# 10.9: Buffers

Buffer solutions, which are of enormous importance in controlling pH in various processes, can be understood in terms of acid/base equilibrium. A buffer is created in a solution which contains both a weak acid and its conjugate base. This creates to absorb excess H+ or supply H+ to replace what is lost due to neutralization. The calculation of the pH of a buffer is straightforward using an ICE table approach.

Example $$\PageIndex{1}$$:

What is the pH of a solution that is 0.150 M in KF and 0.250 M in HF?

Solution

The reaction of interest is

$HF \rightleftharpoons H^+ + F^-$

Let’s use an ICE table!

$$HF$$ $$H^+$$ $$F^-$$
Initial 0.250 M 0 0.150 M
Change -x +x +x
Equilibrium 0.250 M - x x 0.150 M + x

$K_a = \dfrac{[H^+][F^-]}{[HF]}$

$10^{-3.17} M = \dfrac{x(0.150 \,M + x)}{0.250 \,M - x}$

This expression results in a quadratic relationship, leading to two values of $$x$$ that will make it true. Rejecting the negative root, the remaining root of the equation indicates

$[ H^+]= 0.00111\,M$

So the pH is given by

$pH = -log_{10} (0.00111) = 2.95$

For buffers made from acids with sufficiently large values of pKa the buffer problem can be simplified since the concentration of the acid and its conjugate base will be determined by their pre-equilibrium values. In these cases, the pH can be calculated using the Henderson-Hasselbalch approximation.

If one considers the expression for $$K_a$$

$K_a = \dfrac{[H^+][A^-]}{[HA]} = [H^+]\dfrac{[H^-]}{[HA]}$

Taking the log of both sides and multiplying by -1 yields

$pK_a= pH - \log_{10} \dfrac{[A^-]}{[HA]}$

An rearrangement produces the form of the Henderson-Hasselbalch approxmimation.

$pH= pK_a - \log_{10} \dfrac{[A^-]}{[HA]}$

It should be noted that this approximation will fail if:

1. the $$pk_a$$ is too small,
2. the concentrations $$[A^-]$$ is too small, or
3. $$[HA]$$ is too small,

since the equilibrium concentration will deviate wildly from the pre-equilibrium values under these conditions.