# How do you calculate Keq from pKa values?

Jun 26, 2017

Well, a ${K}_{a}$ value is a measure of the following equilibrium.......

$H A \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {A}^{-}$

#### Explanation:

$H A \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {A}^{-}$

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

Now $p {K}_{a} = - {\log}_{10} {K}_{a}$......

And thus ${K}_{a} = {10}^{- p {K}_{a}}$

And thus where $p {K}_{a}$ is small (negative), this specifies a strong acid where the given equilibrium lies to the RIGHT.............

Given ${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{A}^{-}\right]}{\left[H A \left(a q\right)\right]}$ we can take ${\log}_{10}$ of both sides to give......

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

On rearrangement.............

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

And thus......

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

This is a form of the $\text{buffer equation}$, which is used when a weak acid is mixed with its conjugate base in appreciable quantities. Protonation (or deprotonation) of the base or the acid moderates GROSS changes in $p H$, and solution $p H$ remains tolerably close to the $p {K}_{a}$ value.