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Buffer Solutions

A buffer solution is one in which the pH of the solution is "resistant" to small additions of either a strong acid or strong base.  Buffers usually consist of a weak acid and its conjugate base, in relatively equal and "large" quantities.  Calculations are based on the equation for the ionization of the weak acid in water forming the hydronium ion and the conjugate base of the acid.  "HA" represents any weak acid and "A-" represents the conjugate base.

HA(aq) + H2O(l) --> H3O+(aq) + A-(aq)

Ka = [H3O+][A-]

                                                                                               [HA]

A buffer system can be made by mixing a soluble compound that contains the conjugate base with a solution of the acid such as sodium acetate with acetic acid or ammonia with ammonium chloride.  The above equation for Ka can be rearranged to solve for the hydronium ion concentration.  By knowing the Ka of the acid, the amount of acid, and the amount of conjugate base, the pH of the buffer system can be calculated.

[H3O+] = Ka[HA]
                                                                                                      [A-]
pH = -log[H3O+]


Calculation of the pH of a Buffer Solution

In order to calculate the pH of the buffer solution you need to know the amount of acid and the amount of the conjugate base combined to make the solution.  These amounts should be either in moles or in molarities.  The Ka of the acid also needs to be known.

Example:  A buffer solution was made by dissolving 10.0 grams of sodium acetate in 200.0 mL of  1.00 M acetic acid.  Assuming the change in volume when the sodium acetate is not significant, estimate the pH of the acetic acid/sodium acetate buffer solution.  The Ka for acetic acid is 1.7 x 10-5.

CH3COOH(aq) + H2O(l) --> H3O+(aq) + CH3COO-(aq)

[H3O+] = Ka[CH3COOH]

                                                                                                [CH3COO-]
(1.00 M CH3COOH)(200.0 mL)(1 L/1000 mL) = 0.200 mol CH3COOH

(10.0 g NaCH3COO)(1 mol/82.03 g) = 0.122 mol NaCH3COO

[H3O+] = (1.7 x 10-5)(0.200/0.122) = 2.79 x 10-5
pH = 4.56

Example:  Calculate the ratio of ammonium chloride to ammonia that is required to make a buffer solution with a pH of 9.00.  The Ka for ammonium ion is 5.6 x 10-10.

NH4+(aq) + H2O(l) --> H3O+(aq) + NH3(aq)

Ka = [H3O+][NH3]

                                                                                              [NH4+]

[H3O+] = Ka[NH4+]

                                                                                                     [NH3]
[H3O+] = 1 x 10-9 M

1 x 10-9 = 5.6 x 10-10(NH4+/NH3)
(NH4+/NH3) = 1.786/1

A ratio of 1.768 moles of ammonium ion for every 1 mole of ammonia or 1.768 M ammonium ion to 1 M ammonia.

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Calculation of the pH of a Buffer Solution after Addition of a Small Amount of Acid

When a strong acid (H3O+) is added to a buffer solution the conjugate base present in the buffer consumes the hydronium ion converting it into water and the weak acid of the conjugate base.

A-(aq) + H3O+(aq) --> H2O(l) + HA(aq)

This results in a decrease in the amount of conjugate base present and an increase in the amount of the weak acid.  The pH of the buffer solution decreases by a very small amount because of this ( a lot less than if the buffer system was not present).  An "ICE" chart is useful in determining the pH of the system after a strong acid has been added.

Example:  50.0 mL of 0.100 M HCl was added to a buffer consisting of 0.025 moles of sodium acetate and 0.030 moles of acetic acid.  What is the pH of the buffer after the addition of the acid?  Ka of acetic acid is 1.7 x 10-5.

CH3COOH(aq) + H2O(l) --> H3O+(aq) + CH3COO-(aq)

[H3O+] = Ka[CH3COOH]

                                                                                                [CH3COO-]
 
CH3COOH(aq)
H3O+(aq)
CH3COO-(aq)
Initial Amount
0.030 moles
(0.0500 L)(0.100 M) = 0.0050 moles
0.025 moles
Change in Amount
+ 0.005 moles
-0.005 moles
- 0.005 moles 
Equilibrium Amount
0.035 moles
x
0.020 moles
[H3O+] = (1.7 x 10-5)(0.035/0.020) = 2.975 x 10-5
pH = 4.53

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Calculation of the pH of a Buffer Solution after Addition of a Small Amount of Strong Base

When a strong base (OH-) is added to a buffer solution, the hydroxide ions are consumed by the weak acid forming water and the weaker conjugate base of the acid.  The amount of the weak acid decreases while the amount of the conjugate base increases.  This prevents the pH of the solution from significantly rising, which it would if the buffer system was not present.

OH-(aq) + HA(aq) --> H2O(l) + A-(aq)

The process for finding the pH of the mixture after a strong base has been added is similar to the addition of a strong acid shown in the previous section.

Example:  Calculate the pH of a buffer solution that initially consists of 0.0400 moles of ammonia and 0.0250 moles of ammonium ion, after 20.0 mL of 0.75 M NaOH has been added to the buffer.  Ka for ammonium ion is 5.6 x 10-10.

NH4+(aq) + H2O(l) --> H3O+(aq) + NH3(aq)

[H3O+] = Ka[NH4+]

                                                                                                    [NH3]
 
NH4+(aq)
H3O+(aq)
NH3(aq)
Initial Amount
0.0250 moles
* not needed
0.0400 moles
Change in Amount
- 0.0015 moles
* not needed
+ 0.0015 moles
Equilibrium Amont
0.0235 moles
x
0.0415 moles
[H3O+] = (5.6 x 10-10)(0.0235/0.0415) = 3.17 x 10-10
pH = 9.50

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Calculation of the Buffer Capacity

The buffer capactity refers to the maximum amount of either strong acid or strong base that can be added before a significant change in the pH will occur.  This is simply a matter of stoichiometry.  The maximum amount of strong acid that can be added is equal to the amount of conjugate base present in the buffer.  The maximum amount of base that can be added is equal to the amount of weak acid present in the buffer.

Example:  What is the maximum amount of acid that can be added to a buffer made by the mixing of 0.35 moles of sodium hydrogen carbonate with 0.50 moles of sodium carbonate?  How much base can be added before the pH will begin to show a significant change?
 

HCO3-(aq) + H2O(l) --> H3O+(aq) + CO32-(aq)
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