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QUESTION

# 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))