Definition: A buffer is a mixture of approximately equal concentrations of a weak acid and its conjugate base. Buffers are important in chemistry because they have the ability to maintain a relatively constant pH. This occurs because the weak acid portion of the buffer will neutralize base added to the buffer mixture, while the conjugate base portion will neutralize additional acid. Since the additional acid or base is largely neutralized, the pH of the solution remains the same. An example of a buffered system would be blood, where H2CO3 and HCO3- form a buffer that keeps the pH of blood realitively constant.
Calculations: A buffer works because the weak acid and its conjugate base are in equilibrium. If we use HA to represent the weak acid, and A- to represent the conjugate base, the equilibrium between the two can be described as
HA <===> H+ + A-
Since the system is at equilbrium, we can come up with an equation for the equilibrium constant. Since this is a special class of equilibrium dealing with the ionization of an acid it is usually abbreviated Ka (as opposed to the more general Keq).
Ka = [H+] [A-] / [HA]
Rearranging the equation to solve for the [H+] yields
[H+] = Ka [HA] / [A-]
Taking the negative log of both sides gives the form of the equation called the Henderson-Hasselbalch equation) which is very usefule since it provides us with a simple way of determining the pH of a buffer mixture.
pH = pKa + log ( [A-] / [HA] )
This equation shows us that the pH of a buffer depends only on the pKa of the acid and the log of the ratio of the conjugate base to the unionized acid. In a well constucted buffer, the concentrations of the conjugate base and the unionized acid would be approximately equal giving a ratio equal to or fairly close to 1. Since the log 1 is zero, the pH of most buffers will be very close to the pKa of the weak acid.
Making a Buffer
Buffers can be constructed by adding equal amounts of a weak acid and a salt containing its conjugate base. For instance, by adding 50 mL of 1.0 M acetic acid to 50 mL of 1.0 M sodium acetate, you would create 100 mL of an acetic acid/acetate ion buffer where the ratio of the conjugate base to the weak acid would be 1. The pH of this buffer would then be equal to the pKa for acetic acid (4.74).
Buffers can also be constructed by taking a weak acid and adding sufficient strong base to neutralize half of it (and vice versa). For instance, the buffer above could be generated by starting with 50 mL of 2.00 M acetic acid and adding 50 mL of 1.00 M NaOH. The NaOH will react with the weak acid according to the following equation:
HA + OH- ---> A- + H2O
Since .100 moles of acid was added originally , and .050 moles of OH- were added, half of the weak acid would be neutralized resulting in the concentraions of both the acid and its conjugate being .5M. Once again the ratio [A-]/[HA] is equal to 1 and the pH will be equal to the pKa (4.74)
Adjusting the pH of a buffer
In some cases, one might wish to make a buffer that has a slightly different pH value than the pKa of the weak acid. This can be done by adjusting the concentrations of the weak acid and its conjugate base so that the ratio [A-] / [HA] is no longer exactly equal to 1. An acetic acid/acetate buffer in which the [A-] = .4M and the [HA] = .5M yields the follow:
pH = pKa + log ( [A-] / [HA]) = 4.74 + log (.4/.5) = 4.64
It is important to remember that changing the ratio of [A-] / [HA] drastically can affect the ability to maintain constant pH. In such a case, finding another weak acid with a pKa closer to the desired pH would be more appropriate.
Problem 1 - What would be the pH of a buffer solution formed from citric acid and sodium citrate when the molarity of both the acid and the citrate ion is 1.0 M?
Solution - Use the Henderson-Hasselbalch equation:
pH = pKa + log ( [A-] / [HA])
Since the Ka of citric acid is 3.2 x 10-7, the pKa of citric acid would be -log(3.2 x 10-7) or 6.49.
Since [A-] / [HA] = 1 and the log of 1 is equal to 0.
pH = 6.49
Problem 2 - 15mL of 1.0 M NaOH is added to 40mL of 1.0 M HC2H3O2. What would be the pH of the buffer that is formed?
Solution - In this case before we can use the Henderson-Hasselbalch equation we need to find the concentrations of the acid and its conjugate base
The strong base NaOH will neutralize a portion of the weak acetic acid forming the conjugate base according to the following equation:
HC2H3O2 + NaOH --> NaC2H3O2 + H2O
The number of moles of C2H3O2- that will be formed will be equal to the number of moles of NaOH added.
moles of C2H3O2- = moles of NaOH added = .015L x 1.0M = .015 moles
The number of moles of acid remaining will be equal to the original moles of acid (.040L x 1.0M) minus the acid neutralized by the NaOH
moles of HC2H3O2 = .040 moles - .015 moles = .025 moles
Strictly speaking the Henderson-Hasselbalch equation calls for the concentration of the acid and its conjugate base. To get the concentrations of each species we divide the volume in liters
Molarity of HC2H3O2 = .025 mol/ .055L = 4.54M
Molarity of C2H3O2- = .015 mol/.055 L = 2.72 M
Now we are ready to plug into the Henderson-Hasselbalch equation
pH = pKa + log ( [A-] / [HA]) = 4.74 + log ( 2.72M / 4.54M)
pH = 4.74 + log ( .600) = 4.74 - .22 = 4.52