18.2 – Buffer Solutions

18.2 – Buffer Solutions

18.2.1 – Describe the composition of a buffer solution and explain its action

Buffer solutions maintain the pH of solutions by resisting changes caused by the addition of an acid or base.

Acidic Buffer Solution

This is formed from a weak acid and salt of its conjugate base.

In this equation, HA is the weak acid and A^– is the conjugate base. The equilibrium of the reaction favours the left hand side, so there is a larger proportion of the weak acid.

The reaction with the salt is shown below:

Here, MA is the salt and MOH is the strong base. The salt will then dissociate into its ions, M⁺ and A^–

Basic Buffer Solution

This is formed from a weak base and a salt of its conjugate acid.

Here, B is the weak base and BH⁺ is the conjugate acid. The negative ion from a strong acid is added:

The salt, BHX, will dissociate completely in solution to form BH⁺ and X^–. As a result, the solution will have high proportions of both B and BH.

If a small amount of acid is added to the solution, it will react with the B to counteract the change in pH. Likewise, if a base is added, the HB⁺ dissociates to provide more H⁺ ions, thereby restoring the pH.

18.2.2 – Solve problems involving the composition and pH of a specified buffer system

Ethanoic Acid and Sodium Ethanoate

The buffer is a mixture of the weak acid CH₃COOH and the salt of the acid with a strong base CH₃COONa:

To do the calculations, we assume:

The weak acid only dissociates a tiny amount, and can therefore be disregarded.

[HA]initial ≈ [HA]equilibrium

The salt completely dissociates into its ions

[MA]initial ≈ [A^–]equilibrium

The equilibrium expression for the acid (using the equilibrium concentration values) is:

Which can be rearranged to give the formula:

Based on the assumptions made above, this can be substituted to give:

When the -log₁₀ is taken of this equation, it gives the equation:

So, continuing with the example of ethanoic acid and sodium ethanoate:

The pK_a of ethanoic acid is 4.76
10 mol dm^-3 ethanoic acid
15 mol dm^-3 sodium ethanoate

Ammonia and Ammonium Chloride

For a basic buffer solution:

All the same principles of an acidic buffer solution apply to basic buffer solutions.

An example of calculations with a basic buffer solution:

You may be given the amounts of weak base and strong acid mixed to form the salt

200cm³ of 0.10 mol dm^-3 NH₃
200cm³ of 0.050 mol dm^-3 HCl
pK_b = 4.75

Using n = CV, we find that the HCl is the limiting reagent, with 0.01mol present. Therefore, the amount of NH₃ remaining after the reaction is 0.01mol.

Since the volume of the solution is 400cm³, then the concentrations would be:

[NH₃] = 0.025 mol dm^-3
[NH₄Cl] = 0.025 mol dm^-3

Note that diluting a buffer solution will not change its pH, but it does reduce its buffering capacity. Also, the temperature will affect the K_a and K_b, and hence the pH of the solution.