Monoprotic Acid-Base Equilibria
• Monoprotic Weak Acids
• Monoprotic Weak Bases
• Fraction of Dissociation-Association
• Salts of Weak Acids
• Buffers

Weak Acids
Weak Acid  an acid that is not completely dissociated in aqueous solution
Weak Acid Equilibrium
HA H + + A weak acid conjugate base
K a
= Acid Dissociation Constant
[H + ][A ]
[HA]
(Available in Appendix G)

Weak Bases
Weak Base  a base in which the hydrolysis reaction does not go to completion in aqueous solution
Weak Base Equilibrium
B + H
2
O BH + + OH weak base conjugate acid
K b
= Base Hydrolysis Constant
[BH + ][OH ]
[B]

Conjugate Relationships
For conjugate acid-base pairs
K a
 K b
=K w pK a
+pK b
=14 (at 25 ° C)
This is a very useful relation to determine K b can look up K a when you

Conjugate Relationships
The conjugate base of a weak acid is a weak base
• Stronger weak acid

Weaker conjugate base
• Weaker weak acid

Stronger conjugate base
The conjugate acid of a weak base is a weak acid
• Stronger weak base

Weaker conjugate acid
• Weaker weak base

Stronger conjugate acid

Monoprotic Weak Acid Problems
Method
Write pertinent equations for all equilibria
Write expression for mass balance
Write expression for the charge balance

Monoprotic Weak Acid Problems
Equilibria
HA H + + A K a
= [H + ][A ]
[HA]
H
2
O H + + OH K
W
= [H + ][OH ]
Mass Balance
F = formal concentration  the total amount of a compound dissolved in 1 L of solution
F = [HA] + [A ]  [HA] = F - [A ]

Monoprotic Weak Acid Problems
Charge Balance
[H + ] = [A ] + [OH ]
Approximation (a very good one)
Assume that the weak acid reaction products will dominate over the water dissociation reaction in terms of product concentrations so [A ] >> [OH ] because the weak acid reaction dominates over the water dissociation reaction
Charge balance equation becomes
[H + ]  [A ]

Monoprotic Weak Acid Problems
Using our equilibrium expression
K a
= [H + ][A ] = [H + ][H + ] = [H + ][H + ]
[HA] [HA] F - [A ] so let x = [H + ] and K a
= x 2
F - x

Monoprotic Weak Acid Example
Consider placing o-hydroxybenzoic acid in aqueous solution at a formal concentration of F = 0.050 F
K a
= 1.07 x 10 -3
What is the [H + ] ?
K a
= x 2
F - x
 x 2 + K a x - K a
F = 0 so using the quadratic equation x

K a

K
2 a

4 K a
F
2
[H + ] = x

(1.07
x 10 3 )  (1.07
x 10
3 ) 2  4 (1.07
x 10 3 )(0.05)
2(1)
= 6.80 x 10 -3 M and this is ~ [A ]

Monoprotic Weak Acid Example
What is the [OH ] ?
[OH ] = 10 -14 = 10 -14 = 1.47 x 10 -12 M
[H + ] 6.80 x 10 -3
Note our approximation [A ] >> [OH ] is justified!
Another useful approximation is to neglect x in the denominator iff x < 0.01 F
K a
= x 2  x 2
F - x F
Try this with the o-hydroxybenzoic acid problem-
Does this meet our criterion ? 0.00680 < (0.01)(0.050)
NO

Monoprotic Weak Base Problems
Weak base - similar to the weak acid approach except we solve for [OH ]
Equilibria
B + H
2
O BH + + OH -
H
2
O H + + OH -
K b
= [BH + ][OH ]
[B]
K
W
= [H + ][OH ]

Monoprotic Weak Base Problems
Mass Balance
F = [BH + ] + [B]  [B] = F - [BH + ]
Charge Balance
[BH + ] + [H + ] = [OH ]
In the weak base case we want to solve for
[OH ]

Monoprotic Weak Base Problems
Approximation (a very good one)
Assume that the weak base reaction products will dominate over the water dissociation reaction in terms of product concentrations so [BH + ] >> [H + ] because the weak base reaction dominates over the water dissociation reaction
Charge balance equation becomes
[BH + ]  [OH ]

Monoprotic Weak Base Problems
Using our equilibrium expression
K b
= [BH + ][OH ] = [OH ][OH ] = [OH] 2
[B] [B] F - [BH + ] so let x = [OH ] and K b
= x 2
F - x

Monoprotic Weak Base Problems
[OH ] = x

K b

K
2 b

4 K b
F
2
Again a useful approximation is to neglect x in the denominator iff x < 0.01 F
K b
= x 2  x 2
F - x F

Fraction of Dissociation (for an acid)
Fraction of Dissociation  the fraction of acid that is present in the form A -
 = [A ] = x
[A ] + [HA] x + (F -x)
 = x
F

Fraction of Association (for a base)
Fraction of Association  the fraction of base that has reacted with water
 = [BH + ] = x
[BH + ] + [B] x + (F -x)
 = x
F

Salt of Weak Acids
Sodium acetate dissolved in water
Salts completely dissociate in solution
Na + CH
3
COO -
In solution
Na + + CH
3
COO -
CH
3
COO + H
2
0 CH
3
COOH + OH -
K a
=1.75 x 10 -5
A + H
2
0 HA + OH -

Salt of Weak Acids
What if we dissolve F of Na + CH
3
OO in solution?
How would we find the pH?
If you dissolve the salt of a weak acid the result is a weak base equilibrium problem.
Equilibria
CH
3
COO + H
2
0 CH
3
COOH + OH -
K b
= [HA][OH ] = K
W
[A ] K a
= 1.01 x 10 -14 = 5.77 x 10 -10
1.75 x 10 -5
H
2
O H + + OH K
W
= [H + ][OH ]

Salt of Weak Acids
Mass Balance
F = [HA] + [A ]  [A ] = F - [HA]
Charge Balance
[H + ] = [OH ] + [A ]
Weak base case  we want to solve for [OH ]
K b
= [HA][OH ] = K
W
= 1.01 x 10 -14 = 5.77 x 10 -10
[A ] K a
1.75 x 10 -5
[HA][OH ] = x 2 = 5.77 x 10 -10
[A ] F - x
Solve for x = [OH ]

Buffer Solutions
Buffer  a solution that resists changes in pH when acid or base is added or when the solution is diluted
Typical Buffer
• A mixture of an acid and base (salt of the acid) conjugate pair which resists changes in pH
• The weak acid and conjugate base should have similar concentrations (10:1)
• Buffers break down at extremes of pH or dilution
• Buffers are most effective near the pK a of the acid

Buffer Solutions
How do mixtures of weak acids and their conjugate bases act as buffers?
Consider a weak acid of F =0.10 M and K a
HA + H
2
O H
3
O + + A -
=1.0 x 10 -4
K a
= [H + ][A ] = 1.0 x 10 -4
0.1-x x x [HA]
Recall K a
= x 2
F - x
Solve for x x = 3.1 x 10 -3 M
 = x/F = (3.1 x 10 -3 )/0.1 = 0.31  3.1 % dissociated
Addition of additional A makes HA dissociate less!

Buffer Solutions
Consider a weak base of F =0.10 M and K b
=1.0 x 10 -10
A + H
2
O HA + OH K b
= [HA][OH ] = 1.0 x 10 -4
0.1-x x x [A ]
Recall K b
= x 2
F - x
Solve for x x = 3.2 x 10 -6 M
 = x/F = (3.2 x 10 -6 )/0.1 = 3.2 x 10 -5 
0.0032 % associated
Addition of additional HA makes A associate less!

Buffer Solutions
The concentration of HA and A in solution are similar to the amounts added to solution
A buffer acts through the reaction of added base with excess HA and the reaction of added acid with excess
A -
This is how it resists changes in pH
As more A- is formed from the former reaction or as more HA is formed from the latter reaction the pH changes very little until we have a greater ratio than
1:10 or 10:1 between the acid and its salt

Henderson-Hasselbach Equation
Henderson-Hasselbach Equation
HA + H
2
O H
3
O + + A K a
= [H + ][A ]
[HA]
[H + ] = K a
[HA]
[A ]
-log [H + ] = -log K a
- log [HA] + log [A ] pH = pK a
+ log [A ] Use formal concentrations for
[HA] all species pH = pK a
+ log [B] Use the K a
[BH + ] for the acid form

Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation
If [HA] = [A ] then pH = pK a
Effect of the [A-]/[HA] ratio on pH (Table 10-1)
[A-]/[HA]
100:1
10:1
1:1
1:10
1:100 pH pK a
+2 pK a
+1 pK a pK a
- 1 pK a
-2
In any mixture of 5 different acids and bases in solution there will only be a single pH and the 5 Henderson-
Hasselbalch Equations must be simultaneously met

Buffers Doing Their Job
Consider the tris buffer system tris(hydroxymethyl)aminomethane
We add the acid form to a solution as the chloride salt
(BH + Cl )
This salt completely dissociates when placed in solution
BH + Cl to form the weak acid
BH + + Cl -

Buffers Doing Their Job
What is the pH of a solution prepared by dissolving
12.43 g of tris (F.W. = 121.136) to 4.67 g of tris hydrochloride (F.W. = 157.597) in 1.00 L of H
2
O?
Our equilibrium looks like
BH + B + H +
[BH + ] = [4.67 g(1 mol/157.597 g)]/ 1L = 0.0296 M
[B] = [12.43 g(1 mol/121.136 g)]/ 1L = 0.1026 M pH = pK a
+ log [B] = 8.075 + log 0.1026 = 8.61
[BH + ] 0.0296
NOTICE : Volume does not matter! You can use the number of moles of acid and conjugate base directly

Buffers Doing Their Job
Weak Base + Strong Acid react completely
B + H + BH + K = [BH + ] = 1
[B] [H + ] K a
Weak Acid + Strong Base react completely
HA + OH A + H
2
O K = [A ] = 1
[HA] [OH ] K b
The reactions written above lie completely to the right
(i.e. we do not need an arrow written to the left) and only cease when one of the reactants are completely exhausted

Buffers Doing Their Job
What is the pH resulting from addition of 12.0 mL of
1.00 M HCl to the tris buffer in the previous problem?
Addition of strong acid reduces the amount of tris base
B + H + BH + weak base strong acid
One of the two reactants is the limiting reagent and the system reacts until that limiting reagent is completely consumed
Here the mol of H + from HCl = (0.012 L)(1.00 M) =
0.012 mol
Recall we have 0.1026 mol of B so HCl is the limiting reagent

Buffers Doing Their Job
B + H + BH +
Initial moles 0.1026 0.0120 0.0296
Final moles 0.0906 -------0.0416
Now we can use Henderson-Hasselbalch pH = pK a
+ log moles
B
= 8.075 + log 0.0906 = 8.41
moles
BH+
0.0416
NOTICE : Again volume does not matter!

Buffer Solutions
A buffer is the most effective at a pH = pK a
How do we prepare a buffer in the lab?
• Calculate how many grams (mol) of weak acid to add to make final volume (V) and dilute with H
2
O to approximately 0.75V
• Monitor the pH
• Add concentrated base dropwise to adjust pH to desired value
• Pour solution into a volumetric flask and rinse initial container with H flask
2
O several times into volumetric
• Dilute with H
2
O to desired volume and mix well

Buffer Capacity
Buffer Capacity  a measure of how well a solution resists changes in pH when adding strong acid or strong base
Buffer Capacity   = ( d C b
/ d pH) = - ( d C a
/ d pH) where C b and C a are the number of moles required to change the pH by 1 unit
0.100 F solution of H a with a pK a
= 5.00
The top curve indicates the concentration of strong base necessary to yield the indicated pH
The bottom curve is the derivative of the top curve (the buffer capacity)

Buffer Solution Facts
• When choosing a buffer you should choose the system with a pKa that is closest to the desired pH
• Buffers are “good buffers” over a range of pH’s corresponding to pK a
- 1 to pK a
+ 1
• Outside of this range there is not enough A to react with strong acid or HA to react with strong base
• Realize that just as with all thermodynamic equilibrium constants the “true” Henderson-
Hasselbalch Equation should contain activities rather than concentrations
• This leads to a dependence of the pKa and solution pH on temperature and the ionic strength of the solution

Buffer Solutions Assignment
In addition to doing all of the assigned problems, look over Table 10-2 on pp. 230-231
Make sure that for each buffer listed you can: identify the acid and base forms of the buffer write the equilibrium expression that would be used in conjunction with the Henderson-Hasselbalch
Equation
You do not need to memorize the structures of individual buffer systems
However, you should familiarize yourself with the buffers listed as indicated above