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

Here’s how you would do that.

The most common form of the Hendeson – Hasselbalch equation allows you to calculate the of a 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))##

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