posted by Nat on .
A 10.0 mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the addition of 20.0 mL of the HCl solution.
(potentially useful info: Ka of NH4+ = 5.6 x 10−10)
millimoles NH3 = 10 x 0.3 = 3
mmols HCl = 20 x 0.1 = 2
NH3 + HCl => NH4Cl
So you have 2 mmols NH4^+
and 1 mmol NH3.
pH = pKa + log (base)/(acid)
Thanks, I got the answer 8.95. So why do I not need to divide the mmols by the total volume?
Are you ready for the full answer? It will take a bit of reading.
The Henderson-Hasselbalch equation is
pH = pKa + log(base)/(acid)
With regard to the (base) and (acid), that means CONCENTRATION of base and CONCENTRATION of acid. Technically, then, one must divide millimoles/milliliter to get concn. But if you will put in the numbers (let's use made up numbers of 2 mmols for base and 3 mmols for acid and a volume of 10 mL. So this log ratio will come up with
(2/10)/(3/10) but notice that the 10 cancels and you have simply 2/3. Since it is the SAME volume in the solution for both acid and base you will note that the volume will ALWAYS cancel so you can simply use mmols base/mmols acid. When I taught this years ago I always counted off a few points because I told the students it was (concn/concn) not (mmoles/mmoles) and never mind that the answer was the same. So if you want to be technically correct you will use (mmols/mL)/(mmols/mL). If you want to do it quick and dirt you can simply use mmols because the answer will be the same. In fact, when I taught this I NEVER actually divided mmols/mL and plugged in that number; I always wrote mmols and mL, canceled the mL and went from there. Fast forward 25 years: I note some of the profs now use mmols directly and don't insist on the concn stage. I think it's ok to use that procedure as long as we understand what we are doing. If I were a student in a class these days I would ask the prof how s/he wanted it done on exams.
Thank you for the explanation!