid.
The same principle applies to salts with the cation of a weak base..
The calculations are the same as the last two chapters.

3. 4 Logical Steps As we approach these problems:
1. Identify the major species in solution, and consider their acidity or basicity.
2. Identify the important equilibrium that is the source of H+ and therefore determines pH.
3. Tabulate the concentrations at equilibrium (ICE chart)
4. Use the equilibrium-constant expression to calculate [H+] and then pH.

4. Example
What is the pH of a solution made by adding 0.30 mole of acetic acid and 0.30 mole of sodium acetate to enough water to make 1.0 Liter of solution? Appendix D gives Ka = 1.8 E -5.

5. Try this one…
Calculate the pH of a solution containing M nitrous acid (Ka = 4.5 E -4) and 0.10 M potassium nitrite.

6. How about these…
Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.
Calculate the formate ion concentration and pH of a solution that is 0.50 M in formic acid (HCOOH, Ka = 1.8 E -4) and 0.10 M nitric acid.

7. 17.2 Buffered solutions A solution that resists a change in pH.
Either a weak acid and its salt or a weak base and its salt.
We can make a buffer of any pH by varying the concentrations of these solutions.
Which of the following conjugate acid-base pairs will not function as a buffer?
HCOOH and HCOO-1?
HCO3-1 and CO3 -2?
HNO3 and NO3-1?

8. How buffers work
Let’s consider a buffer composed of a weak acid (HX) and one of its salts (MX).
Write the equation for the dissociation of this weak acid:
Write the corresponding acid-dissociation constant expression:
Rearrange, solving for [H+]:

9. What do we have? [H+], and thus pH, is determined by two factors:
The value of Ka of the weak acid
The ratio of the concentrations of the conjugate acid-base pair: [HX]/[X-]
As long as [HX]/[X-] in the buffer are large compared to the amount of H+ or OH- added, the ratio doesn’t change much, and thus the change in pH is small.

10. Give it some thought…
What happens when NaOH is added to a buffer composed of HC2H3O2 and C2H3O2-1?
What happens when HCl is added to this buffer?

11. Calculating the pH of a Buffer
Ka = [H+] [A-] [HA]
so [H+] = Ka [HA] [A-]
The [H+] depends on the ratio [HA]/[A-]
taking the negative log of both sides
pH = -log(Ka [HA]/[A-])
pH = -log(Ka)-log([HA]/[A-])
pH = pKa + log([A-]/[HA])

12. This is called the Henderson-Hasselbach equation
pH = pKa + log([A-]/[HA])
pH = pKa + log(base/acid)

13. Let’s try one
What is the pH of a buffer that is 0.12 M in lactic acid (HC3H5O3) and 0.10 M in sodium lactate? For lactic acid, Ka = 1.4 E -4.
We can use the “ICE CHART” or the Henderson-Hasselbalch equation.

14. Preparing a Buffer
How many moles of ammonium chloride must be added to 2.0 Liters of a 0.10 M ammonia solution to form a buffer whose pH is 9.00? Assume that the addition of ammonium chloride does not change the volume of the solution.

15. Another one
Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (HC7H5O2) to produce a pH of 4.00.

16. Buffer Capacity
The pH of a buffered solution is determined by the ratio [A-]/[HA].
As long as this doesn’t change much the pH won’t change much.
The more concentrated these two are the more H+ and OH- the solution will be able to absorb.
Larger concentrations bigger buffer capacity.

17. pH Range The pH range over which the buffer acts effectively.
Resist a change in pH in either direction when [HX] = [X-].
Optimal pH of a buffer is when pH = pKa.
Try to select a buffer whose acid form has a pKa close the desired pH.
This will give us a usable pH range of + or – one pH unit of pKa.

18. Give this some thought
What is the optimal pH buffered by a solution containing acetic acid and sodium acetate? Ka for acetic acid is 1.8 E -5.

19. Adding a strong acid or base
Do the stoichiometry first.
A strong base will grab protons from the weak acid reducing [HA]0
A strong acid will add its proton to the anion of the salt reducing [A-]0
Then do the equilibrium problem, most easily done with the Henderson-Hasselbalch equation.

20. Calculating pH Changes in Buffers
A buffer is made by adding 0.30 mole acetic acid and 0.30 mole sodium acetate to enough water to make 1.0 L solution. The pH of the buffer is 4.74.
Calculate the pH of this solution after 0.20 mol of NaOH is added
For comparison, calculate the pH that would result if 0.20 mole of NaOH was added to 1.0 Liter of pure water.

21. Practice This Determine
The pH of the original buffer described in the problem above after the addition of 0.02 mol HCl
To compare, calculate the pH of the solution what would result from the addition of 0.02 mol HCl to 1.0 L pure water.

22. 17.3 Titrations Millimole (mmol) = 1/1000 mol
Molarity = mmol/mL = mol/L
Makes calculations easier because we will rarely add Liters of solution.
Adding a solution of known concentration until the substance being tested is consumed.
This is called the equivalence point.
Graph of pH vs. mL is a titration curve.

23. Titration Curves
Can be used to determine the equivalence point in the titration
Can also be used to determine the Ka of a weak acid or the Kb of the weak base being titrated.
3 Kinds of Titrations:
Strong acid - strong base
Weak acid - strong base
Polyprotic acid – strong base.

24. Strong acid with Strong Base
Do the stoichiometry.
There is no equilibrium .
They both dissociate completely.
The titration of 50.0 mL of M HNO3 with M NaOH
Analyze the pH

25. Strong Acid with Strong Base
Equivalence at pH 7 7 pH
mL of Base added

26. Strong base with strong acid
Equivalence at pH 7 7 pH
mL of Base added

27. 4 Regions of Curve
1. The initial pH – determined by the initial concentration of the strong acid.
Example: 0.1 M HCl has pH =1
2. Between the initial pH and the equivalence point: as NaOH added, pH rises slowly at first, then rapidly when it nears the equivalence point. In this case, the pH is determined by how much HCl has not yet been neutralized.

28. 4 Regions of Curve
3. The equivalence point: An equal number of moles of NaOH and HCl have reacted, leaving only a solution of their salt, NaCl. pH = 7 because the cation of a strong base and the anion of a strong acid do not hydrolyze and therefore do not affect pH.
4. After the equivalence point: pH of the solution is determined by the concentration of the excess NaOH in the solution.

29. Sample Problem
Calculate the pH when the following quantities of 0.10 M NaOH solution have been added to 50 mL of 0.10 M HCl solution:
49.0 mL
51.0 mL

30. Weak acid with Strong base
Four Areas of the Graph:
Initial pH – found by calculations involving Ka (Section 16.6)
Between the initial pH and the equivalence point:
Consider the neutralization of the acid by OH- to determine [HX] and [X-].
Calculate the pH of this buffer pair (Section 17.2)

31. 3. The equivalence point: The X- ion is a weak base and therefore affects the pH at the equivalence point. >7
4. After the equivalence point – pH is determined by the concentration of the excess NaOH. The X- ion is negligible compared to OH- in this titration.

32. Weak acid with strong Base
Equivalence at pH >7
>7 pH
mL of Base added

33. Weak base with strong acid
Equivalence at pH <7
<7 pH
mL of Base added

34. Summary Strong acid and base just stoichiometry.
Determine Ka, use for 0 mL base
Weak acid before equivalence point
Stoichiometry first
Then Henderson-Hasselbach
Weak acid at equivalence point Kb
Weak base after equivalence - leftover strong base.

35. Summary Determine Ka, use for 0 mL acid.
Weak base before equivalence point.
Stoichiometry first
Then Henderson-Hasselbach
Weak base at equivalence point Ka.
Weak base after equivalence - leftover strong acid.

36. Sample Problems
Calculate the pH of the solution formed when 45.0 mL of 0.10 M NaOH is added to 50 mL of 0.10 M HC2H3O2.
Ka = 1.8 E -5.
Calulate the pH in the solution formed by adding 10.0 mL of 0.05 M NaOH to 40 mL of M benzoic acid. Ka = 6.3 E -5.
Calculate the pH in the solution formed by adding 10 mL of 0.10 M HCl to 20 mL of
0.1 M NH3.

37. Try these:
Calculate the pH at the equivalence point in the titration of 50 mL of 0.1 M acetic acid with 0.1 M sodium hydroxide.
Calculate the pH at the equivalence point when
40 mL of M benzoic acid is titrated with 0.05 M NaOH
40 mL of 0.10 M ammonia is titrated with 0.10 M HCl

38. Comparison of Curves
The pH titration curves for weak acid-strong base titrations differ from those for strong acid-strong base titrations in 3 ways:
1. The solution of the weak acid has a higher initial pH than a solution of a strong acid of the same concentration.
2. The pH change at the rapid-rise portion of the curve near the equivalence point is smaller for the weak acid than it is for the strong acid.
3. The pH at the equivalence point is above 7.0 for the weak acid-strong base titration.

39. Effect of Ka on Curve
The weaker the acid, the smaller the Ka value, the higher the initial pH and the smaller the pH change at the equivalence point.

40. Use of Indicators
Optimally, an indicator would change color at the equivalence point.
Not really necessary in practice.
pH changes by several units quickly near the equivalence point by a single drop of titrant– the indicator needs to change color where the rise is steep.
Called its “END POINT”

41. Indicators Weak acids that change color when they become bases.
weak acid written HIn
Weak base
HIn H+ + In clear red
Equilibrium is controlled by pH
End point - when the indicator changes color.

42. Indicators Since it is an equilibrium the color change is gradual.
It is noticeable when the ratio of [In-]/[HI] or [HI]/[In-] is 1/10
Since the Indicator is a weak acid, it has a Ka.
pH the indicator changes at is.
pH=pKa +log([In-]/[HI]) = pKa +log(1/10)
pH=pKa - 1 on the way up

43. Indicators pH=pKa + log([HI]/[In-]) = pKa + log(10)
pH=pKa+1 on the way down
Choose the indicator with a pKa 1 less than the pH at equivalence point if you are titrating with base.
Choose the indicator with a pKa 1 greater than the pH at equivalence point if you are titrating with acid.

44. Titrations of Polyprotic Acids
Neutralization occurs in 2 stages.
Exhibits a titration curve with multiple equivalence points.

45. Solubility Equilibria
Will it all dissolve, and if so, how much?

46. All dissolving is an equilibrium.
If there is not much solid it will all dissolve.
As more solid is added the solution will become saturated.
Solid dissolved
The solid will precipitate as fast as it dissolves .
Equilibrium

47. General equation M+ stands for the cation (usually metal).
Nm- stands for the anion (a nonmetal).
MaNmb(s) aM+(aq) + bNm- (aq)
K = [M+]a[Nm-]b/[MaNmb]
But the concentration of a solid doesn’t change.
Ksp = [M+]a[Nm-]b
Called the solubility product for each compound.

48. Watch out Solubility is not the same as solubility product.
Solubility product is an equilibrium constant.
it doesn’t change except with temperature.
Solubility is an equilibrium position for how much can dissolve.
A common ion can change this.

49. Calculating Ksp from Solubility
Solid silver chromate is added to pure water at 25 C. Some of the solid remains undissolved at the bottom of the flask. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved silver chromate and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 E -4M. Assuming that silver chromate completely dissociates in water and that there are no other important equilibria involving the two ions in the solution, calculate the Ksp for the compound.

50. Another Practice
A saturated solution of magnesium hydroxide in contact with undissolved solid is prepared at 25 C. The pH of the solution is found to be Assuming that magnesium hydroxide dissociates completely in water and that there are no other simultaneous equilibria involving the two ions in the solution, calculate the Ksp for this compound.

51. Calculating Solubility from Ksp
The Ksp for calcium fluoride is 3.9 E -11 at 25 C. Assuming that calcium fluoride dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of calcium fluoride in grams per liter.

52. Another Try…
The Ksp for LaF3 is 2 E What is the solubility of LaF3 in water in moles per liter?

53. 17.5 Factors that Affect Solubility
Solubility is not only affected by temperature, but also by the presence of other solutes.
There are 3 factors that affect the solubility of ionic compounds:
1. The presence of common ions
2. The pH of the solution
3. The presence of complexing agents
Amphoterism affects both pH and complexing agents.

54. Common Ion Effect
If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve.
Sample Exercise: Calculate the molar solubility of calcium fluoride at 25 C in a solution that is
0.010 M in calcium nitrate
0.010 M in sodium fluoride
Appendix D shows a Ksp for CaF2 = 3.9 E -11

55. Practice Some More
The value for Ksp for manganese II hydroxide is 1.6 E Calculate the molar solubility of manganese II hydroxide in a solution that contains M NaOH.

56. pH and solubility OH- can be a common ion. More soluble in acid.
For other anions if they come from a weak acid they are more soluble in a acidic solution than in water.
CaC2O4 Ca+2 + C2O4-2
H+ + C2O HC2O4-
Reduces C2O4-2 in acidic solution.

57. General Rule
The solubility of slightly soluble salts containing basic anions increases as [H+] increases (as pH is lowered).
Carbonates, phosphates, cyanides, sulfides, etc.
The more basic the anion, the more the solubility is influenced by pH.
Salts with anions of negligible basicity (the anions of strong acids) are unaffected by pH changes.

58. Predicting the Effect of Acid on Solubility
Which of the following substances will be more soluble in acidic solution than in basic solution?
Nickel II hydroxide
Calcium carbonate
Barium fluoride
Silver chloride

59. Practice this
Write the net ionic equation for the reaction of the following copper (II) compounds with acid:
CuS
Cu(N3)2

60. Formation of Complex Ions
Metal ions have the ability to act as Lewis acids, (electron pair acceptors), towards water molecules, which act as Lewis bases, (electron pair donors).
Transition-metal ions can interact with other molecules, such as NH3.
An assembly of a metal ion and the Lewis bases bonded to it, such as Ag(NH3)2+, is called a complex ion.

61. Complex ion Equilibria
A charged ion surrounded by ligands.
Ligands are Lewis bases using their lone pair to stabilize the charged metal ions.
Common ligands are NH3, H2O, Cl-,CN-
Coordination number is the number of attached ligands.
Cu(NH3)4+2 has a coordination # of 4

62. Kf
Formation Constant – the equilibrium constant for the formation of a complex ion.
The magnitude of this number indicates the ion’s stability.
Larger the number, the more stable the ion.

63. The addition of each ligand has its own equilibrium
Usually the ligand is in large excess.
And the individual K’s will be large so we can treat them as if they go to equilibrium.
The complex ion will be the biggest ion in solution.

64. Practice Exercise
Calculate the concentration of Ag+ present in solution at equilibrium when concentrated ammonia is added to a M solution of silver nitrate to give an equilibrium concentration of [NH3] = 0.20 M. Neglect the small volume change that occurs when ammonia is added.

65. Try this one… Calculate the [Cr+3] in equilibrium with
Cr(OH)4 -1 when mol of chromium III nitrate is dissolved in a liter of solution buffered at pH 10.0.
General Rule: The solubility of metal salts increases in the presence of suitable Lewis bases, such as ammonia, cyanide ion, and hydroxide ion.

66. Amphoteric Oxides and Bases
Some metal oxides and hydroxides that are relatively insoluble in neutral water dissolve in strongly acidic and strongly basic solutions.
Because they themselves act as either an acid or base.
Include those of Al3+, Cr 3+, Ag2+, and Sn2+

67. 17.6 Precipitation and Separation of Ions
Use the reaction quotient, Q, to determine the direction in which a reaction must proceed to reach equilibrium.

68. Precipitation Ion Product, Q =[M+]a[Nm-]b
If Q>Ksp a precipitate forms.
If Q<Ksp No precipitate.
If Q = Ksp equilibrium.

69. Predicting Whether a Precipitate forms
Will a precipitate form when 0.10 L of 8 E -3 M lead II nitrate is added to 0.40 L 5 E -3 M sodium sulfate?
Will a precipitate form when L of 2.0 E -2 M NaF is mixed with L of 1.0 E -2 M calcium nitrate?

70. Selective Precipitation of Ions
Used to separate mixtures of metal ions in solutions.
Add anions that will only precipitate certain metals at a time.
Used to purify mixtures.

71. Sample Problem
A solution contains 1.0 E -2 M Ag+ ion and 2.0 E -2 Pb+2 ion. When Cl-1 is added to the solution, both AgCl (Ksp = 1.8 E -10) and PbCl2 (Ksp = 1.7 E -5) precipitate from the solution. What concentration of Cl-1 is necessary to begin the precipitation of each salt? Which salt precipitates first?

72. Try This One…
A solution consists of M Mg+2 and M Cu+2. Which ion will precipitate first as OH- is added to the solution? What concentration of OH- is necessary to begin the precipitation of each cation? [Ksp = 1.8 E -11 for magnesium hydroxide, and Ksp = 2.2 E -20 for copper II hydroxide]

73. Selective Precipitation
Often use H2S because in acidic solution Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will precipitate.
In Basic adding OH-solution S-2 will increase so more soluble sulfides will precipitate.
Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3, Al(OH)3

74. Selective precipitation
Follow the steps first with insoluble chlorides (Ag, Pb, Ba)
Then sulfides in Acid.
Then sulfides in base.
Then insoluble carbonate (Ca, Ba, Mg)
Alkali metals and NH4+ remain in solution.

75. Putting it all together
A sample of 1.25 L of HCl gas at 21 C and atm is bubbled through L of 0.15 M ammonia solution. Assuming that all the HCl dissolves and that the volume of the solution remains L, calculate the pH of the resulting solution.