Chapter 15 Applications of Aqueous Equilibria

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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.
• Let’s start with common ions.
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.
Think about what would have
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
the OH- ions added
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
add
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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