3
$\begingroup$

In my titration curve for HCl being titrated with NaOH, the initial pH-value was plotted as 2.1 according to a pH-meter. When I found the equivalence point and calculated the concentration of HCl, it was 0.075 M. I calculated how much the initial pH-value should be from that by -lg(0.075) which is around 1.12. That is way off from the original pH-value I got.

Why did this happen? Is it because the risk of me calibrating the pH-meter incorrectly? If that is the case, will the x-value of the equilibrium point also be wrong?? For the latter, I feel like the answer would be no, because what is wrong is that if the pH-meter was calibrated wrong, then all of the values would be 0.9 pH higher than what they should be, so the curve would only differ in the y-direction, but would be exactly the same in the x-direction. Is that correct?

Thanks in advance!

$\endgroup$
3
$\begingroup$

Maybe your electrode is incorrectly calibrated. But it may be another effect : you should know that at high acidic concentration, the pH values is not obtained by taking the logarithm of the concentration of $\ce{H3O+}$. At high acidic concentration, the concentration should be replaced by the activity, which could be rather different from the concentration. Just for the fun try this experimentation : Take your $\ce{HCl}$ solution with an electrode showing pH $2.1$. Add an equal volume of saturated $\ce{NaCl}$ solution (which is neither acidic nor basic). You will not believe your eyes by reading your electrode : The pH values goes down to $1$, as if diluting the initial solution produces an increase of concentration !

The activity is sometimes described as the amount of $\ce{H3O+}$ ions divided by the volume of the so-called "free water" (and not by the volume of the solution). In concentrated solutions the "free water" is this small amount of water that is free to move, and not attracted and fixed around the ions. One Liter $1 M$ $\ce{HCl}$, should contain much less that $1$ Liter free water, maybe about $0.1$ Liter free water, or less.

$\endgroup$
  • $\begingroup$ Intreresting! I did not know about that. Thank you! $\endgroup$ – Csharpyikes Nov 10 '20 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.