Applications of Aqueous
Equilibria
Chapter 15
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)
adding


NaCl( aq ) shifts equilibrium position
The amount of Ag+ ion decreases.
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) <---> H+(aq) + F-(aq)
Ka = 7.2 x 10-4 = [H+][F-]/[HF]
Common Ion Effect Calculations
Continued
ICE
[HF]
Initial (mol/L)
1.0
Change (mol/L)
-x
Equil. (mol/L)
1.0 - x
[F-]
1.0
+x
1.0 + x
[H+]
0
+x
x
Common Ion Effect Calculations
Continued
Ka = 7.2 x 10-4 = [H+][F-]/[HF]
7.2 x 10-4 = [x][1.0 + 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 = [x][1.0]/[1.0]
x = 7.2 x 10-4 M
Common Ion Effect Calculations
Continued
% Dissociation =( [H+]/[HF]o)(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 %.
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.
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
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
Henderson-Hasselbalch
Equation
-
Useful for calculating pH when the
[A]/[HA] ratios are known.
pH  pKa  log( A  / HA ) 
pKa  log( base / acid )
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.
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) ---> C2H3O2-(aq) + HOH(l)
The stoichiometry of the neutralization reaction
must be done first, then the equilibrium
calculation.
NaOH Added to Buffered Solution
Continued
ICE (Stoichiometry)
Initial (mol)
Change (mol)
End (mol)
[HC2H3O2]
[OH-]
[C2H3O2-]
0.50
0.010
0.50
- 0.010
-0.010
+0.010
0.49
0.00
0.51
NaOH Added to Buffered Solution
Continued
HC2H3O2(aq) <---> H+(aq) + C2H3O2-(aq)
ICE (Equilibrium)
[HC2H3O2]
[H+]
[C2H3O2-]
Initial (mol)
0.49
0
0.51
Change (mol)
-x
+x
+x
0.49 - x
x
Equil. (mol)
0.51 + x
NaOH Added to Buffered Solution
Continued
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.
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.
Buffered Solution Characteristics
-
For a particular buffering system (acidconjugate base pair), all solutions that have
the same ratio [A-]/[HA] will have the same
pH.
Henderson-Hasselbach Equation.
pH  pKa  log( A

/ HA ) 
pKa  log( base / acid )
Buffering Capacity
. . . represents the amount of H+
or OH the buffer can absorb
without a significant change
in pH.
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
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
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.
15_327
pH
13.0
Equivalence
point
7.0
1.0
0
50.0
100.0
Vol NaOH added (mL)
Titration curve for a strong base added to a strong acid
-- the equivalence point has a pH of 7.
15_328
14.0
pH
Equivalence
point
7.0
50.0 mL
Vol 1.0 M HCl added
Titration curve for the addition of strong acid to a strong
base -- pH at equivalence point is 7.00.
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.
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.
15_329
12.0
Equivalence
point
pH
9.0
3.0
25
50
Vol NaOH added (mL)
Titration curve for the addition of a strong base to a weak
acid-- pH is above 7.00.
15_330
pH
Weak acid
Strong acid
Vol NaOH
The equivalence point is defined by the stoichiometry, not
the pH.
15_331
12.0
Ka = 10– 10
10.0
Ka = 10– 8
8.0
pH
Ka = 10– 6
6.0
Ka = 10– 4
4.0
Ka = 10– 2
2.0
Strong acid
0
10
20
30
40
50
60
Vol 0.10 M NaOH added (mL)
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.
15_332
12
pH
10
Equivalence
point
8
6
4
2
0
10
20
30 40 50 60 70
Vol 0.10 M HCl (mL)
Titration curve for the addition of a strong acid to a weak
base -- the pH at equivalence is below 7.00.
Determining the End Point in a
Titration
Two methods are used:
pH meter
acid-base indicator
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:
[In - ] 1  pH = pK  1

a
[HIn] 10
15_333
OH
HO
C
C
–
O
O
OH
C
O–
C
O
(Colorless acid form, HIn)
O–
O
(Pink base form, In– )
The acid and base forms of the indicator phenolphthalein.
pH
15_334
0
1
2
3
4
5
6
7
8
9
10
11
12
13
Crystal Violet
Cresol Red
Thymol Blue
Erythrosin B
2,4-Dinitrophenol
Bromphenol Blue
Methyl Orange
Bromcresol Green
Methyl Red
Eriochrome* Black T
Bromcresol Purple
Alizarin
Bromthymol Blue
Phenol Red
m - Nitrophenol
o-Cresolphthalein
Phenolphthalein
Thymolphthalein
Alizarin Yellow R
* Trademark
CIBA GEIGY CORP.
The pH ranges shown are approximate. Specific transition ranges depend on the indicator solvent chosen.
The useful pH ranges of several common indicators -- the
useful range is usually pKa  1. Why do some indicators
have two pH ranges?
15_335AB
14
14
12
12
10
10
Phenolphthalein
pH
Equivalence
point
6
Methyl red
6
4
4
2
2
0
Equivalence
point
8
pH
8
Phenolphthalein
0
20 40
60 80 100 120
Vol 0.10 M NaOH added (mL)
0
Methyl red
0
20 40 60
80 100 120
Vol 0.10 M NaOH added (mL)
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.
Solubility
•
Allows us to flavor foods -- salt & sugar.
•
Solubility of tooth enamel in acids.
•
Allows use of toxic barium sulfate for
intestinal x-rays.
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) > CuI(s) > AgI(s)
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) > Ag2S(s) > CuS(s)
Why does this order from most to least
soluble appear to be out of order?
Solubility Product
For solids dissolving to form aqueous solutions.
Bi2S3(s)  2Bi3+(aq) + 3S2(aq)
Ksp = solubility product constant
and
Ksp = [Bi3+]2[S2]3
Why is Bi2S3(s) not included in the solubilty
product expression?
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.
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) <---> Cu2+(aq) + 2 IO3-(aq)
3.3 x 10-3 M ---> 3.3 x 10-3 M + 6.6 x 10-3 M
Ksp = [Cu2+][IO3-]2
Ksp = [3.3 x 10-3][6.6 x 10-3]2
Ksp = 1.4 x 10-7
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) ---> Cu(IO3)2(s) + 2 NaNO3(aq)
Cu(IO3)2(s) <---> Cu2+(aq) + 2 IO3-(aq)
Qsp = [Cu2+][IO3-]2
Qsp = [1.0 x 10-3][1.0 x 10-2]2
Qsp = 1.0 x 10-7
Qsp < Ksp  no precipitate forms.
Solubility Product Calculations
Cu(IO3)2(s) <---> Cu2+(aq) + 2 IO3-(aq)
Ksp = [Cu2+][IO3-]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.
Common Ion Effect
CaF2(s) <---> 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
Chatelier’s 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.
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).
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
Ion Product, Qsp
Continued
Qsp = [Ce3+]0[IO3-]o3
Qsp = [2.86 x 10-3][5.72 x 10-3]3
Qsp = 5.32 x 10-10
Ksp = 1.9 x 10 -10
Qsp > Ksp  Ce(IO3)3 will precipitate.
Progressive
Precipitation
A Solubility Experience
An experiment to show the effect of
solubility on an equilibrium system!
Solutions of:
AgNO3
Na2SO4
K2CrO4
(NH4)2S
NaCl
If AgNO3 is mixed with Na2SO4 what ions are most
abundant in the solution?
AgNO3
Na2SO4
K2CrO4
(NH4)2S
NaCl
With what ions is the solution saturated?
2AgNO3(aq) + Molecular
Na2SO4(aq)
 2NaNO3 (aq) +
Equation
Ag2SO4(s)
+
2Ag+(aq) + 2NO3-(aq) +Overall
2Na+(aq) Ionic
+ SO42-Equation
(aq)  2Na (aq) + 2NO3 (aq) + Ag2SO4(s)
+
2AgNet
SO42-(aq)
 Ag2SO4(s)
Equation
(aq) +Ionic
First Precipitation
Silver Sulfate
Precipitate
Ksp = 1.2  10-5
Ag2SO4(s)  2Ag+(aq) + SO42-(aq)
Ksp = [Ag+]2 [SO42-] = 1.2  10-5
Ksp = [2x]2  [x] = 1.2  10-5
Molar Solubility = 1.4  10-2 mol/liter
Silver Sulfate
Precipitate
What ions will be most abundant in the solution
when these are mixed?
Silver
Sulfate
Precipitate
Potassium
Chromate
solution
With what ions will the solution be saturated?
Second Reaction
Silver
Potassium
Sulfate
Chromate
Precipitate solution
Ag2SO4(s) + K2CrO
K2SO4(aq) + Ag2CrO4(s)
Molecular
4(aq) Equation
2Ag2SO4(s) + 2K+Overall
2K+(aq) + SO42-(aq) + Ag2CrO4(s)
Equation
(aq) + CrO4Ionic
(aq) 
2Ag2SO4(s) + Net
CrO4Ionic
Ag2CrO4(s) + SO4 2-(aq)
Equation
(aq) 
Silver
Chromate
precipitate
Ksp = 9.0  10-12
Ag2CrO4(s)  2Ag+(aq) + CrO42-(aq)
Ksp = [Ag+]2 [CrO42-] = 9.0  10-12
Ksp = [2x]2  [x] = 9.0  10-12
Molar Solubility = 1.3  10-4 mol/liter
Silver
Chromate
precipitate
Ksp = 9.0  10-12
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?
Third Reaction
Ag2CrO4(s) + 2NaCl
Na2CrO4(aq) + 2AgCl(s)
Molecular
(aq) Equation
+
Ag2CrO4(s) + 2Na
2Na+(aq) + CrO42-(aq) + 2AgCl(s)
Overall
Ionic
Equation
(aq) + 2Cl
(aq) 
Ag2CrO4(s)
+ 2Cl
 2AgCl(s) + CrO4 2Net
Ionic
(aq) Equation
Silver
Chloride
precipitate
Ksp = 9.0  10-12
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
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?
Fourth Reaction
2AgCl(s) + (NHMolecular
2NH4Cl(aq) + Ag2S(s)
Equation
4)2S(aq) 
+
22AgCl(s) + 2NH
2NH4+ (aq) + 2Cl-(aq) + Ag2S(s)
Overall
Equation
4 (aq) + SIonic
(aq) 
22AgClNet
Ag2S(s) + 2Cl-(aq)
Equation
(s) + SIonic
(aq) 
Silver
Chloride
precipitate
Ksp = 9.0  10-12
Ag2S(s)  2Ag+(aq) + S2-(aq)
Ksp = [Ag+]2 [S2-] = 1.6  10-49
Ksp = [2x]2  [x] = 1.6  10-49
Molar Solubility = 3.4  10-17 mol/liter
Silver
Chloride
precipitate
Ksp = 1.6  10-49
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
Of the four “insoluble” compounds, which one is the
least 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
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.
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.
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.
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) <---> Ag+(aq) + Cl-(aq)
Ksp = 1.6 x 10-10
Ag+(aq) + NH3(aq) <---> Ag(NH3)+(aq)
K1 = 2.1 x 103
Ag(NH3)+(aq) + NH3(aq) <---> Ag(NH3)2+(aq) K2 = 8.2 x 103
Ag(NH3)2+ forms because its K value is greater than AgCl.
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