Step-by-step 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