### Chapter 15 Applications of Aqueous Equilibria

```Chapter 15 Applications of
Aqueous Equilibria
• Some of what we do in this chapter will
be the same as what we did in Chapter
14.
What do the weak acid HF and the salt NaF
have in common?
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 % dissociation of the acid
• This same principle applies to salts with
the cation of a weak base as well.
• Calculations are the same as last chapter
15.2 Buffered solutions
• When an acid-base solution contains a
common ion, it’s called a buffered solution.
• A buffered solution is one that resists a
change in pH with the addition of hydroxide
ions or protons.
• Often buffered solutions contain a weak
acid and it’s salt (HF and NaF) OR a weak
base and it’s salt (NH3 and NH4Cl).
• We can make a buffer of any pH by varying
the concentrations of these solutions.
Finding the pH of a buffered solution
• Calculate the pH of a 1L buffered solution
of 0.20M acetic acid solution and 0.30M
sodium acetate. Ka=1.8x10-5
1 .8 x1 0
5
1 .8 x1 0
5
 H    C 2H 3 O 2  
x  0 .3  x 
 



0 .2  x
 H C 2H 3 O 2 

0 .3 x
0 .2
x  1 .2 x1 0
p H   lo g (1 .2 x1 0
5
)  4 .9 2
5
pH changes in a buffered solution
• We will use the solution from the last problem,
but add 0.01M NaOH. Compare the pH change
that occurs with the addition of the NaOH
solid.
• There are 2 major step to proceed with these
types of problems!!
– 1. The stoichiometry
– 2. The equilibrium
The stoichiometry
1L of 0.20M acetic acid solution and 0.30M sodium
acetate (Ka=1.8x10-5 ) [H+]=1.2x10-5 Buffered with .01M
NaOH
• Use moles not molarity
Before
reaction
After
reaction
HC2H3O2
OH-
↔
H 2O
0.20 mol
0.01
0.30 mol
0.20 – 0.01
=0.19mol
0
0.30+0.01
0.31mol
+
C2H3O2-
• We will assume that all of the NaOH will be
consumed
Now for the equilibrium - ICE
Before
reaction
After
reaction
HC2H3O2
OH-
↔
H 2O
0.20 mol
0.01
0.30 mol
0.20 – 0.01
=0.19mol
0
0.30+0.01
0.31mol
+
C2H3O2-
• Since there is 1L of solution
Concentration
Initial
Change
Equilibrium
HC2H3O2
H+
↔
+
C2H3O2-
0.19M
0
0.31M
-x
+x
+x
0.19-x
X
0.31+x
1 .8 x1 0
5
1 .8 x1 0
5
 H    C 2H 3 O 2  
x  0 .3 1  x 
 



0 .1 9  x
 H C 2H 3 O 2 

0 .3 1x
0 .1 9
pH   log(1.1x10
x  1 .1x1 0
5
5
)  4.96
pH change = 4.96 – 4.92 (from previous problem)
Change in pH = .04 with the addition of a buffer.
happened….
• If base was added to just water


K w   H   O H 

  H   0 .0 1
 H    1x1 0  1 2
 
1x1 0
14
• That means the pH of 0.01M NaOH in water is 12
• Minus the pH of water (7) means a change in
pH of 5.
• Compare adding NaOH to a buffered solution
vs. an unbuffered solution….5 vs 0.04…
Remember
• Buffered solutions are simply solutions of
weak acids and weak bases containing a
common ion
• When a strong acid is added to a
buffered soltuion…You know how to do
these problems. Just remember the two
steps…
– Stoichiometry first
– ICE (equilibrium) second
How does the buffer work?
• Solve the equilibrium expression for [H+]
H    A  
 

Ka 
H A 

H  
 
K a H A 
A  


• This means the [H+] depends on the ratio
of [HA]/[A-]…if you take the –log of both
sides…(I won’t bore you with the math)…
you get…
 A   


p H  p K a  lo g 
 H A  


It has a name!
 A   


p H  p K a  lo g 
 H A  


• It’s called the Henderson-Hasselback
equation!
• It also works for an acid and its salt, like
HNO2 and NaNO2
• Or a base and its salt, like NH3 and NH4Cl,
but you must remember to convert Ka to
Kb in the equation
An assumption
 A   


p H  p K a  lo g 
 H A  


• One assumption of this formula is that
the initial concentrations and the
equilibrium concentrations are
equivalent. (<5% validity)
• It is a pretty safe assumption since the
initial concentrations of HA and A- are
relatively large in a buffered system.
Let’s try
• What is the pH of a solution containing
0.30M HCOOH and 0.52M KCOOH
(formic acid and potassium formate)
Ka=1.8x10-4
  0 .5 2  
4
p H   lo g (1 .8 x1 0 )  lo g 
  0 .3 0  


pH  3.77  0.24  4.01
Given base, salt, and Kb
• Calculate the pH of 0.25M NH3 and
0.40M NH4Cl. (Kb = 1.8 x 10-5)
• Don’t forget to find Ka FIRST!! And the
ratio is base over acid..
Ka 
1x10
 14
1.8 x10
5
p H   lo g (5 .6 x1 0
 5.6 x10
10
 10
  0 .2 5  
)  lo g 
  0 .4 0  


p H  9 .0 5
15.3 Buffer Capacity
• The pH of a buffered solution is
determined by the ratio of [A-]/[HA].
• If the ratio doesn’t change much…then
the pH won’t change much either
• The more concentrated these two are,
the more H+ and OH- the solution will be
able to absorb.
• Larger concentrations = bigger buffer capacity
Try this problem.
• Calculate the change in pH that occurs when
0.040 mol of HCl(g) is added to 1.0 L of each of
the following: Ka= 1.8x10-5
5.00 M HAc and 5.00 M NaAc
0.050 M HAc and 0.050 M NaAc
• Calculate initial pH for each solution
(henderson-hasselbalch)
p H   lo g (1 .8 x1 0
5
 5  
)  lo g 
 4 .7 4

 5  


Now find the change after adding
0.04mol of HCl to 1.0L of solution
HC2H3O2
Before
reaction
After
reaction
H+
↔
+
C2H3O2-
5.0 mol
0.04
5.0 mol
5.0 + 0.04
=5.04
0
5.0 - 0.04
=4.96
p H   lo g (1 .8 x1 0
5
  4 .9 6  
)  lo g 
 4 .7 4
  5 .0 4  


• There’s virtually NO change. Now look at
solution B. We already know the initial pH.
Solution B is 1L of
0.050 M HAc and 0.050 M NaAc
HC2H3O2
Before
reaction
After
reaction
H+
↔
+
C2H3O2-
0.050 mol
0.04mol
0.050 mol
0.050 + 0.04
=0.09
0
0.050 - 0.04
=0.01
p H   lo g (1 .8 x1 0
5
  0 .0 1 
)  lo g 
 3 .7 9

  0 .0 9  


• Compared to 4.74 before the acid was added.
Solution A contains much larger quantities of
buffering components and has a larger
buffering capacity than solution B
Larger concentrations =
bigger buffer capacity
• Here’s why
 A   


p H  p K a  lo g 
 H A  


Buffer Capacity
• The best buffers have a ratio [A-]/[HA] = 1
• This is the most resistant to change and
true only when [A-] = [HA]
• Makes pH = pKa (since the log of 1=0)
10.5 Titrations and pH Curves
• Titration is commonly adding a solution
of known concentration until the
substance being tested is consumed.
• This is often noted by a color change and
is called the equivalence point.
• This is often observed in a graph of pH vs
mL of titrant and is called a titration
curve.
Strong acid with a strong base
titration
• Net ionic equation
H+ + OH- H2O
• To know the amount of H+ at any point in the
titration, we must determine the amount of H+
remaining and divide by the total volume of the
solution.
• Let’s first consider a new unit for molarity that is
smaller as most titrations usually deal with mL
Millimole (mmol) = 1/1000 mol
Molarity = mmol/mL = mol/L
The pH Curve for the Titration of 50.0 mL of
0.200 M HNO3 with 0.100 M NaOH
Where all the H+
ions originally
present, have
reacted with all
Things to note
• You need to do the stoichiometry for each step
• mL x Molarity = mmol
• There is no equilibrium. Both the acid and base
dissociate completely
• Use [H+] or [OH-] to figure pH or pOH
• The equivalence point is when you have an
equal number of moles of H+ and OH• In other words enough H+ is present to react
exactly with the OH• Before the equivalence point H+ is in excess
• After the equivalence point OH- is in excess
The pH Curve for the Titration of 100.0
mL of 0.50 M NaOH with 1.0 M HCI
Strong base titrated
with strong acid.
• Very similar to strong acid with a strong
base except before the equivalence point
the OH- is in excess and H+ is in excess
after the equivalence point.
The pH Curve for the Titration of 50.0 mL of
0.100 M HC2H3O2 with 0.100 M NaOH
weak acid – strong base
At equivalence point (pH > 7): A- is basic
Titrating a weak acid
with a strong base
• Again, there is no equilibrium
• Do the stoichiometry – the reaction of OHwith the weak acid is assumed to run to
completion & the concentrations of the acid
remaining & conjugate base are
determined.
• Determine the major species
• Since HA is stronger than H2O it is the
dominant equilibrium
• Then do the equilibrium (HendersonHasselbach)
The pH Curve for the Titrations of 100.0mL of
0.050 M NH3 with 0.10 M HCl
weak base – strong acid
At equivalence point (pH < 7): NH4+ is acidic
In a titration curve
• Equivalence point is defined by the
stoichiometry NOT by the pH.
• Remember it is where the mol of H+ are
equal to the moles of OH-
Summary
• Strong acid and base just stoichiometry.
• Weak acid with 0 ml of base - Ka
• Weak acid before equivalence point
–Stoichiometry first
–Then Henderson-Hasselbach
• Weak acid at equivalence point- Kb
-Calculate concentration
• Weak acid after equivalence - leftover
strong base.
-Calculate concentration
Summary
• Weak base with 0 ml of acid - Kb
• Weak base before equivalence point.
–Stoichiometry first
–Then Henderson-Hasselbach
• Weak base at equivalence point Ka.
-Calculate concentration
• Weak base after equivalence – left over
strong acid.
-Calculate concentration
15.5 Acid-Base Indicators
• Two common methods for monitoring
the pH
– 1. pH meter
– 2. acid-base indicator (this is not a good
choice for finding the equivalence point.)
Careful selection of indicator can help get
results close.
Endpoint IS change in color
Equivalence point IS moles acid = moles base
HIn
• Most indicators are themselves a weak
acid. They are one color with the proton
attached and another without the
proton.
• Phenolphthalein, the most common
indicator, is colorless as an acid and pink
as a base (In-)
Did the color change?
• Typically, 1/10 of the initial form must be
converted for the human eye to see a
new color. When titrating an acidic
solution…
Ka
 H    In  
 


 H In 
 In  
1



 H In  1 0
 pH  pKa  1
If a basic solution is titrated
• Indicator will initially exist as In- and
more HIn will form when acid is added
Ka
 H    In  
 


 H In 
 In  
10



 H In  1
 pH  pKa  1
Which indicator to use?
• It is best to choose an indicator whose
endpoint is closest to our equivalence
point.
• It is easier to choose an indictor if there
is a large change in pH near the
equivalence point (vertical area of pH curve)
• The weaker the acid, the smaller the vertical
area around the equivalence point, giving
less flexibility in indicator choice
15.6 Solubility Equilibria and Ksp
• Will it dissolve, and if not, how much?

 dissolved
solid 

• If everything dissolves, it is an
equilibrium position
• Partial dissolving -- The solid will
precipitate as fast as it dissolves
(equilibrium)
• Surface area changes the rate at which
something dissolves, but NOT the
amount (no change in equilibrium position)
Generic equation
M aN m b
s
 aM

( aq )
 bNm
• M+ is the cation (usually metal)
• Nm- is the anion (a nonmetal)
• The solubility product for each
compound is found below

a

K sp   M   N m 
b

( aq )
Solubility vs. solubility product
• 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.
Let’s try a couple…
first an easy one!
• What is the Ksp value of copper (I)
bromide with a measured solubility of
2.0 x 10-4 mol/L.
C u B r( s )  C u
CuBr
Cu+
↔
+
Br-
I
2x10-4
0
0
C
solid
+2x10-4
+2x10-4
E
solid
2x10-4
2x10-4

(aq )
 Br
K sp  (2 x1 0

(aq )
4 1
) (2 x1 0
K sp  4 x1 0
8
4 1
)
Now a bit more difficult
• Calculate the Ksp for bismuth sulfide,
which has a solubility of 1.0x10-15M at
25oC
B i 2 S 3 ( s )  2B i
Bi2S3
I
1x10-15
C
solid
E
solid
↔
2Bi+3
0
+ S-2
0
+2(1x10-15) +3(1x10-15)
2x10-15
3x10-15
3
(aq )
 3S
K sp  (2 x1 0
 15
2
(aq )
2
) (3 x1 0
K sp  1 .1x1 0
 73
 15
)
3
Solubility from Ksp
• Find the solubility of Silver Chloride.
Ksp=1.6x10-10.
A g C l( s )  A g

( aq )


K sp   A g   C l 
1 .6 x1 0
10
1 .6 x1 0
  x  x 
10
1 .3 x1 0
5

 x
x
2
 Cl

( aq )
One more…a bit tougher
• Find the solubility of copper (II) iodate,
Cu(IO3)2. The Ksp for Cu(IO3)2 = 1.4x10-7
C u(IO 3 ) 2 ( s )  C u
2


K sp   C u   2(IO 3 ) 
1 .4 x1 0
7
1 .4 x1 0
3
  x  2 x 
7
3 .5 x1 0
 4x
8
3 .3 x1 0
3

3
 x
 2IO 3
(aq )

2
2
3
x
3
The solubility of
copper (II) iodate is
3.3x10-3
(aq )
Comparing Ksp values
• Ksp values give info about solubility
(relative solubility).
1. When salts have the SAME number of ions:
the larger the Ksp the more soluble
2. When salts have different number of ions:
you CANNOT compare directly.
Common ion effect and solubility
• The presence of a common ion
DECREASES the solubility of the salt.
Let’s compare:
What is the solubility of AgBr in both pure
water & in 0.0010M NaBr. (pure H2O first)
A g B r( s )  A g
K sp  7 .7 x1 0
7 .7 x1 0
13
7 .7 x1 0
(aq )
13
  x  x 
13
8 .8 x1 0

7

 x
x
2
 Br

(aq )
Now with a common ion
What is the solubility of AgBr in 0.0010M NaBr
N a B r( s )  N a

A g B r( s )  A g
(aq )
 Br

(aq )

B r   


(aq )
 Br
0 .0 0 1M

(aq )
K sp  7 .7 x1 0
13
13
  x  x  .0 0 1
AgBr
↔
Ag+
I
7.7x10-13
0
0.0010
a ssu m e x is n e g lig ib le
C
solid
+x
+x
in x + 0 . 0 0 1 0
E
solid
x
.0010x
+
Br-
7 .7 x1 0
7 .7 x1 0
13
 0 .0 0 1 0 x
7 .7 x 1 0
10
 x
Compare the two results
In Water
K sp  7 .7 x1 0
7 .7 x1 0
13
7 .7 x1 0
13
K sp  7 .7 x1 0
  x  x 
13
8 .8 x1 0
Common ion solution
7

 x
x
2
7 .7 x1 0
13
13
  x  x  .0 0 1
a ssu m e x is n e g lig ib le
in x + 0 . 0 0 1 0
7 .7 x1 0
13
 0 .0 0 1 0 x
7 .7 x 1 0
10
 x
Notice that the common ion reduces the solubility as stated earlier
(the reverse reaction occurs faster: equilibrium lies to the left)
pH and solubility
• Presence of a common ion decreases solubility
Insoluble bases dissolve in acidic solutions
• Insoluble acids dissolve in basic solutions
remove
2



M g(O H ) 2 ( s ) 
M
g

2O
H

( aq )
( aq )
K s p  1 .2 x1 0
K s p   M g
1 .2 x1 0
1 .2 x1 0
2
11

 O H 


11
 s( 2 s )
11
 4s
s  1 .4 x1 0
2
2
3
4
 O H    2 s  2 .8 x1 0  4


p O H  3 .5 5 & p H  1 0 .4 5
At pH less than 10.45
Lower [OH-]
Increase solubility of Mg(OH)2
At pH greater than 10.45
Raise [OH-]
Decrease solubility of Mg(OH)2
15.7 Precipitation – mixing two
solutions to get a solid
• Let’s look at the reverse process of
dissolving salts to form a solution.
• Ion product is Q   M   a  N m   b

 

• Initial concentrations are used
Q < Ksp Unsaturated solution No precipitate
Q = Ksp Saturated solution
Q > Ksp Supersaturated solution Precipitate
will form
Try this
• A solution of 750.0 mL of 4.00 x 10-3M
Ce(NO3)3 is added to 300.0 mL of
2.00 x 10-2M KIO3. Will Ce(IO3)3
(Ksp=1.9x10-10M) precipitate and if so,
what is the concentration of the ions?
• Step 1: find the concentration of each of
the ions expected to precipitate
• Step 2: Find Q
• Step 3: compare to Ksp
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