# What is the effect of buffer concentration on pH?

Sep 20, 2017

Well, it depends on how you compose your buffer....

#### Explanation:

${\log}_{10} {K}_{a} = {\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

(Why? Because ${\log}_{10} A B = {\log}_{10} A + {\log}_{10} B$.)

Rearranging,

$- {\log}_{10} \left[{H}_{3} {O}^{+}\right] - {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\} = - {\log}_{10} {K}_{a}$

But BY DEFINITION, $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$
, and $- {\log}_{10} {K}_{a} = p {K}_{a}$.

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

When we have EQUAL CONCENTRATIONS of ${A}^{-}$ and $H A$, clearly $p H = p {K}_{a}$. Why? Because $\left[{A}^{-}\right] = \left[H A\right]$, and thus ${\log}_{10} \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right) = {\log}_{10} 1 = 0$.

Most of the time, the $p H$ remains tolerably close to the $p {K}_{a}$ of the weak acid used in the buffer.