Transcript Aqueous Equilibria Applications

Chemistry 100 - Chapter 17
Applications of Aqueous Equilibria
The Common Ion Effect
Add sodium formate (HCOONa) to
a solution of formic acid (HCOOH)
that has already established
equilibrium?
HCOOH (aq) ⇄ H+ (aq) + HCOO- (aq)

weak acid
HCOONa (aq)  HCOO- (aq) + Na+
(aq)
strong electrolyte
Another Example of the Common Ion
Effect
What would happen if we added
HCOOH to a solution of a strong
acid?
HCl (aq)  H+ (aq) + Cl- (aq)
strong acid
HCOOH (aq) ⇄ H+ (aq) + HCOO- (aq)
 The ionization of the weak acid
would be supressed in the presence
of the strong acid!
 By Le Chatelier’s Principle, the 
value of the weak acid is decreased!


How would we calculate the pH of
these solutions?

Ka 
 log K a
[H ][ HCOO

]
HCOOH 
 
 H  HCOO 
  log 
 HCOOH 
 


Define pKa = -log (Ka )
pK

a
  log[ H ]  log[ HCOO

]  log[ HCOOH ]
note pH = -log [H+]
The Buffer Equation
• Substituting and rearranging
pK
a
 [HCOO  ] 
 pH  log 

 [HCOOH ] 
pH  pK
a
 [HCOO  ] 
 log 

 [HCOOH ] 
The Generalized Buffer Equation
pH  pK
a
 [ conj . base ] 
 log 

 [ weak acid ] 
The solution pH is determined by the
ratio of the weak acid to the conjugate
base at equilibrium.
Henderson-Hasselbalch equation
The Definition of a Buffer



Buffer  a reasonably concentrated
solution of a weak acid and its
conjugate base.
Buffer solutions resists pH changes
when additional strong acid or
strong base are added to the
solutions.
Note: The Henderson-Hasselbalch
equation is really only valid for pH
ranges near the pKa of the weak
acid!
How Do We Use the Buffer (the H-H)
Equation?


The pH of the buffer is determined
by the concentration ratio of weak
acid to conjugate base at
equilibrium.
How different are the equilibrium
concentrations of weak
acid/conjugate base from the initial
concentrations?

Buffer  CH3COONa (aq) and CH3COOH
(aq))
CH3COOH (aq) ⇄ CH3COO- (aq) + H+ (aq)
The Equilibrium Data Table
+
-
[CH3COOH] [H ] [CH3COO ]
Start
A
0
B
Change
-x
+x
+x
m
(A-x)
(x)
(B+x)

According to the Henderson-Hasselbalch Equation, the
pH of the solution is calculated as follows
pH   log x   pK a

 [B  x ] 
 log 

[A  x ] 
What if the original concentrations of acid and base
([A] and [B], respectively) are much larger than x (i.e.,
the  value of the weak acid is very small)?

The pH of the solution will be almost
entirely due to the original
concentrations of acid and base!!
pH  pK
a
 [B ] 
 log 

[A ] 
The pH of the solution changes very little after
adding strong acid or base (it is buffered)
Examples of Buffer Calculations


How do we calculate the pH of a
buffer solution?
How would we prepare a buffer solution
of a specified pH?
The pH of a Buffer Solution

Major task


obtain the ratio of the concentrations
of conjugate base to weak acid!
Using the Ka of the appropriate acid,
the pH of the solution is obtained
from the Henderson-Hasselbalch
equation.
Preparing a Buffer Solution of a
Specific pH



The first step in the process is to
choose a suitable weak acid.
The H-H equation is really only valid
in a pH range near the pKa of the
weak acid;
Second Step  Calculate the
required ratio of conjugate base to
weak acid from the HendersonHasselbalch Equation

From the previous problem, there are a
number of concentrations where the ratio
of the conjugate base to the weak acid
will be acceptable
[CH3CH2COONa] = 0.13 M; [CH3CH2COOH] = 0.10 M
[CH3CH2COONa] = 0.065 M;[CH3CH2COOH] =
0.050 M
[CH3CH2COONa] = 0.39 M; [CH3CH2COOH] = 0.30
M
[CH3CH2COONa] = 0.65 M; [CH3CH2COOH] = 0.50 M
[CH3CH2COONa] = 1.3 M; [CH3CH2COOH] = 1.0 M
Buffer Capacity


Buffer Capacity  refers to the
amount of strong acid/base that can
be added to the buffer solution
Buffer capacity is directly related to
the concentrations of the weak acid
and conjugate base in the buffer
solution
Choosing Concentrations for a High
Buffer Capacity
[CH3CH2COONa] = 0.13 M;
[CH3CH2COOH] =
0.10 M
NOT ACCEPTABLE
[CH3CH2COONa] = 0.065 M;
[CH3CH2COOH] =
0.050 M
NOT ACCEPTABLE
[CH3CH2COONa] = 0.39 M; [CH3CH2COOH] = 0.30
M
REASONABLE
[CH3CH2COONa] = 0.65 M; [CH3CH2COOH] = 0.50 M
GOOD CHOICE
[CH3CH2COONa] = 1.3 M; [CH3CH2COOH] = 1.0 M
GOOD CHOICE
Adding Acid to Buffer Solutions



What happens when we add strong acid
solutions to the buffer?
[H+] increased and the basic part of the
buffer goes to work
H+ (aq) + CH3COO- (aq)  CH3COOH (aq)
This is the reverse of the usual acid
dissociation equilibrium, hence, the reaction
essentially goes to completion
Adding Base to Buffer Solutions
What happens when we add the base to
the buffer?
 The [OH-] increases and the acid part of
the buffer goes to work
OH- (aq) + CH3COOH (aq)  CH3COO- (aq)
+ H2O (l)
 This is the reverse of the usual base
dissociation equilibrium, hence, the
reaction essentially goes to completion

Acid-base Titration Curves

Chapter 4 – acid-base titrations



A known (standard) basic solution is
slowly added to an unknown acid
solution
What if we monitored the pH of the
solution as a function of added
titrant  acid base titration curve is
generated
Three cases to consider
Strong Acid/strong Base
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
Net ionic
H+ (aq) + OH- (aq)  H2O (l)

When n(H+) = n(OH-), we are at the
equivalence point of the titration

Product of reaction is a strong acid/strong
base salt. The pH at the equivalence point
is 7.00.
The Titration Curve
Weak Base /Strong Acid
NH3 (aq) + HCl (aq)  NH4Cl (aq)
Net ionic
NH3 (aq) + H+ (aq)  NH4+ (aq)


Equivalence point, n(NH3) = n(H+).
pH of the solution < 7.00. Determined by the
ionization of the conjugate acid
NH4+ (aq) ⇌ NH3 (aq) + H+ (aq)
The Titration Curve
11.00
pH
9.00
7.00
Eq. point
5.00
3.00
V (strong acid) / mL
Weak Acid/Strong base.
NaOH (aq) + CH3COOH (aq)  CH3COONa
(aq) + H2O (l)
Net ionic
OH- (aq) + CH3COOH (aq)  CH3COO- (aq)
+ H2O (l)
The Titration Curve


Equivalence point when n(CH3COOH) =
n(OH-),
At the equivalence point
pH of the solution is determined by the
ionization of the conjugate base of the weak
acid
CH3COO- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH(aq)


Therefore, the pH at the equivalence
point is > 7.00!
Comparison Between Strong/Weak
Acid Titrations
Indicators


Indicators are used to detect the
endpoint of the acid-base titration.
Indicators are weak acids. Their
ionization can be represented by the
following reaction.
HIn (aq) ⇌ H+ (aq) + In- (aq)
Usually coloured
Also usually coloured, but the
colour is different than for the
acid form of the indicator.



Choose the indicator whose transition range
(i.e., the pH range where it changes colour)
matches the steep part of the titration curve
Note: we can use the following ratios as a
guide.
[HIn] / [In-] > 10  acid colour dominates.
[In-] / [HIn] > 10  base colour dominates.



Strong Acid/Strong Base 
steep part
of titration curve pH 4-10. A number of
indicators change colour in this range
Weak Acid/Strong Base  steep part of
titration curve pH >7.0. The indicator
colour change must occur in this range
Strong Acid/Weak Base  steep part of
titration curve pH <7.0. The indicator
colour change must occur in this range
Indicators in Titrations
Solubility Equilibria


Examine the following systems
AgCl (s) ⇄ Ag+ (aq) + Cl- (aq)
BaF2 (s) ⇄ Ba2+ (aq) + 2 F- (aq)
Using the principles of chemical
equilibrium, we write the
equilibrium constant expressions as
follows

Ag Cl 


K eq

AgCl 
note AgCl   constant


 10
 K sp  K eq AgCl   Ag Cl   1 . 8 x 10

K sp  K eq BaF 2   Ba
2
F 
 2
 1 . 0 x 10
6
The Definition of the Ksp


Ksp  the solubility product
constant.
The product of the molar
concentrations of the dissolved ions
in equilibrium with the undissolved
solid at a particular temperature.
Don’t confuse the solubility of the solid
with the Ksp. These quantities are
related, but they are not the same.
Examples of Ksp Calculations

Calculate the solubility of a sparingly
soluble solid in water.

Calculate the solubility of a solid in the
presence of a common ion.

Calculate the solubility of a solid as a
function of the pH of the solution.
Solubility of Sparingly Soluble
Solids in Water
AgCl (s) ⇄ Ag+ (aq) + Cl- (aq)

We approach this using the
principles of chemical equilibrium.

equilibrium data table, establish and
solve for our unknown quantity!
The Common Ion Effect

What about the solubility of AgCl in
solution containing NaCl (aq)?
AgCl (s) ⇄ Ag+ (aq) + Cl- (aq)
NaCl (aq)  Na+ (aq) + Cl- (aq)
AgCl (s) ⇄ Ag+ (aq) + Cl- (aq)
Equilibrium is displaced to the left by LeChatelier’s
principle (an example of the common ion effect).
Solubility and pH


What happens when we try to
dissolve a solid like Mn(OH)2 in
solutions of varying pH?
We first calculate the pH of the
saturated solution of the Mn(OH)2.
Suppose that we try to dissolve
Mn(OH)2 in an acidic solution
Mn(OH)2 (s) ⇄ Mn2+ (aq) + 2 OH(aq)
H+ (aq) + OH- (aq)  H2O (l)
Mn(OH)2 (s) ⇄ Mn2+ (aq) + 2 OH(aq)
Equilibrium is displaced to the right by
LeChatelier’s principle.

Solubility of CaF2 vs. pH

Increase the [OH-] in the solution

looking at an example of the common ion effect
Mn(OH)2 (s) ⇄ Mn2+ (aq) + 2 OH- (aq)
NaOH (aq)  Na+ (aq) + OH- (aq)
Mn(OH)2 (s) ⇄ Mn2+ (aq) + 2 OH- (aq)
Equilibrium is displaced to the left by
LeChatelier’s principle.

Any solid that produces a
moderately basic ion on dissociation
(e.g., CaF2, MgCO3).


solubility will decrease as the pH is
increased (i.e., the [OH-] in the
solution is increased).
solubility will increase as the pH is
decreased (i.e., the [H+] in the solution
is increased).

What about solids whose anions do
not exhibit basic tendencies?


PbCl2, the Cl- has no tendency to react
with added acid.
solubility of PbCl2, does not depend on the
solution pH!
Predicting Precipitation: the Qsp Value

Let’s examine the following equilibrium
system.
AgCl (s) ⇄Ag+ (aq) + Cl- (aq)
Let’s say that we two solutions so that
they would have the following
concentrations.
[NaCl]o = 1.0x 10-5 M  [Cl-]o = 1.0 x 10-5
M
[AgNO3] = 1.0 x 10-6 M  [Ag+]o = 1.0 x
10-6 M
 Would we be able to predict whether or
not a precipitate will occur?

The QspValue
•
Define the solubility product quotient 
Qsp.



Q sp  [ Ag ] o [ Cl ] o  1 . 0 x 10
6
1 . 0 x10   1 .0 x10
5
We now examine the magnitude of the
solubility product quotient (Qsp) with respect
to the Ksp.
Qsp < Ksp no precipitate will form
Qsp > Ksp a precipitate will form
Qsp = Ksp saturated solution.
 11
The Formation of Metal Complexes
We see that a number of metal anions
can act as Lewis acids; therefore, these
ions can react strongly with Lewis bases
and form complex ions.
AgCl (s) ⇄ Ag+ (aq) + Cl- (aq)
Ag+ (aq) + 2 NH3 (aq) ⇄ Ag(NH3)2+ (aq)
 For the NH3 (aq) to increase the solubility
of the metal salt, the NH3 (aq) must be a
stronger Lewis base than the water
molecules that it displaces.


Solubility of many metal containing
compounds increases markedly in
the presence of suitable complexing
species



NH3 (aq)
OH- (aq)
CN- (aq)

The species Ag(NH3)2+ (aq) is known
as a complex ion. The equilibrium
constant for the second reaction, Kf
Kf 
Ag NH 3 2 
2

Ag  NH 3 
 1 . 7 x 10
7
is known as the formation constant for
the complex ion.
The Magnitudes of Kf values


The complexation reaction
effectively removes all the Ag+ (aq)
from the solution.
For the original equilibrium system
AgCl (s) ⇄ Ag+ (aq) + Cl- (aq)
the equilibrium position is strongly displaced to the right
by LeChatelier’s principle. The solubility of AgCl is
increased significantly in the presence of the complexing
agent!
Precipitation and
Separation of Ions
• Consider a mixture of Zn2+(aq) and Cu2+(aq).
CuS (Ksp = 610-37) is less soluble than ZnS
(Ksp = 210-25), CuS will be removed from
solution before ZnS.
• As H2S is added to the green solution, black
CuS forms in a colorless solution of Zn2+(aq).
• When more H2S is added, a second
precipitate of white ZnS forms.
Precipitation and
Separation of Ions
• Ions can be separated from each other
based on their salt solubilities.
• Example: if HCl is added to a solution
containing Ag+ and Cu2+, the silver
precipitates (Ksp for AgCl is 1.8  10-10)
while the Cu2+ remains in solution.
• Removal of one metal ion from a solution is
called selective precipitation.
• Qualitative analysis is
designed to detect the
presence of metal
ions.
• Quantitative analysis
is designed to
determine how much
metal ion is present.
Qualitative Analysis for
Metallic Elements
• We can separate a complicated
mixture of ions into five groups:
– Add 6 M HCl to precipitate insoluble chlorides
(AgCl, Hg2Cl2, and PbCl2).
– To the remaining mix of cations, add H2S in 0.2 M
HCl to remove acid insoluble sulfides (e.g. CuS,
Bi2S3, CdS, PbS, HgS, etc.).
– To the remaining mix, add (NH4)2S at pH 8 to
remove base insoluble sulfides and hydroxides
(e.g. Al(OH)3, Fe(OH)3, ZnS, NiS, CoS, etc.).
Qualitative Analysis for
Metallic Elements
– To the remaining mixture add (NH4)2HPO4 to
remove
insoluble
phosphates
(Ba3(PO4)2,
Ca3(PO4)2, MgNH4PO4).
– The final mixture contains alkali metal ions and
NH4+.