# How would you rearrange the Henderson–Hasselbalch equation to find out [a/ha] from pH=pKa + log[a/ha] ?

Jan 2, 2016

Here's how you would do that.

#### Explanation:

The most common form of the Hendeson - Hasselbalch equation allows you to calculate the pH of a buffer solution that contains a weak acid and its conjugate base

color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))" "

Here $p {K}_{a}$ is equal to

$\textcolor{b l u e}{p {K}_{a} = - \log \left({K}_{a}\right)} \text{ }$, where

${K}_{a}$ - the acid dissociation constant of the weak acid.

So, for a generic weak acid - conjugate base buffer

${\text{HA"_text((aq]) + "H"_3"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "A}}_{\textrm{\left(a q\right]}}^{-}$

The pH of the solution will be

"pH" = pK_a + log( (["A"^(-)])/(["HA"]))

Now, in order to determine the ratio that exists between the concentration of the conjugate base, ${\text{A}}^{-}$, and the concentration of the weak acid, $\text{HA}$, you will need to isolate the log term on one side of the equation

log( (["A"^(-)])/(["HA"])) = "pH" - pK_a

Now, can say that if $x = y$, then

${10}^{x} = {10}^{y}$

This means that the above equation will be equivalent to

10^(log( (["A"^(-)])/(["HA"]))) = 10^("pH" - pK_a)

But since

$\textcolor{b l u e}{{10}^{{\log}_{10} \left(x\right)} = x}$

you will end up with

color(green)((["A"^(-)])/(["HA"]) = 10^("pH" - pK_a))