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?