Applications of Aqueous Equilibria

1
Applications of Aqueous Equilibria

• Chapter 15

2
Common Ion Effect

• The shift in equilibrium that occurs because of
  the addition of an ion already involved in the
  equilibrium reaction.
• AgCl(s) ? Ag(aq) Cl?(aq)
• The amount of Ag ion decreases.

3
Common Ion Effect Calculations

• A 1.0 M HF solution has an H of 2.7 x 10-2 M
  and a 2.7 dissociation. What is the H and
  the dissociation for a solution of 1.0 M HF and
  1.0 M NaF?
• Major Species HF, F-, Na, HOH
• HF(aq) lt---gt H(aq) F-(aq)
• Ka 7.2 x 10-4 HF-/HF

4
Common Ion Effect CalculationsContinued

• ICE
• HF F-
  H
• Initial (mol/L) 1.0 1.0 0
• Change (mol/L) - x x
  x
• Equil. (mol/L) 1.0 - x 1.0 x x
  

5
Common Ion Effect CalculationsContinued

• Ka 7.2 x 10-4 HF-/HF
• 7.2 x 10-4 x1.0 x/1.0 - x
• Ka is more than 100 x smaller than concentration,
  x is neglected in denominator and numerator.
• Ka 7.2 x 10-4 x1.0/1.0
• x 7.2 x 10-4 M

6
Common Ion Effect CalculationsContinued

• Dissociation ( H/HFo)(100)
• Dissociation (7.2 x 10-4/1.0) (100)
• Dissociation 0.072
• The 1.0 M HF solution was 2.7 dissociated while
  in the mixture of 1.0 M HF and 1.0 M NaF, the HF
  is only dissociated 0.072 .

7
A Buffered Solution

• . . . resists change in its pH when either H or
  OH? are added.
• 1.0 L of 0.50 M H3CCOOH
• 0.50 M H3CCOONa
• pH 4.74
• Adding 0.010 mol solid NaOH raises the pH of the
  solution to 4.76, a very minor change.

8
Preparation of Buffered Solutions

• Buffered solution can be made from
• 1. a weak acid and its salt (e.g. HC2H3O2
  NaC2H3O2).
• 2. a weak base and its salt (e.g. NH3 NH4Cl).
• Other examples of buffered pairs are
• H2CO3 NaHCO3 H3PO4 NaH2PO4
• NaH2PO4 Na2HPO4 Na2HPO4 Na3PO4

9
Buffer Calculations

• A buffered solution contains 0.50 M acetic acid
  and 0.50 M sodium acetate. Calculate the pH of
  this solution. See Sample Exercise 15.2 on pages
  723-724 for the long solution. The short
  solution is
• pH pKa log(A-/HA)
• pH -log (1.8 x 10-5) log (.50/.50)
• pH 4.74 0 4.74

10
Henderson-Hasselbalch Equation

• Useful for calculating pH when the
  A?/HA ratios are known.

11
Key Points on Buffered Solutions

• 1. They are weak acids or bases containing a
  common ion.
• 2. After addition of strong acid or base, deal
  with stoichiometry first, then equilibrium.

12
NaOH Added to Buffered Solution

• Calculate the change in pH that occurs when 0.010
  mol of solid NaOH is added to 1.0 L of buffered
  solution from the previous example.
• Stoichiometry Problem
• HC2H3O2(aq) OH-(aq) ---gt C2H3O2-(aq) HOH(l)
• The stoichiometry of the neutralization reaction
  must be done first, then the equilibrium
  calculation.

13
NaOH Added to Buffered SolutionContinued

• ICE (Stoichiometry)
• HC2H3O2 OH-
  C2H3O2-
• Initial (mol) 0.50 0.010 0.50
• Change (mol) - 0.010 -0.010 0.010
• End (mol) 0.49 0.00
  0.51

14
NaOH Added to Buffered SolutionContinued

• HC2H3O2(aq) lt---gt H(aq) C2H3O2-(aq)
• ICE (Equilibrium)
• HC2H3O2 H
  C2H3O2-
• Initial (mol) 0.49 0 0.51
• Change (mol) - x x x
• Equil. (mol) 0.49 - x x 0.51 x

15
NaOH Added to Buffered SolutionContinued

• pH pKa log(A-/HA)
• pH -log (1.8 x 10-5) log (.51/.49)
• pH 4.74 0.017
• pH 4.76
• Note The pH only changed 0.02 pH units. The
  addition of the same amount of NaOH to water
  would have changed the pH by 5.00 units.

16
Buffered Solution Characteristics

• Buffers contain relatively large amounts of weak
  acid and corresponding base.
• Added H reacts to completion with the weak base.
• Added OH? reacts to completion with the weak
  acid.
• The pH is determined by the ratio of the
  concentrations of the weak acid and weak base.

17
Buffered Solution Characteristics

• For a particular buffering system (acid-conjugate
  base pair), all solutions that have the same
  ratio A-/HA will have the same pH.
• Henderson-Hasselbach Equation.

18
Buffering Capacity

• . . . represents the amount of H or OH? the
  buffer can absorb without a significant change in
  pH.

19
Buffering Capacity

• The pH of a buffered solution is determined by
  the ratio A-/HA
• The buffering capacity of a buffered solution is
  determined by the magnitudes of HA and A-.
• When the ratio A-/HA equals 1, then the
  system is said to have optimal buffering. pH pKa

20
Buffering Capacity

• The pKa of a weak acid in the buffer should be as
  close as possible to the desired pH.
• A chemist needs a solution buffered at pH 4.30.
  Which of the following acids and their sodium
  salts would be best?
• chloroacetic acid Ka 1.35 x 10-3
• propanoic acid Ka 1.4 x 10-5
• benzoic acid Ka 6.4 X 10-5
• hypochlorous acid Ka 3.5 x 10-8

21
Titration (pH) Curve

• A plot of pH of the solution being analyzed as a
  function of the amount of titrant added.
• Equivalence (stoichiometric) point Enough
  titrant has been added to react exactly with the
  solution being analyzed.

22
Titration curve for a strong base added to a
strong acid -- the equivalence point has a pH of
7.
23
Titration curve for the addition of strong acid
to a strong base -- pH at equivalence point is
7.00.
24
Strong Acid - Strong Base Titration

• Before equivalence point, H is determined by
  dividing number of millimoles of H remaining by
  total volume of solution in mL.
• At equivalence point, pH is 7.00.
• After equivalence point OH- is calculated by
  dividing number of millimoles of excess OH- by
  total volume of solution in mL.
• See Example on pages 737-740.

25
Weak Acid - Strong Base Titration

• Know understand example on pages 741-745.
• Step 1 - A stoichiometry problem - reaction is
  assumed to run to completion - then determine
  remaining species.
• Step 2 - An equilibrium problem - determine
  position of weak acid equilibrium and calculate
  pH.
• At halfway to the equivalence point the
  concentration of A- HA are equal H Ka
  pH pKa.

26
Titration curve for the addition of a strong base
to a weak acid-- pH is above 7.00.
27
The equivalence point is defined by the
stoichiometry, not the pH.
28
The pH curves for the titrations of 50.0 mL
samples of 0.10 M acids with various Ka values
with 0.10 M NaOH.
29
Titration curve for the addition of a strong acid
to a weak base -- the pH at equivalence is below
7.00.
30
Determining the End Point in a Titration

• Two methods are used
• pH meter
• acid-base indicator

31
Acid-Base Indicator

• . . . marks the end point of a titration by
  changing color. The color change will be sharp,
  occurring with the addition of a single drop of
  titrant.
• The equivalence point is not necessarily the same
  as the end point.
• Indicators give a visible color change will occur
  at a pH where
• ? pH pKa ? 1

32
15_333
OH

HO
O
O
C
C
OH

C
O

C
O
O
O

(Colorless acid form, HIn)
(
Pink base form, In
)
The acid and base forms of the indicator
phenolphthalein.
33
The useful pH ranges of several common indicators
-- the useful range is usually pKa ? 1. Why do
some indicators have two pH ranges?
34
On the left is the pH curve for the titration of
a strong acid and a strong base. On the right
is the curve for a weak acid and a strong base.
35
Solubility

• Allows us to flavor foods -- salt sugar.
• Solubility of tooth enamel in acids.
• Allows use of toxic barium sulfate for intestinal
  x-rays.

36
Solubility Product

• See Table 15.4 on page 759 for common solubility
  products.
• Relative solubilities can be predicted by
  comparing Ksp values only for salts that produce
  the same total number of ions.
• AgI(s) Ksp 1.5 x 10-16
• CuI(s) Ksp 5.0 x 10-12
• CaSO4(s) Ksp 6.1 x 10-5
• CaSO4(s) gt CuI(s) gt AgI(s)

37
Solubility Product

• CuS(s) Ksp 8.5 x 10-45
• Ag2S(s) Ksp 1.6 x 10-49
• Bi2S3(s) Ksp 1.1 x 10-73
• Bi2S3(s) gt Ag2S(s) gt CuS(s)
• Why does this order from most to least soluble
  appear to be out of order?

38
Solubility Product

• For solids dissolving to form aqueous solutions.
• Bi2S3(s) ? 2Bi3(aq) 3S2?(aq)
• Ksp solubility product constant
• and
• Ksp Bi32S2?3
• Why is Bi2S3(s) not included in the solubilty
  product expression?

39
Solubility Product

• Solubility s concentration of Bi2S3 that
  dissolves. The Bi3 is 2s and the S2? is 3s.
• Note Ksp is constant (at a given temperature)
• s is variable (especially with a common
• ion present)
• Solubility product is an equilibrium constant
  and has only one value for a given solid at a
  given temperature. Solubility is an equilibrium
  position.

40
Solubility Product Calculations

• Cupric iodate has a measured solubility of 3.3 x
  10-3 mol/L. What is its solubility product?
• Cu(IO3)2(s) lt---gt Cu2(aq) 2 IO3-(aq)
• 3.3 x 10-3 M ---gt 3.3 x 10-3 M 6.6 x 10-3 M
• Ksp Cu2IO3-2
• Ksp 3.3 x 10-36.6 x 10-32
• Ksp 1.4 x 10-7

41
Solubility Product Calculations

• If a 0.010 M solution of sodium iodate is mixed
  with a 0.0010 M cupric nitrate, will a
  precipitate form?
• 2 NaIO3(aq) Cu(NO3)2(aq) ---gt Cu(IO3)2(s) 2
  NaNO3(aq)
• Cu(IO3)2(s) lt---gt Cu2(aq) 2 IO3-(aq)
• Qsp Cu2IO3-2
• Qsp 1.0 x 10-31.0 x 10-22
• Qsp 1.0 x 10-7 Qsp lt Ksp ? no precipitate forms.

42
Solubility Product Calculations

• Cu(IO3)2(s) lt---gt Cu2(aq) 2 IO3-(aq)
• Ksp Cu2IO3-2
• If solid cupric iodate is dissolved in HOH
  double square the iodate concentration.
• If mixing two solutions, one containing Cu2 and
  the other IO3-, then use the concentration of
  iodate and only square it.

43
Common Ion Effect

• CaF2(s) lt---gt Ca2(aq) 2F-(aq)
• What will be the effect on this equilibrium if
  solid sodium fluoride is added? Explain.
• Equilibrium will shift to the left, due to Le
  Chateliers Principle. Solubility product must
  stay constant, so the amount of Ca2 F- must
  decrease by forming solid CaF2.
• See Sample Exercise 15.15 on pages 764-765.

44
pH Solubility

• If a solid precipitate has an anion X- that is
  an effective base (HX is a weak acid), then the
  salt MX will show increased solubility in an
  acidic solution.
• Salts containing OH-, S2-, CO32-, C2O42-,
  CrO42- are all soluble in acidic solution.
• Limestone caves are made up of insoluble CaCO3,
  but dissolve in acidic rain water (H2CO3).

45
Ion Product, Qsp

• If 750.0 mL of 4.00 x 10-3 M Ce(NO3)3 is added to
  300.0 mL of 2.00 x 10-2 M KIO3, will Ce(IO3)3
  precipitate?
• Ce3 (750.0 mL)(4.00 x 10-3 mmol/mL)
• (750.0 mL 300.0 mL)
• Ce3 2.86 x 10-3 M
• IO3- (300.0 mL)(2.00 x 10-2 mmol/mL)
• (750.0 mL 300.0 mL)
• IO3- 5.71 x 10-3 M

46
Ion Product, QspContinued

• Qsp Ce30IO3-o3
• Qsp 2.86 x 10-35.72 x 10-33
• Qsp 5.32 x 10-10
• Qsp gt Ksp ? Ce(IO3)3 will precipitate.

Ksp 1.9 x 10 -10
47
Progressive Precipitation

• A Solubility Experience

48

• An experiment to show the effect of solubility on
  an equilibrium system!

49
Solutions of
AgNO3
K2CrO4
(NH4)2S
Na2SO4
NaCl
50
If AgNO3 is mixed with Na2SO4 what ions are most
abundant in the solution?
AgNO3
K2CrO4
(NH4)2S
Na2SO4
NaCl
With what ions is the solution saturated?
51
2AgNO3(aq) Na2SO4(aq) ? 2NaNO3 (aq)
Ag2SO4(s)
Molecular Equation
2Ag(aq) 2NO3-(aq) 2Na(aq) SO42-(aq) ?
2Na(aq) 2NO3- (aq) Ag2SO4(s)
Overall Ionic Equation
2Ag(aq) SO42-(aq) ? Ag2SO4(s)
Net Ionic Equation
First Precipitation
Silver Sulfate Precipitate
Ksp 1.2 ? 10-5
52
Ag2SO4(s) ? 2Ag(aq) SO42-(aq)
Ksp Ag2 SO42- 1.2 ? 10-5
Ksp 2x2 ? x 1.2 ? 10-5
Molar Solubility 1.4 ? 10-2 mol/liter
Silver Sulfate Precipitate
53
What ions will be most abundant in the solution
when these are mixed?
Potassium Chromate solution
Silver Sulfate Precipitate
With what ions will the solution be saturated?
54
Second Reaction
Silver Sulfate Precipitate
Potassium Chromate solution
55
Ag2SO4(s) K2CrO4(aq) ? K2SO4(aq)
Ag2CrO4(s)
Molecular Equation
Ag2SO4(s) 2K(aq) CrO42-(aq) ? 2K(aq)
SO42-(aq) Ag2CrO4(s)
Overall Ionic Equation
Ag2SO4(s) CrO42-(aq) ? Ag2CrO4(s) SO4
2-(aq)
Net Ionic Equation
Silver Chromate precipitate
Ksp 9.0 ? 10-12
56
Ag2CrO4(s) ? 2Ag(aq) CrO42-(aq)
Ksp Ag2 CrO42- 9.0 ? 10-12
Ksp 2x2 ? x 9.0 ? 10-12
Molar Solubility 1.3 ? 10-4 mol/liter
Silver Chromate precipitate
Ksp 9.0 ? 10-12
57
What ions will be most abundant in the solution
when NaCl solution is added to Ag2CrO4
precipitate?
With what ions will the solution be saturated?
58
Third Reaction
59
Ag2CrO4(s) 2NaCl(aq) ? Na2CrO4(aq)
2AgCl(s)
Molecular Equation
Ag2CrO4(s) 2Na(aq) 2Cl-(aq) ? 2Na(aq)
CrO42-(aq) 2AgCl(s)
Overall Ionic Equation
Ag2CrO4(s) 2Cl-(aq) ? 2AgCl(s) CrO4 2-
Net Ionic Equation
Silver Chloride precipitate
Ksp 9.0 ? 10-12
60
AgCl(s) ? Ag(aq) Cl-(aq)
Ksp Ag Cl- 1.6 ? 10-10
Ksp x ? x 1.6 ? 10-10
Molar Solubility 1.3 ? 10-5 mol/liter
Silver Chloride precipitate
Ksp 1.6 ? 10-10
61
What ions will be most abundant in the solution
when (NH4)2S solution is added to AgCl
precipitate?
With what ions will the solution be saturated?
62
Fourth Reaction
63
2AgCl(s) (NH4)2S(aq) ? 2NH4Cl(aq) Ag2S(s)
Molecular Equation
2AgCl(s) 2NH4 (aq) S2-(aq) ? 2NH4 (aq)
2Cl-(aq) Ag2S(s)
Overall Ionic Equation
2AgCl(s) S2-(aq) ? Ag2S(s) 2Cl-(aq)
Net Ionic Equation
Silver Chloride precipitate
Ksp 9.0 ? 10-12
64
Ag2S(s) ? 2Ag(aq) S2-(aq)
Ksp Ag2 S2- 1.6 ? 10-49
Ksp 2x2 ? x 1.6 ? 10-49
Molar Solubility 3.4 ? 10-17 mol/liter
Silver Chloride precipitate
Ksp 1.6 ? 10-49
65
Of the four insoluble compounds, which one is
the most insoluble?
Molar Solubility of Ag2SO4 1.4 ? 10-2 mol/liter
Molar Solubility of Ag2CrO4 1.3 ? 10-4
mol/liter
Molar Solubility of AgCl 1.3 ? 10-5 mol/liter
Molar Solubility of Ag2S 3.4 ? 10-17 mol/liter
Molar Solubility of Ag2S 3.4 ? 10-17 mol/liter
66
Of the four insoluble compounds, which one is
the least insoluble?
Molar Solubility of Ag2SO4 1.4 ? 10-2 mol/liter
Molar Solubility of Ag2SO4 1.4 ? 10-2 mol/liter
Molar Solubility of Ag2CrO4 1.3 ? 10-4
mol/liter
Molar Solubility of AgCl 1.3 ? 10-5 mol/liter
Molar Solubility of Ag2S 3.4 ? 10-17 mol/liter
67
Qualitative Analysis

• The separation of ions by selective
  precipitation.
• Much descriptive chemistry can be learned from
  qualitative analysis.
• Qualitative analysis can be done for both cations
  and anions.

68
Qualitative Analysis

• Group I -- Insoluble chlorides -- Ag, Pb2,
  Hg22
• Group II -- Sulfides insoluble in acid solution
  -- Hg2, Cd2, Bi3, Cu2, Sn4
• Group III -- Sulfides insoluble in basic solution
  -- Co2, Zn2, Mn2, Ni2, Fe3
• Group IV -- Insoluble carbonates --Ba2, Ca2,
  Mg2
• Group V -- alkali metal and ammonium ions --
  soluble so must be identified by flame tests,
  etc.

69
Equilibria Involving Complex Ions

• Complex Ion A charged species consisting of a
  metal ion surrounded by ligands (Lewis bases).
• Coordination Number Number of ligands attached
  to a metal ion. (Most common are 6 and 4.)
• Formation (Stability) Constants The equilibrium
  constants characterizing the stepwise addition of
  ligands to metal ions.
• See Example 15.19 on pages 775-776.

70
Complex Ions Solubility

• Many insoluble solids can be dissolved by
  complexing one of the ions to make it soluble.
• AgCl is very insoluble but easily goes into
  solution in the presence of concentrated NH3 by
  forming the complex ion Ag(NH3)2.
• AgCl(s) lt---gt Ag(aq) Cl-(aq)
  Ksp 1.6 x 10-10
• Ag(aq) NH3(aq) lt---gt Ag(NH3)(aq)
  K1 2.1 x 103
• Ag(NH3)(aq) NH3(aq) lt---gt Ag(NH3)2(aq) K2
  8.2 x 103
• Ag(NH3)2 forms because its K value is greater
  than AgCl.

71
Aqua Regia

• Aqua regia is a mixture of concentrated HCl and
  concentrated HNO3. Either one of these acids
  alone will not affect Au.
• The mixture will dissolve gold--one of the most
  inactive metals.
• What does aqua regia mean?
• Royal Water