# What is the Henderson-Hasselbach equation?

Nov 22, 2015

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

#### Explanation:

Consider a weak acid in water:

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

Now the equilbrium constant, ${K}_{a}$, describes this equilibrium:

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

Now, as long as we do it to both sides, we can manipulate this equation. We take logarithms to the base 10 to make it a bit easier:

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

(Note, when we write ${\log}_{a} b = c$, we ask to what power do we raise the base $a$ to get $b$; so ${a}^{c} = b$; likewise ${\log}_{10} 100 = 2 ,$ and ${\log}_{10} 1000 = 3$. Before the days of ready access to electronic calculators, everyone would use log tables to perform multiplication and division.)

Rearranging this equation:

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

But by definition $- {\log}_{10} \left(\left[{H}_{3} {O}^{+}\right]\right) = p H$, and $- {\log}_{10} {K}_{a} = p {K}_{a}$.

So finally, we get:

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

Note that this tells us that at the point of half equivalence of a titration, when $\left[{A}^{-}\right] = \left[H A\right]$, ${\log}_{10} \left(1\right) = 0$ (clearly), and thus $p H = p {K}_{a}$ at half equivalence.