# How does a buffer affect pH?

$p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$
So when $\left[{A}^{-}\right] = \left[H A\right]$, $p H = p {K}_{a}$ because ${\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\} = {\log}_{10} \left(1\right) = 0$