# The pH of 0.02 M solution of an unknown weak acid is 3.7, how would you find the pka of this acid?

Jan 21, 2016

Knowing the pH, you know the concentration of protons:

-log["H"^(+)] = "pH" = 3.7

$\left[{\text{H}}^{+}\right] = {10}^{- 3.7}$ $\text{M}$

Now, since the weak (monoprotic) acid dissociates into its conjugate base and a proton, the $\text{mol}$s of protons are equimolar with the $\text{mol}$s of conjugate base---the protons came FROM the weak acid, so the conjugate base that forms must be equimolar with the protons given out to the solvent.

${\text{HA" rightleftharpoons "A"^(-) + "H}}^{+}$

Hence, $\left[{\text{A"^(-)] = ["H}}^{+}\right]$ in the same solution volume. Using the equilibrium constant expression, we get:

K_a = (["H"^(+)]_"eq"^2)/(["HA"]_"eq"]

Don't forget that the $\text{HA}$ form of $\text{HA}$ had given away protons, so the $\text{mol}$s of protons given away to generate ${\text{A}}^{-}$ is subtracted from the $\text{mol}$s of (protons in) $\text{HA}$.

= (["H"^(+)]_"eq"^2)/(["HA"]_i - ["H"^(+)]_"eq")

$= \left({10}^{- 3.7} \text{M")^2/(0.02 "M" - 10^(-3.7) "M}\right)$

${K}_{a} = 2.0105 \times {10}^{- 6}$ $\text{M}$

Thus:

$\textcolor{b l u e}{{\text{pKa") = -log("K}}_{a}}$

$= - \log \left(2.01059 \times {10}^{-} 6\right)$

$\approx \textcolor{b l u e}{5.70}$

where the logarithm of any number is unitless.