# Why do solutions that contain buffers tend to resist pH changes?

Sep 2, 2016

Because $\left[{H}_{3} {O}^{+}\right]$ or $\left[H {O}^{-}\right]$ is moderated by the equilibrium expression.

#### Explanation:

We can represent the acid base equilibrium of a weak acid as:

$H A + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {A}^{-}$

$p H = p {K}_{a} + {\log}_{10} \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right)$

Because the ratio ([A^-])/[[HA] is moderated by the logarithmic term, $p H$ remains tolerably close to $p {K}_{a}$, the negative logarithm of the acid dissociation constant. Where $\left[{A}^{-}\right] = \left[H A\right]$, $p H = p {K}_{a}$, because ${\log}_{10} \left(1\right) = 0$.