How do you solve chemistry buffer problems?

Aug 18, 2016

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

Explanation:

See this old answer..

As you have probably heard ad nauseaum, at the point of half equivalence in the titration weak acid with a strong base, $\left[{A}^{-}\right] = \left[H A\right]$, and the expression ${\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$ $=$ ${\log}_{10} \left(1\right) = 0$.

Going back to the equation, this means that $p H = p {K}_{a}$ at the point of half equivalence in a strong base/weak acid or strong acid/weak base titration.

Problems are usually composed so that the formation of a buffer is not immediately obvious. For example, we may start with some concentration of say, acetic acid, $H O A c$, and $\frac{1}{2}$ an equiv of sodium hydroxide is added. If the $p {K}_{a}$ of acetic acid $=$ $4.76$ (which it does), then what is the $p H$ of the given buffer?