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?

  • $\begingroup$ 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


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?


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