# Experimental determination of pKa

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?

• 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?!? – MaxW Dec 17 '15 at 23:18
• @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 :/ – ParaH2 Dec 17 '15 at 23:22
• @Shadock - You're correct. I actually did the calculation and the solution would be more acidic than I thought. – MaxW Dec 17 '15 at 23:29

Do I use the formula $\mathrm{p}K_\mathrm{a}= 2 \mathrm{pH} + \log(c (\mathrm{acid}))$ ?
$$K_\mathrm{a} = \frac{([\ce{H+}])^2}{[\ce{HA}]_\mathrm{initial}-[\ce{H+}]}$$