14.6 Buffers
What is a buffer?
A buffer solution is a solution made of a combination of a weak acid and its conjugate base or a weak base with its conjugate acid. This combination solution resists the change in the pH of the solution and maintains constant pH of the solution throughout. It neutralizes any incoming acid or base to the original solution.
Examples of such combinations:
A buffer made of acetic acid and sodium acetate:
CH3COOH (aq) + H2O (l) 
H3O + (aq) + CH3COO – (aq)
Acid base Conjugate acid Conjugate base
A buffer made of ammonia and ammonium nitrate
NH3 (aq) + H2O (l) ß-à OH – (aq) + NH4 + (aq)
Base acid Conjugate base Conjugate acid
How does the buffer work?
The conjugate acid Conjugate acid or conjugate base Conjugate base can react and neutralize any incoming acid H3O + (aq) or incoming base OH – (aq).
Example:
CH3COOH (aq) + H2O (l) H3O + (aq) + CH3COO – (aq)
In case of incoming acid H3O + (aq):
The equilibrium will shift to the left which means more CH3COOH (aq) will be produced on the expense of the amount of CH3COO – (aq). The ratio of CH3COO – (aq) concertation to CH3COOH (aq) concentration does not change drastically and it remains almost constant.
In case of incoming acid OH – (aq):
The equilibrium will shift to the right because OH – (aq) will react with H3O + (aq) and hence more CH3COO – (aq) concentration will be produced on the expense of CH3COOH (aq). The ratio of CH3COO – (aq) concertation to CH3COOH (aq) concentration does not change drastically and it remains almost constant.
This buffer function can be illustrated in the graph below:
It can be seen from the graph; the change of pH is minimal and not drastic over the Buffer Range. When the buffering capacity is exceeded, then buffering feature is broken and a quick pH change is observed. This observation is based on the face that the conjugate acid is totally consumed. Therefore a larger amount of conjugate acid is expected to have greater buffering capacity.
pH Changes in Buffered and Unbuffered Solutions
Example:
The buffer solution made of acetic acid and sodium acetate has a pH of 4.7 to 5. It used very common in the undergraduate laboratories.
The example below is taken from OpenStax eBook second edition: Chapter 14 | Acid-Base Equilibria
a. Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.
b. Calculate the pH after 1.0 mL of 0.10 NaOH is added to 100 mL of this buffer.
c. For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an
unbuffered solution with a pH of 4.74.
Solution
- Following the ICE approach to this equilibrium calculation yields the following:
Substituting the equilibrium concentration terms into the Ka expression, assuming x <<< 0.10, and solving
the simplified equation for x yields
x = 1.8 × 10 − 5 M
[H3 O +] = 0 + x = 1.8 × 10 − 5 M
pH = −log[H3O +] = − log(1.8 × 10 − 5) = 4.74
- Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer.
Adding strong acid will neutralize some of the acetic acid, yielding the conjugate base acetate ion. Compute
the new concentrations of these two buffer components, then repeat the equilibrium calculation of part a
using these new concentrations.
0.0010 L × [ 0.010 mol NaOH ] / L] = 1.0 × 10−4 mol NaOH0.10 mol NaOH = 1.0 × 10 −4 mol NaOH
The initial molar amount of acetic acid is
[ 0.100 L × 0.100 mol CH3COOH / L] = 1.00 × 10 −2 mol CH3COOH
The amount of acetic acid remaining after some is neutralized by the added base is
(1.0 × 10 −2) − (0.01 × 10 −2) = 0.99 × 10 −2 mol CH3COOH
The newly formed acetate ion, along with the initially present acetate, gives a final acetate concentration of
(1.0 × 10 −2) + (0.01 × 10 −2) = 1.01 × 10 −2 mol NaCH3COO
Compute molar concentrations for the two buffer components:
[CH3COOH] = 9.9 × 10 −3 mol / 0.101 L = 0.098 M
[NaCH3COO] = 1.01 × 10 −2 mol / 0.101 L = 0.100 M
Using these concentrations, the pH of the solution may be computed as in part a above, yielding pH = 4.75
(only slightly different from that prior to adding the strong base).
- For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an
unbuffered solution with a pH of 4.74.
The amount of hydrogen ion initially present in the solution is
[H3O +] = 10 − 4.74 = 1.8 × 10 −5 M
mol H3O + = (0.100 L) x (1.8 × 10 −5 M) = 1.8 × 10 −6 mol H3O +
The amount of hydroxide ion added to the solution is mol OH − = (0.0010 L) x (0.10 M) = 1.0 × 10−4 mol OH −
The added hydroxide will neutralize hydronium ion via the reaction
H3O+ (aq) + OH− (aq) 2 H2O (l)
The 1:1 stoichiometry of this reaction shows that an excess of hydroxide has been added (greater molar
amount than the initially present hydronium ion).
The amount of hydroxide ion remaining is
1.0 × 10 −4 mol − 1.8 × 10 −6 mol = 9.8 × 10 −5 mol OH−
corresponding to a hydroxide molarity of
9.8 × 10 −5 mol OH − /0.101 L = 9.7 × 10 −4 M
The pH of the solution is then calculated to be
pH = 14.00 − pOH = 14.00 − − log (9.7 × 10−4) = 10.99
In this unbuffered solution, addition of the base results in a significant rise in pH (from 4.74 to 10.99)
compared with the very slight increase observed for the buffer solution in part (b) (from 4.74 to 4.75).
Buffer Capacity:
The buffer solution will resist the change of the pH when small amount of incoming acid or base is added externally to the buffered solution. When the incoming acid or base added externally to the buffered solution in larger amount, the buffer effect or capacity will be broken and the resulted solution will act as a pure acid or a base.
The buffer capacity is the amount of acid or base that can be added externally to a specific volume of a buffer solution before the pH changes drastically.
Selecting proper components for desired pH:
When the pKa conjugate weak acid is matching the desired buffer range, then the buffer functions very well in resisting the change of the pH of the desired buffered solution.
For example, acetic acid Ka is 1.8 x 10 -5 and the pKa is 4.74. and acetic acid buffer would work best in a buffer range of around pH = 4.74
In general, the buffer functions well, when the buffer pair less than 10% of the other
Weak acid and their salts buffers work well for buffering solution with the pH around 7 or less and weak base and their salts buffers work well for buffering solution within a basing range of the pH: 8 to 14.
The Henderson-Hasselbalch Equation:
Let us consider the ionization of acetic acid:
CH3COOH (aq) + H2O (l) CH3COO – (aq) + H3O + (aq)
Ka = { [CH3COO – (aq)] x [H3O + (aq)] } / [CH3COOH (aq)]
Solving for [H3O + (aq)]:
[H3O + (aq)] = Ka x { [CH3COOH (aq)] / [CH3COO – (aq)]
Taking the – log of both sides:
- log [H3O + (aq)] = – log {Ka x { [CH3COOH (aq)] / [CH3COO – (aq)] }
- log [H3O + (aq)] = – log Ka – log [CH3COOH (aq)] / [CH3COO – (aq)]
pH = pKa + log { [CH3COO – (aq)] / [[CH3COO – (aq)] }
This equation is called: The Henderson-Hasselbalch Equation which can be written in a general formula:
pH = pKa + log {[ base] / [acid] }
Example:
How much sodium formate (HCOONa, with molar mass of 68.0069 g/mol ) do you need to add to 400. mL of 1.00 M formic acid for a pH 3.500 buffer. Ka = 1.77 x 10¯4
Solution:
pH = pKa + log [base/acid]
3.500 = 3.752 + log (X / 1.00) ===è [1.00 M is formic acid concentration]
log X = − 0.252
X = 0.560 M =è [this is the molarity of the formate that is required]
Number of moles = molarity multiply by volume = M * V
(0.560 mol/L) * (0.400 L) = # moles = X / 68.0069 (g/mol)
X = 15.2 g