Question

The pH of a 0.1 M solution of an unknown weak acid is 3.7. What is the pKa of this acid?

Answer 1

Explanation:

Knowing the pH, you know the concentration of protons:

−log[H+]=pH=3.7

[H+]=10−3.7 M

Now, since the weak (monoprotic) acid dissociates into its conjugate base and a proton, the mols of protons are equimolar with the mols 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.

HA⇌A−+H+

Hence, [A−]=[H+] in the same solution volume. Using the equilibrium constant expression, we get:

Ka=[H+]2eq[HA]eq

Don't forget that the HA form of HA had given away protons, so the mols of protons given away to generate A− is subtracted from the mols of (protons in) HA.

=[H+]2eq[HA]i−[H+]eq

=(10−3.7M)20.02M−10−3.7M

Ka=2.0105×10−6 M