The use of HH becomes somewhat superfluous except for quick, easy ideas of the "region" of the pH, since if you have even a few more seconds available you only need to plug in six more numbers (two times [H+] and one times Kw, on both the top and bottom) to get whatever result you want. Literally, if HH is Ka=([H+]*[salt])/[weak acid], then
[weak acid]=Ca-[H+]+(Kw/[H+])
[salt]=Cs+[H+]-(Kw/[H+])
From the exact Charlot equation. Which means that for anything you'd use the HH for, you need only to plug in 3 more numbers and you'll get the exact result! Thus the HH is ok for a very very approximate guess, e.g. an order of magnitude estimate, but shouldn't even be discussed when you're talking about exact calculations - because it's so easy to just use the more exact version! Of course if you were calculating H+ it would be perfectly excusable just to solve the HH equation because a cubic is not great.
Please explain - in more details - the procedure you are proposing. Give some example and solve it.
OK, let's take a high school problem (to which one would usually apply the Henderson-Hasselbalch). In this particular example please note that a very similar value could perhaps have been reached simply using the Henderson-Hasselbalch approximation, but as you will see if you look at the Wikipedia page, the Charlot equation does not rely on the same assumptions (the reason this problem probably yields the same value both ways is because it is a high school problem and thus designed to be solved by pupils with no idea of any solution beyond the HH).
Because of my inability to manipulate the mathematical font on this forum everything will be expressed in plain text but the brackets will make it clear what goes where.
"Calculate the mass of CH3COONa that has to be added to 100 cm3 of a 1.00 mol/dm3 solution of CH3COOH to make a buffer solution of pH=4.38. Ka[CH3COOH]=1.8*10
-5 mol/dm3."
(Assume that the volume of the solution does not change on addition of the salt; that Kw=1.00*10
-14; etc., etc.)
Now first we'll describe what we know, in the terms of our Charlot equation:
Ca=1.00 mol/dm
3[H+]=10
-4.38 mol/dm
3Ka=1.8*10
-5 mol/dm3
Kw=1.00*10
-14We appear only to lack a Cs from our Charlot equation, and that's what we want to find anyway. Rearrange the Charlot equation for Cs (can be done in your head):
Cs=((Ka*(Ca-[H+]+(Kw/[H+])))/[H+])-[H+]+(Kw/[H+])
The volume of the solution does not change so n
s=V*Cs. m
s=n
s*Mr[CH3COONa]. And now our solution is ready:
m
s=V*Cs*Mr[CH3COONa]
m
s=V*Mr[CH3COONa]*((Ka*(Ca-[H+]+(Kw/[H+])))/[H+])-[H+]+(Kw/[H+])
Let's substitute in the numbers (V=0.1, Mr=82, and the rest as written above):
m
s=3.54019 g
Which is, of course, correct. Now in this case using the Henderson-Hasselbalch would have worked fine (to my calculation, it results in an answer of 3.54068 g), but that's not as accurate as it could be. We've increased the accuracy and limited our approximations (to an extent - many are still being made but at least none in the concentration-solving stage; now our approximations are in the conversion from pH to [H+], which is vastly more complicated to even attempt properly than just [H+]=10-
pH), and there is virtually no added work - just three more added numbers!
But why are we discussing this again? Neither the Charlot equation nor the Henderson-Hasselbalch, as far as I can see, can come anywhere near to the task at hand. We need a more serious polyprotic method (don't ask me to type one up).