buffer solutions

Analytical Chemistry Dr. Mohamed E. I. Badawy

Buffer Solutions

1. Introduction:

A buffer solution is a solution that resists a change in its pH upon the addition of

small quantities of either an acid or a base. An example of such a solution is one

containing a weak acid and the salt of the weak acid, or a solution of a weak base and

the salt of the weak base. A solution containing NH4Cl with NH3 will be a buffer

solution.

We will look at a buffer consisting of a weak acid and its conjugate base,

CH3COOH(aq) and CH3COO- (aq). The addition of a small amount of a strong acid

will increase the hydronium ion present in the solution. This amount of hydronium ion

will react with an equal amount of the anion of the weak acid to form un-ionized weak

acid.

CH3COO- (aq) + H3O+ (aq) ↔ CH3COOH(aq) + H2O This removes the added hydronium ion and changes the equilibrium amounts of the

weak acid and its anion.

The addition of a small amount of strong base will increase the hydroxide ion

present in the solution.

CH3COOH(aq) + OH- (aq) ↔ CH3COO- (aq) + H2O This amount of hydroxide ion will react with an equal amount of the weak acid to

form an increase in the amount of the anion of the weak acid (which is a weaker base

than the hydroxide). This removes the added hydroxide ion and changes the

equilibrium amounts of weak acid and its anion. In this way the increased amounts of

hydrogen or hydroxide ion are replaced by the weaker acid molecule or weaker base

ion.

Such solutions are called buffers, and their buffering action is a consequence of

the relationship between pH and the relative concentrations of the conjugate weak

acid/weak base pair.

Buffer: A solution containing a conjugate weak acid/weak base pair that is resistant

to a change in pH when a strong acid or strong base is added.

A mixture of acetic acid and sodium acetate is one example of an acid/base buffer.

The equilibrium position of the buffer is governed by the reaction:

CH3COOH(aq) + H2O(l) ↔ H3O+(aq) + CH3COO–(aq)

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and its acid dissociation constant is:

The relationship between the pH of an acid–base buffer and the relative amounts

of CH3COOH and CH3COO– is derived by taking the negative log of both sides of

equation and solving for the pH as follows:

Buffering occurs because of the logarithmic relationship between pH and the ratio

of the weak base and weak acid concentrations. For example, if the equilibrium

concentrations of CH3COOH and CH3COO– are equal, the pH of the buffer is 4.76. If

sufficient strong acid is added such that 10% of the acetate ion is converted to acetic

acid, the concentration ratio [CH3COO–]/[CH3COOH] changes to 0.818, and the pH

decreases to 4.67.

2. Systematic Solution to Buffer Problems:

Equation of pH calculation for buffer acetate is written in terms of the

concentrations of CH3COOH and CH3COO– at equilibrium. A more useful

relationship relates the buffer’s pH to the initial concentrations of weak acid and weak

base. A general buffer equation can be derived by considering the following reactions

for a weak acid, HA, and the salt of its conjugate weak base, NaA.

NaA(aq) → Na+(aq) + A–(aq)

HA(aq) + H2O(l) ↔ H3O+(aq) + A–(aq)

2H2O(l) ↔ H3O+(aq) + OH–(aq)

Since the concentrations of Na+, A–, HA, H3O+, and OH– are unknown, five

equations are needed to uniquely define the solution’s composition. Two of these

equations are given by the equilibrium constant expressions:

The remaining three equations are given by mass balance equations on HA and Na+:

CHA + CNaA ↔ [HA] + [A–]

CNaA = [Na+]

and a charge balance equation:

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[H3O+] + [Na+] ↔ [OH–] + [A–]

Substituting equation into the charge balance equation and solving for [A–] gives

[A–] = CNaA – [OH–] + [H3O+]

which is substituted into equation to give the concentration of HA as follows:

[HA] = CHA + [OH–] – [H3O+]

Finally, substituting these equations into the Ka equation for HA and solving for pH

gives the general buffer equation:

If the initial concentrations of weak acid and weak base are greater than [H3O+] and

[OH–], the general equation simplifies to the Henderson–Hasselbalch equation:

[ ]⎣ ⎦Acid

SaltLogpKpH a +=

or

[ ]⎣ ⎦BaseSaltLogpKpOH b +=

The Henderson–Hasselbalch equation provides a simple way to calculate the pH of a

buffer and to determine the change in pH upon adding a strong acid or strong base.

Henderson–Hasselbalch equation

Equation showing the relationship between a buffer’s pH and the relative amounts of

the buffer’s conjugate weak acid and weak base.

Example:

Calculate the pH of a buffer that is 0.020 M in NH3 and 0.030 M in NH4Cl. What

is the pH after adding 1.00 mL of 0.10 M NaOH to 0.10 L of this buffer?

Solution:

The acid dissociation constant for NH4+ is 5.70×10–10; thus the initial pH of the

buffer is:

Adding NaOH converts a portion of the NH4

+ to NH3 due to the following reaction:

NH4+(aq) + OH–(aq) ↔ NH3(aq) + H2O(l)

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Since the equilibrium constant for this reaction is large, we may treat the reaction

as if it went to completion. The new concentrations of NH4+ and NH3 are therefore:

Substituting the new concentrations into the Henderson–Hasselbalch equation gives a

pH of:

3. Questions:

1. Calculate the pH of a buffer solution, which contains 0.200 M formic acid (Ka =

2.1×10−4) and 0.150 M sodium formate.

2. Calculate the change in pH of the buffer solution in Question 1 when 0.01000 M of

sodium hydroxide is added to the solution.

3. Calculate the volume of 0.200 M of sodium hydroxide which must be added to

100.0 ml of 0.150 M of acetic acid (CH3COOH, Ka= 1.8×10-5) in order to

prepare a buffer solution with pH = 5.00.

4. The pH of a buffer solution containing 0.0100 M benzoic acid (C6H5COOH, Ka =

6.6×10−5) and 0.0100 M sodium benzoate is:

a. 5.00, b. 4.18, c. 9.82, d. 9.00

* In the problems below equal volumes of the following solutions A and B are mixed:

5. A: 0.100 M CH3COOH (Ka= 1.8×10−5), B: 0.0500 M NaOH,

(i) The final solution:

a. contains a weak acid,

b. contains a strong base,

c. is a buffer solution,

d. none of the above

(ii) The pH of the final solution is:

a. 3.02, b. 4.44, c. 3.17, d. 7.00 6

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A: 0.100 M CH3COOH (Ka= 1.8×10−5), B: 0.150 M NaOH







a. 12.00, b. 12.70, c. 13.18, d. 12.40 7.

A: 0.150 M CH3COOH, B: 0.100 M NaOH







a. 3.17, b. 3.02, c. 2.78, d. 3.22 8.

A: 0.100 M CH3COOH, B: 0.100 M NaOH,







a. 7.00, b. 13.00, c. 2.87, d. 3.02

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buffer solutions

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