Chemical Forums

Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: owen61 on May 03, 2009, 01:54:31 AM

Title: Calculating pH before equilibrium point in titration
Post by: owen61 on May 03, 2009, 01:54:31 AM
Hi, I'm really stumped by this problem so I thought I'd seek help.

We're titrating a 10 mL sample of 0.1M Na3AsO4 with 7.0 mL HCl. The Na3AsO4 has pK's of: 11.6, 6.77 and 2.25

What is throwing me off about this problem is the amount of HCl we are adding is after the 1/2 equivalence point of 5.0 mL but before the equivalence point of 10 mL.  My professor says Henderson Hassel. is not a valid equation in this case (the reasons for which are also unclear to me) and that the pH value we should get is 11.17.

I tried setting up an ICE box

    Na3AsO4         +        HCl -->          HNa2AsO4 +       NaCl
I   (0.1M)(0.01L)     (0.1M)(0.007L)         ----                  ----
C  -0.0007 mols           -0.0007 mols    +0.0007 mols     ------
E   0.0003 mols            0                     0.0007 mols

Now, I know in HHb we would then divide the mol values for Na3AsO4 and its conjugate acid by the total volume (0.017 mL) and set pH = pKa + log ([A-]/[HA]) and would get a pH of 11.23. However, I don't know what to do since we cannot use HHb.

If anyone has any suggestions, I would love them eternally. Well, maybe not eternally, but for a little while  :)
Title: Re: Calculating pH before equilibrium point in titration
Post by: Borek on May 03, 2009, 05:07:06 AM
The difference between 11.23 that you got using HH and the real value of 11.17 (or 11.16 which I got) is about 0.05. Stating that HH can't be used in this case is a nitpicking, especially in the solution of ionic strength close to 0.3, where thermodynamic effects are much more important (calculating pH using Davies equation for activity coefficients I got pH 10.35, thats a one unit difference, not 0.05).

Trick is, HH equation describes situation when you know exact concentrations of acid (HAsO42- in this case) and conjugate base (AsO43-). We use it for calculation of pH assuming that these concentrations are given just by the stoichiometry of neutralization - in this case we assume that protonation was complete. However, reactions in the solution doesn't stop here - both AsO43- and HAsO42- will react further (the first one hydrolyzing, the latter dissociating), slightly shifting pH till the equilibrium is reached.

I draw a blank at the moment - no easy idea how to calculate the pH taking these things into account. Rigorous approach will probably yield at least 3rd degree polynomial.