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

1 Answer
Jan 2, 2016

Answer:

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))#