Question

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

Answer 1

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 `pK_a` is equal to

`color(blue)(pK_a = - log(K_a))" "`, where

`K_a` - the acid dissociation constant of the weak acid.

So, for a generic weak acid - conjugate base buffer

`"HA"_text((aq]) + "H"_3"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "A"_text((aq])^(-)`

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, `"A"^(-)`, and the concentration of the weak acid, `"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

`color(blue)(10^(log_10(x)) = x)`

you will end up with

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