Additional Aspects of Equilibria

Salt Solutions in Water
Salt solutions completely ionize in H
2
O. If their ions can react with H
2
O to form H + of OH , the pH of the solution will change.

Anions that are the conjugate bases of weak acids will react with H
2
O to form
OH ions.
Anions that are conjugate bases of strong acids are very weak bases and aren’t strong enough to react with H
2
O. No pH change.

Anions that have ionizable protons (HSO
4
) are amphoteric. Can act as an acid or a base and must be determined based on the K a and K b for the ion.

3-
+2 +1
Will K
2
HC
6
H
5
O
7 form an acidic or basic solution in H
2
O?
or
HC
6
H
5
O
7
2 -
HC
6
H
5
O
7
2 -

H
2
O


H
2
O

H
3
O

C
6
H
5

C
6
H
5
O
3 -
7
O
3 -
7

OH
-
Must look at K a and K b
.
From table 16.3 in book get K a3
= 4.0x10
-7
Must calculate K b from appropriate K ax
.
K a2

HC
6
H
5
O
7
2 -
H
2
C
6
H
5
O
-
7

H
 
H
2
C
6
H
5
O
-
7

H
 
HC
6
H
5
O
3 -
7


K w
= K a x K b
K w
1 x 10
-14
K a2
1.7
x 10
5

5 .
9 x 10
10 
K b
Now compare for our situation
K a
= 4.0x10
-7
K b
= 5.9x10
-10
K a is

700 times larger than K b so the acid reaction will predominate and solution will be acidic.

All cations (except alkali and heavy alkaline earth metal ions) act as weak acids.
Rules
• Salts derived from strong acid and strong base from neutral solutions in H
2
O.
ex. NaCl Na + from NaOH
Cl from HCl
• Salt derived from strong base/weak acid. Anion will be relatively strong conjugate base.
 pH > 7.
ex. Ba(C
2
H
3
O
2
)
2

• Salt derived from weak base/strong acid.
Strong conjugate acid.
 pH < 7.
ex. Al(NO
3
)
3
Al 3+ + 3H
2
O

Al(OH
3
) + 3H +
• Salt derived from weak base/weak acid. ex. NH
3
C
2
H
3
O
2
. Both conjugate acid and base will be fairly strong in solution. pH depends on which ion hydrolyzes H
2
O better. Must compare K a and K b
.

Additional aspects of Aqueous Equilibria
1. Common Ion Effect
Remember LeChatlier
HC
2
H
3
O
2(aq) equil
 
H
(aq)

C
2
H
3
O
-
2(aq)
What happens if I add more C
2
H
3
O
2
the solution?
to

The dissociation of a weak electrolyte is decreased by adding a strong electrolyte that has an ion in common with the weak electrolyte.
Ex. ? pH of solution of 0.30 mol acetic acid (HC
2
H
3
O
2
) and 0.30 mol NaC
2
H
3
O
2
(sodium acetate) in 1.0L?

Steps
1. Identify major species and are they acid or base.
HC
2
H
3

O
2
Weak acid
Won’t dissociate completely so:
C
2
H
3
O
2
-
H + +
+
HC
2
H
3
O
2
C
2
H
3
O
2
-
HC
2
H
3
O
2
H +
C
2
H
3
O
2
-
H +
Na +
C
2
H
3
O
2
-
C
2
H
3
O
2
-
Na +
C
2
H
3
O
2
-
HC
2
H
3
O
2
H +
Na +
C
2
H
3
O
2
-
NaC
2

H
3
O
2
Strong * electrolyte so:
Na + +
C
2
H
3
O
2
-
* How do I know that?

2) Identify important equilibrium reaction.
HC
2
H
3
O
2(aq)

H
2
O

H
3

O
(aq)

C
2
H
3
O
-
2(aq)
H + is the same thing as H
3
O + in aqueous solution
(at this point in your chemistry life)
3) Calculate concentrations of equilibrium species.
Remember: C
2
H
3
O
2
-
NaC
2
H
3
O
2 comes from HC
2
H
3
O
2 and

I

Eq
K a

HC
-
2
H xM
3
O
2
0.30M
(0.30
x)M

H
2
O
N
A

0
H
3
O

N
A
 xM
N
A xM

C
2
H3O
-
2
0 .
30 M
 xM
(0.30
 x)M
1.8x10
5 
[H
3
0

][C
2
H
3
O
-
2
]
[HC
2
H
3
O
2
]

( x)(0.30
(0.30

x) x)
Can we make the assumption that ‘x’ can be ignored here? How do you know?

2) Acid-Base Titrations
Quantitatively add an acid (or base) to a base (or acid).
Strong acid/strong base titrations:
The graph of pH vs. ml of titrant is called a titration curve.

pH before equivalence point is determined by concentration of acid NOT
YET NEUTRALIZED .
pH at equivalence point is pH of the salt solution.
pH past equivalence point is determined by concentration of excess base.

? pH when 0.100M NaOH solution is added to 20.0 ml
0.100M HBr a) 10.0 ml NaOH
0 .
020 L HBr



0.100
mol
L 



0 .
002 mol HBr
0.002
mol HBr



1 mol H

1 mol HBr 



0 .
002 mol H

NaOH

Na + + OH in water
So ALL these will react with H + until one or the other runs out!

So 10 ml
0 .
010 L NaOH
0.100
mol
L NaOH

0 .
001 mol NaOH
0.001 mol NaOH yields 0.001 mol OH in H
2
O.
OH + H +

H
2
O
0 .
001 mol OH
-


1 mol H

1 mol OH
-



0 .
001 mol H

Consumed by OH -
Started with 0.002 mol H +
That leaves 0.001 mol H + pH = 3


0.001
mol H

0.030
L



0 .
033 M H
 pH

1.5

B) 20.0 ml NaOH total
0 .
020 L



0.100
mol
L NaOH 



0 .
002 mol NaOH
0 .
002 mol H


0 .
002 mol H
 consumed by
0.000
mol H

So, is our pH 14?
NO
Autoprotolysis H
2
O

H + + OH pH would be that of the salt + autoprotolysis

7

C) 30.0 ml NaOH
0 .
003 mol OH


0 .
002 mol OH

0.001
mol OH

Consumed by
0.002 mol H + in initial solution pOH = 3 pH = 11
0.001
mol OH
-

0 .
02 M OH
-
0.050
L pOH

1.70
pH

12.3

Buffered Solutions
Solutions that resist change in pH when small amounts of acid or base are added.
Many neutral systems are buffered systems.

Buffers contain both an acidic species and a basic species.
A weak acid/base conjugate pair is common.
NaC
2
H
3
O
2 and HC
2
H
3
O
2
NH
4
Cl and NH
3

Choose appropriate components and adjust concentrations to get buffer at any pH.

HX

H
 
X
-
K a

[H

][X
-
]
[HX]
[ H

]

K a
[HX]
[X
-
]
Ratio of conjugate acid/base pair
Na HCl
OH + HX

H
2
NaCl
O + X -
[HX]

[X ]

But if [HX] and [X ] are large enough, a small amount of [OH ] won’t matter.

H + + X -

HX
Same goes for a small amount of acid.
When [HX] and [X ] are approximately equal, buffer is most effective for pH change in either direction. Choose buffer whose acid has a pKa close to desired pH.

Buffer Capacity and pH
In General: pH
 pK a
 log
[base]
[acid]
Henderson-Hasselbach equation
Use starting concentration of acid and base components of the buffer.

? pH of Buffer
[H

]
0.15M NaHCO
3
0.10M Na
2
CO
3
H
2
CO
HCO
-
3
3

K a1

H
HCO
 
-
3
CO

K a2
2 -
3
CO
3
2 -

K a
[acid]
[base] pH
 pK a
 log
[base]
[acid]

-log(5.6x1
0
11
)
 log
(0.10)
(0.15)
10 .
07

( 10 .
25 )

(

0 .
176 )
10 .
1
 pH

Strong Acid-Weak Base and
Weak Acid-Strong Base
Titrations

* pH > 7 at stoichiometric endpoint for weak acid/strong base titration.
* pH < 7 at stoichiometric endpoint for strong acid/weak base titration.

Example
30.0ml sample of 0.20M C
6
H
5
COOH is titrated with
0.30M KOH.
K a
C
6
H
5
COOH = 6.5x10
-5
Calculate pH at stoichiometric endpoint.
C
6
H
5
COOH weak acid
C
6
H
5
COOH

C
6
H
5
COO + H +
K a
= 6.5x10
-5 in water
But what about with a strong base?

OH will deprotonate all of the C
6
H
5
COOH
So, at the endpoint, enough OH has been added to react with all the C
6
H
5
COOH.
Up to this point, the problem looks just like a strong acid/strong base titration!

What’s different?
Weak acid yields a strong conjugate base.
So,
C
6
H
5
COO + H
2
O

C
6
H
5
COOH + OH -
That’s the logic.

0 .
030

L


0.20
0.006
mol mol
C
6
C
L
6
H
5
COOH
H
5
COOH
Must have same moles of OH -
0 .
006 mol OH
-
1 L
0.30
mol KOH

0.02
L

vol KOH needed
Total volume = 0.03L + 0.02L
= 0.05L = 50ml

[C
6
H
5
COO
-
]

0.006
mol
0.050
L

0 .
12 M
What is happening in solution?
Relatively strong base in water:
C
6
H
5
COO + H
2
O

C
6
H
5
COOH + OH -
K b

[C
6
H
5
COOH][OH
-
]
[C
6
H
5
COO
-
]
K b

?
K w
K a

K b

I

K b
C
6
H
5
COO -
1x10
-14

6.5x10

H
2
O
5
0.12M
N
A
x N
A

1 .
54 x10
10

C
6
H
5
COOH
0
 x x Eq (0.12
x)M N
A
K b

( x)(x)
(0.12
x)

1 .
54 x10
10

OH
-
0
 x x

x
2

1 .
54 x10
10
0.12
x 2 = 1.85x10
-11 x = 4.3x10
-6 x = [C
6
H
5
COOH] = [OH ] pOH = 5.37
pH = 8.6

? pH
10.0g KCH
3
CO
2 solution.
dissolved in 250ml
K a
(CH
3
CO
2
) = 1.8x10
-5
10.0g KCH
3
CO
2
= 0.093mol KCH
3
CO
2
0.093mol KCH
3
CO
2
CH
3
CO
2
yields 0.093mol

I

Eq
0.093mol
CH
3
CO
-
2

0 .
37 M CH
3
CO
-
2
CH
3
0.250L
CO
-
2

H
2
O

CH
3
CO
2
H

OH
-
K b

K w
K a
1 x10
14

1.8x10
5

5 .
6 x10
10
  
CH
3
CO
-
2
0.37M
x
(0.37
x)M
H
2
O
N
A
N
A
N
A
CH
3
CO
2
H
0
 x x
OH
-
0
 x x

K b

(x)(x)
(0.37
x )

5 .
6 x10
10 x
2

5 .
6 x10
10
0.37
x
2 
1.68x10
10 x

1.3x10
5 pOH

4.89
pH

9.11

[OH
-
]

Solubility Equilibria
K sp
= the degree to which a solid is soluble in water.
BaSO
4(s)
 2

Ba
(aq)
 2 -
SO
4(aq)
K sp

[Ba
2

][SO
4
2 -
]

Table for K sp at 25ºC
Appendix D
K sp
BaSO
4
= 1.1x10
-10
The smaller K sp dissolve in H
2
O is means less will
K sp
Ca
3
(PO
2.0x10
-29
4
)
2
= [Ca 2+ ] 3 [PO
4
3] 2 =
Ca
3
(PO
4
)
2

3Ca 2+ + 2PO
4
3-

Converting between Solubility and K sp
Solubility of compound
(g/L)
Molar solubility of compound
(mol/L)
Molar
[ions] K sp

17.34) PbBr
2 molar solubility = 1.0x10
-2 mol/L
? K sp
PbBr
2(s)
 2

Pb
(aq)
 -
2Br
(aq)
[Pb 2

(aq)
]

1.0x10
2 mol
L
-
[Br
(aq)
]

2.0x10
2 mol
L
K sp
= [Pb 2+ ][Br ] 2
K sp
= (1.0x10
-2 )(2.0x10
-2 ) 2
K sp
= 4.0x10
-6

Common Ion Effect
Remember Le Châtlier!
CaF
2(s)
 2

Ca
(aq)

2F
-
(aq)
17.36) ? solution of CaF
2
CaF
2(s)
 in g/L in 0.15M KF solution
2

Ca
(aq)

2F
-
(aq)
K sp
= [Ca 2+ ][F ] 2 molar solution CaF
2
= [Ca 2+ ] = X
[F ] = 2X another source of F = 0.15M F -

So [Ca 2+ ] = X
[F ] = 2X + 0.15
K sp
= (X)(2X + 0.15) 2
* assume X is small compared to 0.15
K sp
= 3.9x10
-11 = (X)(0.15) 2 = 0.0225X
X = 1.7x10
-9 M Ca
1.7x10
-9 mol CaF
2
1L
1.3x10
7 g CaF
2
L
 x
78.1g
CaF
2
1 mol
0.13
 g

L

0 .
13 ppb CaF
2

Precipitation will happen when
Q > K sp
Q = ion product if Q = K sp
, equilibrium exists, saturated solution if Q < K sp
, solid dissolves until Q = K sp

Solubility and pH
Hg 2+
Very important in environmental systems
Consider Hg(OH)
2
K sp
= 3.0x10
-26
Hg(OH)
2

Hg 2+ + 2OH -
K sp
= [Hg 2+ ][OH ] 2 = 3.0x10
-26 at pH = 3, pOH = 11

[OH] = 1.0x10
-11
[Hg 2+ ][1.0x10
-11 ] 2 = 3.0x10
-26
[Hg 2+ ] = 3x10 -4 M = 6.0x10
-2 g/L = 60mg/L = 60ppm

Selective Precipitation
Say you have Cu 2+ and Zn 2+ in solution
You can ppt one out of solution without the other use S 2by adding H
2
S
(g)
CuS K sp
= 6.3x10
-36
ZnS K sp
= 1.1x10
-21
Must adjust pH to accomplish this.

[Zn
2

][S
2 -
]

K sp

1.1x10
-21
[S
2 -
]

1.1x10
21
[Zn
2

]
Say the solution in 0.10M in Zn 2+ and 0.10 in Cu 2+
1.1x10
-21
[Zn
2

]

1.1x10
-21
0.10

1.1x10
20
M
Now look at how H
2
S affects pH
[H + ] 2 [S 2] = 7x10 -22

[H

]
2 
7x10
-22
[S
2 -
]
7x10
-22

1.1x10
20

6x10
2
[H

]

6x10
2 
0.24M
pH

-log(2.4x1
0
1
)

0.6
ZnS won’t precipitate at pH below 0.6

Complex formation
Ions in solution come together and form a soluble complex so they can no longer be directly measured

12H
(aq)

3Fe
(s)

2C
2
HCl
3(aq)

6HCl

3Fe
2
 
2C
2
H
4
Should be able to use Cl analysis to monitor reaction
NO WAY MAN
Can (and does) form a soluble FeCl
2
/FeCl
3 complex