# A buffer is prepared by mixing 1*mol of acetic acid with 1*mol of sodium acetate...what is the concentration of the buffer?

Nov 12, 2015

The molarity is $1.0$ $m o l \cdot {L}^{- 1}$ with respect to both acetic acid, and to sodium acetate. On the other hand, if you have mixed equal volumes of acetic acid and sodium acetate, then the concentration is $0.5$ $m o {l}^{- 1}$

#### Explanation:

As you know, this is a buffer solution, whose $p H$ would remain very close to the $p {K}_{a}$ of acetic acid (which is from memory $4.76$).

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

$H O A c$ $=$ ${H}_{3} C - C \left(= O\right) O H$

As I said before, if you have taken equal volumes of $1.0$ $m o l \cdot {L}^{- 1}$ acetic acid and sodium acetate, because you ADDED these volumes, (i.e doubled the volume), the concentration of each reagent is HALVED.