The Common Ion Effect and
Acid/Base Equilibria
Additional Aspects of Acid/Base Equilibria
 So far, we have examined the equilibrium
concentrations of ions in solutions containing a weak
acid or weak base
 Now, we will consider solutions that contain a weak
acid and a soluble salt of that acid
Additional Aspects of Acid/Base Equilibria
 An example of a solution containing a weak acid and
soluble salt is acetic acid and sodium acetate

Sodium acetate is a soluble ionic compound and therefore, a
strong electrolyte (dissociates completely):
CH3 COONa aq → Na+ aq + CH3 COO− aq

Acetic acid is a weak electrolyte that ionizes only partially, and
is represented by a the dynamic equilibrium:
CH3 COOH aq ↔ H + aq + CH3 COO− aq
So, What is the Common-Ion Effect?
 If we add sodium acetate to a solution of acetic acid in water, the
CH3COO- from CH3COONa causes the equilibrium
concentrations of the substances in the acetic acid system to shift
to the left, thereby decreasing the equilibrium concentration of
H+
CH3COONa (s)
CH3COOH (aq)
Na+ (aq) + CH3COO- (aq)
H+ (aq) + CH3COO- (aq)
Common
Ion
 In other words, whenever a weak electrolyte and a strong
electrolyte containing a common ion are together in solution,
the weak electrolyte ionizes less than it would if it were alone
in solution
 We call this observation the common-ion effect
A General Overview of the Common-Ion Effect
 The common-ion effect is generally defined as the
shift in equilibrium caused by the addition of a
compound having an ion in common with the
dissolved substance
 The presence of a common ion suppresses the
ionization of a weak acid or a weak base
Real-Life Applications of the Common-Ion Effect
 The most important application of acid-base solutions
containing a common ion is for buffering
 A buffered solution is one that resists a change in its
pH when either hydroxide ions or protons are added

This is because it contains both an acid to neutralize added OH- ions
and a base to neutralize added H+ ions
 HUGE EXAMPLE OF A BUFFERED SOLUTION– Our
blood!

Blood can absorb the acids and bases produced in biologic reactions
without changing its pH


Vital because cells can survive only in a very narrow pH range!
Buffering system in blood involves HCO3- and H2CO3
Composition of a Buffer Solution
 A buffer solution can consist of:

A weak acid and its conjugate base (its salt)


HF and NaF
A weak base and its conjugate acid (its salt)

NH3 and NH4Cl
Ways to Make a Buffered Solution
 There are two ways to make a buffer:

Mix a weak acid or weak base with a salt of that acid or base


For example, the CH3 COOH/CH3 COO− buffer can be prepared by
adding CH3 COONa to a solution of CH3 COOH
Make the conjugate acid or base from a solution of a weak base
or acid by the addition of a strong acid or base to neutralize
about half of the weak acid or weak base

For example, the CH3 COOH/CH3 COO− buffer can be prepared by
adding some NaOH to the solution – enough to neutralize about
half of CH3 COOH according to the reaction:
CH3 COOH aq + OH − aq ↔ CH3 COO− aq + H2 O (l)
Practice!
#4 AND #6 ON PAGE
pH and Buffer Solutions
 Because a buffer resists changes in pH, it is useful to
calculate the initial pH of the solution
 So, how do we do this?
Steps for Calculating pH of Buffer Solutions
THINK – Which solutes are strong electrolytes and which
are weak electrolytes?
THINK – Major species in solution
THINK – What is the important equilibrium reaction that is
the source of H+ (therefore, controls pH)?
WRITE – Equilibrium equation
THINK – Do not know equilibrium conditions so use ICE
table and Ka expression to solve for [H+] ([H3O+] )
1.
2.
3.
4.
5.
 Note – there is an initial concentration of conjugate base in
the ICE table!

1.
Make assumption that [HA]int
WRITE – pH = -log[H+]
– x = [HA]eq = [HA]int if Ka << 1
Practice!
#8 ON PAGE
There’s an Even Easier Way to Calculate pH of a
Buffer!
 The acid dissociation equilibrium expression can
be rearranged to provide a useful equation when
calculating pH:
[ H ][ A ]
[ HA]
Ka 
 [ H ]  Ka
[ HA]
[A ]
[ HA]
  log[ H ]   log Ka  log
[A ]
Derivation Continued…
pH = pK a + log
A−
HA
= pK a + log
base
acid
 This log form of the expression for Ka is called the
Henderson-Hasselbach equation
 It is useful for calculating the pH of solutions when the
ratio [HA]/[A-] is known AND Ka <<< 1
Practice!
#9 ON PAGE
Preparing a Buffer
 In the laboratory, it is useful to know the amounts of
the acid and its conjugate base or weak base and
conjugate acid needed to achieve a specific pH
 The most effective buffer occurs when the ratio of
[A-] to [HA] or [B] to [BH+] is 1

If this ratio is 1, then according to the Henderson-Hasselbalch
equation:
pH = pKa
Preparing a Buffer with a Certain pH
Method #1
 To do this:
1. THINK – Major species in buffer solution
2. THINK – What conjugate acid-base pair
determines pH?
3. WRITE – Equilibrium equation between conjugate
acid-base pair and water
4. WRITE – Equilibrium expression
5. WRITE – Obtain [H+] or [OH-] using given pH:
[H+] = 10-pH or [OH-] = 10-pOH
WRITE – Plug and chug using all known information
into equilibrium expression and solve for unknown!
1.
1.
Usually, this includes [HA] or [B] and Ka/Kb value
Preparing a Buffer Solution with a Specific pH
Method #2
 Choose a weak acid whose pKa is close to the desired
pH

You can:
Calculate [H+] from pH
 Substitute this value with Ka value of weak acid into the equation:
[HA]
+
H = Ka
[A− ]
 Choose the ratio [HA]/[A-] that is closest to 1


OR:
Substitute pKa (-log Ka) and pH values into the HendersonHasselbach Equation
 This will give a ratio of [A-]/[HA]
 Then, convert ratio to molar quantities

Practice!
#14 ON PAGE
So, Why are Buffers So Important?
 Buffers find many important applications in the
laboratory and in medicine

Many biological reactions occur at the optimal rates only when
properly buffered
 Remember, buffered solutions resist pH change with
addition of a STRONG acid or base

The amount of acid or base that buffer can neutralize before
the pH begins to change significantly is called the buffer
capacity
What Exactly Happens When a Strong Acid or
Base is Added to a Buffer?
 For a weak acid buffer
(pH < 7)


Added strong acid (H+)
reacts with conjugate
base
Added strong base (OH-)
reacts with weak acid
 For a weak base buffer
(pH > 7):


Added strong acid (H+)
reacts with weak base
(B)
Added strong base (OH-)
reacts with conjugate
acid (BH+)
Buffer Action
Acetic Acid/Acetate Buffer
Buffer Action
Buffer Response to Strong Acid
Buffer Response to Strong Base
Calculating How pH of a Buffer Responds to
Addition of Strong Acids and Bases
 Solution #1
 Do a “Before Reaction” – “Change” – “After Reaction” table using
MOLES ONLY for stoichiometry calculations to determine new moles of
buffer components




Assume reaction with H+/OH- goes to completion
Volume of solution changed with addition of SA/SB so calculate new
concentrations by assuming volumes are additive
Set-up an ICE table with new concentrations to determine
equilibrium concentrations
Calculate pH from [H+] using Ka expression
 Solution #2
 Do a “Before Reaction” – “Change” – “After Reaction” table for
stoichiometry calculations to determine new moles of buffer components



Assume reaction with H+/OH- goes to completion
Volume of solution changed with addition of SA/SB so calculate new
concentrations by assuming volumes are additive
Use Henderson-Hasselbach equation to solve for pH
Steps to Calculate the pH of a Buffer After
Addition of Strong Acid or Base
Buffer Tutorial
How to Do Buffer Calculations
Practice!
#13 ON PAGE
Making a Buffer Calculations
You want to prepare 500.0 mL of a buffer with a pH = 10.00, with
both the acid and conjugate base having molarities between 0.10 M to
1.00 M. You may choose from any of the acids listed below:
Name
acetic acid
tartaric acid
citric acid
malonic acid
carbonic acid
phosphoric acid
ammonium
Formula
HC2H3O2
H2C4H4O6
H3C6H5O7
H2C3H2O4
H2CO3
H3PO4
NH41+
Ka1
1.8 x 10-5
1.0 x 10-3
7.4 x 10-4
1.5 x 10-3
4.3 x 10-7
7.5 x 10-3
5.6 x 10-10
Ka2
4.6 x 10-5
1.7 x 10-5
2.0 x 10-6
5.6 x 10-11
6.2 x 10-8
Ka3
4.0 x 10-7
4.2 x 10-13
You must select an acid with a Ka value close to 10- assigned pH.
The only two options are ammonium or the hydrogen carbonate
ions.
2
+
+
H
NH
H
CO
 3


3
-10
-11
K a = 5.6 x 10 =
K a2 = 5.6 x 10 =
+
1NH4
HCO3
 






Making a Buffer Calculations
K a = 5.6 x 10 -10 =
+
H
 NH3 

NH4
5.6 x 10 -10 NH3 
=
-10
10  NH4 +

+


5.6 x 10 -10 NH3 
=
+
+
H
NH
 
4


5.6 NH 3 
=
+
1
NH 4


0.56 NH3 
Divide both [ ] by 10.....
=
+
0.1
NH4



0.1 mol NH + 53.5 g NH Cl 
4
4
x g NH 4 Cl = 500 mL 
= 2.68 g NH 4 Cl

+ 
 1000 mL  1 mol NH 4 

0.56 mol NH 3 1000 mL conc NH 3 
x mL conc NH 3 = 500 mL buffer 


1000 mL buff  14.8 mol NH 3 
=18.9 mL 14.8 M NH3
Making a Buffer Calculations
Place ~250 mL of distilled water in a 500 mL volumetric flask.
Add 2.68 g NH4Cl and dissolve. Then mix in 18.9 mL of 14.8 M
NH3. Fill with distilled water to the 500 mL mark on the flask.
If there is no concentrated NH3 available, the NH3 can be
produced by neutralizing additional NH4Cl with 1.00 M NaOH.
0.56 mol NH 3 1 mol NH 4 Cl 58.5 g NH 4 Cl 
x g NH 4 Cl = 500 mL buffer 



1000 mL buff  1 mol NH 3 1 mol NH 4 Cl 
=16.38 g NH 4 Cl
0.56 mol NH 3 1 mol NaOH 1000 mL NaOH 
x mL NaOH = 500 mL buffer 



1000 mL buff  1 mol NH 3 1.00 mol NaOH 
= 280. mL NaOH
Dissolve 19.06 g NH4Cl (2.68 g + 16.38 g) in 280. mL of 1.00 M NaOH.
Then dilute with distilled water and fill to the 500 mL mark on the flask.
Making a Buffer Calculations
Place ~250 mL of distilled water in a 500 mL volumetric flask.
Add 2.68 g NH4Cl and dissolve. Then mix in 18.9 mL of 14.8 M
NH3. Fill with distilled water to the 500 mL mark on the flask.
If there is no NH4Cl available, the NH4+ can be produced by
neutralizing additional NH3 with 1.00 M HCl.
0.10 mol NH  1 mol NH 1000 mL NH 
4
3
3
x mL NH 3 = 500 mL buffer 





1000 mL buff 1 mol NH 4 14.8 mol NH 3 
= 3.38 mL NH 3
0.10 mol NH   1 mol HCl 1000 mL HCl 
4
x mL HCl = 500 mL buffer 



 
1000 mL buff 1 mol NH 4 1.00 mol HCl 
= 50.0 mL HCl
Place ~250 mL distilled water in volumetric flask. Add 50.0 mL of 1.00 M
HCl and mix. Then add 22.3 mL of concentrated NH3 (18.9 mL + 3.4 mL).
Mix and fill with distilled water to the 500 mL mark on the flask.
Making a Buffer Calculations
K a2 = 5.6 x 10
-11
-11
=



H +  CO 3

1-
HCO3
CO 3
2
2


5.6 x 10
=
-10
110
HCO
 

3


5.6 x 10
+
H
 


-11
CO 3


=
HCO 
CO 3
2
13
2


0.56
=
11
HCO 3


CO 3
2-


0.100
Multiply both [ ] by 0.179.....
=
0.179 HCO 31
Prepare 500. mL of the buffer that has [CO32] = 0.100 M and [HCO31-] = 0.179 M.

Making a Buffer Calculations
Prepare 500. mL of the buffer that has [CO32] = 0.100 M and [HCO31-] = 0.179 M.
0.100 mol CO 2- 106 g Na CO 
3
2
3
x g Na 2CO 3 = 500 mL 

2- 
 1000 mL
 1 mol CO 3 
= 5.30 g Na 2CO 3
0.179 mol HCO 1- 84.0 g NaHCO 
3
3
x g NaHCO 3 = 500 mL 

1- 
1000 mL

 1 mol HCO 3 
= 7.52 g NaHCO 3
Place ~250 mL distilled water in a 500 mL volumetric flask.
Add 5.30 g Na2CO3 and 7.52 g NaHCO3 and dissolve. Fill
with distilled water to the 500 mL mark on the flask.