How Buffer Solutions Work

ACIDS AND BASES
BUFFER SOLUTIONS
Common Ion Effect
Adding a common ion to an equilibrium system will shift
the equilibrium forward or reverse.
For a weak acid equilibrium this can affect the pH of the
solution!
For example: Adding NaF to a solution of HF.
HF H + + F-
Adding F- ions will shift the equilibrium to reactants and

produce more HF and reduce [H+] ions in the process.
Remember, NaF is a salt and ionizes completely in
solution.
This makes the pH increase.
The Common Ion Effect
• Consider a solution that contains a weak acid, such as
Acetic Acid (CH3COOH), and a soluble salt of that acid,
such as Sodium Acetate (CH3COONa).
• Both components of the solution contain a common ion

in this case – CH3COO-.
Common Ion Effect
• Sodium Acetate (CH3COONa) is a soluble ionic compound, and is
a strong electrolyte. Consequently it dissociates completely in
aqueous solution:
CH3COONa (aq)  Na+ (aq) + CH3COO- (aq)
• Acetic Acid (CH3COOH) is a weak electrolyte and partially ionizes.
CH3COOH(aq)  H+ (Aq) + CH3COO- (aq)
Using Le Chataliers Principle, what do we predict happens

to the Ka CH3COOH?
Common Ion Effect
• According to Le Chatalier’s Principle, CH3COO- from
CH3COONa causes the equilibrium to shift to the left ,
thereby decreasing the equilibrium concentrations of H+.
This would increase the pH of the solution
CH3COOH(aq)  H+ (Aq) + CH3COO- (aq)
Whenever a weak electrolyte and a strong electrolyte

contain a common ion, the weak electrolyte ionizes less
than it would it if were alone in the solution.
Calculating pH when a Common Ion is Involved
• What is the pH of a solution made by adding 0.30 mol of

acetic acid,CH3COOH, and 0.30 mol of sodium
acetate,CH3COONa to enough water, to make a 1.0L
solution?
o This solution contains a weak electrolyte, and a strong

electrolyte that share a common ion, CH3COO-
Solution:
CH3COOH (aq) H+ (aq) + CH3COO- (aq)
Initial 0.30 0 0.30
Change -x +x +x
Equilibrium 0.30 – x x 0.30 + x
Ka= 1.8 x 10 -5 = [H+][CH3COO-]= (x)(0.30 + x) = x(0.30)

[CH3COOH] (0.30 – x) (0.30)
From the
dissociation of
CH3COONa
x = 1.8 x 10-4 M = [H+]
pH= - log( 1.8 x 10-4) = 4.74

Buffered Solutions
• Solutions which contain a weak conjugate acid-base
pair, can resist drastic changes to pH upon the addition of
strong acids or base. These solutions are called buffered
solutions.
• A buffer resists changes to pH because it contains both an

acid to neutralize OH- ions, and a base to neutralize H+
ions.
• The acid and base cannot consume each other during the
reaction.
10
Explaining a Buffer Solution

• Ex: acetic acid + sodium acetate = buffer
[CH3COOH] and [CH3COO-] are both high
• adding acid or base has little effect because the added H3O+ or OH- ions are
removed by one of the components in the buffer solution
• Acetic acid consumes any added hydroxide ion
CH3COOH(aq) + OH-(aq)  CH3COO-(aq) + H2O(l)
• Acetate ion consumes any added hydronium ion
H3O+(aq) + CH3COO-(aq)  H2O(l) + CH3COOH(aq)
• equilibrium shifts as predicted by Le Chatelier’s principle

http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/buffer12.swf
Examples of Buffer Solutions
Acid or Salt
• Buffers are often prepared by mixing a
Base
weak acid or a weak base, with a salt of
Acetic acid Sodium
that acid or base. acetate
Phosphoric Potassium
• CH3COOH and CH3COO (acidic buffer)
- acid phosphate
(Weak Acid) Salt such as CH3COONa Oxalic acid Lithium

oxalate
• NH3 and NH4+ (basic buffer) Carbonic Sodium

acid carbonate
Salt such as NH4Cl
Ammonium Ammonium
hydroxide nitrate
There are two ways to make buffered solutions !

Trivia Time !
• Which of the following conjugate acid-base pairs will not
function as a buffer. Explain Why!
a) C2H5COOH and C2H5COO-
b) HCO3- and CO3 2-
c) HNO3 and NO3-

1. Calculating the pH of a buffer
What is the pH of a buffer that is 0.12 M in lactic acid
[CH3CH(OH)COOH, or HC3H5O3] and 0.10 M in sodium
lactate [CH3CH(OH)COONa or NaC3H5O3]?
For lactic acid, Ka = 1.4 × 10–4.

Calculating the pH of a buffer
Alternatively, we can use the

Henderson–Hasselbalch equation
with the initial concentrations of acid
and base to calculate pH directly:
** AP**
AP
Henderson-Hasselbach Equation
• Derived from the equilibrium expression of a weak acid
and the pH equation.
[A-]
•
pH = pKa + log[HA]
This equation allows you to determine the pH of a buffer
solution
• *** pKa = -log Ka
AP
• Example: Determine the pH in which 1.00 mole of H2CO3
(Ka = 4.2 x 10-7) and 1.00 mole NaHCO3 dissolved in
enough water to form 1.00 Liters of solution.
[A-]
pH = pKa + log
[HA]
AP
• Example: Determine the pH in which 1.00 mole of H2CO3
(Ka = 4.2 x 10-7) and 1.00 mole NaHCO3 dissolved in
enough water to form 1.00 Liters of solution.
[A-]
pH = pKa + log
[HA]
pH = -log (4.2 x 10-7) + log (1.00M)/(1.00M)
pH = pKa = 6.4
AP
Buffer Capacity
• Buffer Capacity: the amount of acid or base the buffer
can neutralize before the pH begins to change to an
appreciable degree.
• This depends on the amounts of acid and base the buffer

is made of.
• Most efficient buffer is when:

[A-]
[HA] =1
AP
pH Range
• pH Range: the range in pH over which the buffer acts
effectively.
• Buffers most effectively resist a change in pH in either

direction when the concentrations of weak acids and weak
bases are approximately the same.
• When the concentration of acids and bases are equal, pH=

pKa…. This is the optimal pH of any buffer.
• If the concentration of one component is 10x more than the

concentration of the other component, the buffering action is
poor.

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ACIDS AND BASES

Adding F- ions will shift the equilibrium to reactants and

• Both components of the solution contain a common ion

CH3COONa (aq)  Na+ (aq) + CH3COO- (aq)

• Acetic Acid (CH3COOH) is a weak electrolyte and partially ionizes.

CH3COOH(aq)  H+ (Aq) + CH3COO- (aq)

Using Le Chataliers Principle, what do we predict happens

CH3COOH(aq)  H+ (Aq) + CH3COO- (aq)

Whenever a weak electrolyte and a strong electrolyte

• What is the pH of a solution made by adding 0.30 mol of

o This solution contains a weak electrolyte, and a strong

Equilibrium 0.30 – x x 0.30 + x

Ka= 1.8 x 10 -5 = [H+][CH3COO-]= (x)(0.30 + x) = x(0.30)

pH= - log( 1.8 x 10-4) = 4.74

• A buffer resists changes to pH because it contains both an

Explaining a Buffer Solution

• Acetic acid consumes any added hydroxide ion

CH3COOH(aq) + OH-(aq)  CH3COO-(aq) + H2O(l)

• Acetate ion consumes any added hydronium ion

H3O+(aq) + CH3COO-(aq)  H2O(l) + CH3COOH(aq)

• equilibrium shifts as predicted by Le Chatelier’s principle

(Weak Acid) Salt such as CH3COONa Oxalic acid Lithium

• NH3 and NH4+ (basic buffer) Carbonic Sodium

There are two ways to make buffered solutions !

a) C2H5COOH and C2H5COO-

b) HCO3- and CO3 2-

c) HNO3 and NO3-

For lactic acid, Ka = 1.4 × 10–4.

Alternatively, we can use the

• This depends on the amounts of acid and base the buffer

• Most efficient buffer is when:

• Buffers most effectively resist a change in pH in either

• When the concentration of acids and bases are equal, pH=

• If the concentration of one component is 10x more than the

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