# Question #96889

Jun 17, 2016

So simply take a weak acid, $H A$, and add approx 1/2 an equiv of sodium hydroxide solution. Alternatively, take the salt of a weak acid, $N {a}^{+} {A}^{-}$, and again add 1/2 an equiv of strong mineral acid. The $p H$ of the solution will remain tolerably close to the $p {K}_{a}$ of the starting acid.
$p H = p {k}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$