2
$\begingroup$

I have $20~\mathrm{cm^3}$ of a $0.1\ \mathrm{M}$ ethanoic acid solution and have to find out the $\mathrm{p}K_\mathrm{a}$.

In experiment this solution had a $\mathrm{pH}$ of $2.91$.

Do I use the formula $\mathrm{p}K_\mathrm{a}= 2 \mathrm{pH} + \log(c (\mathrm{acid}))$ ?

What I do not understand is that my $\mathrm{p}K_\mathrm{a}$ value varies. I measured many solutions with 18:2, 15:5, 10:10 sodium ethanoate/ethanoic acid composition. Which I calculated with the formula $\mathrm{p}K_\mathrm{a} = \mathrm{pH} - \log \frac{c(\mathrm{salt})}{c(\mathrm{acid})}$

I get close values but not the equal. So the one above would be $\mathrm{p}K_\mathrm{a}=4.82$ and another value I obtained is $4.62$ (for the 15:5 solution with $\mathrm{pH}\ 5.11$).

Is this deviation purely due to experimental inaccuracy or does the composition matter. Would it make more sense to use the literature $\mathrm{p}K_\mathrm{a}$ value for further calculation rather than those experimental ones?

Improve this question
$\endgroup$
3
  • $\begingroup$ A pH of 2.91 is far more acidic than a 0.1 molar solution should be. How did you measure pH - I assume a pH meter. So how was the meter calibrated?!? $\endgroup$
    – MaxW
    Dec 17, 2015 at 23:18
  • 2
    $\begingroup$ @MaxW no it's not... ! Adam, can you be a bit more precise in what you did, I don't really understand what you did during your experiment :/ $\endgroup$
    – ParaH2
    Dec 17, 2015 at 23:22
  • $\begingroup$ @Shadock - You're correct. I actually did the calculation and the solution would be more acidic than I thought. $\endgroup$
    – MaxW
    Dec 17, 2015 at 23:29

1 Answer 1

Reset to default
1
$\begingroup$

Do I use the formula $\mathrm{p}K_\mathrm{a}= 2 \mathrm{pH} + \log(c (\mathrm{acid}))$ ?

No, for the pure acide use:

$$K_\mathrm{a} = \frac{([\ce{H+}])^2}{[\ce{HA}]_\mathrm{initial}-[\ce{H+}]}$$

Then, if you are still having problems, see this question and its answer:

Difference in calculated pH and the real pH of a phosphate buffer?

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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