Question

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

Answer 1

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 `"mol"`s of protons are equimolar with the `"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.

`"HA" rightleftharpoons "A"^(-) + "H"^(+)`

Hence, `["A"^(-)] = ["H"^(+)]` in the same solution volume. Using the equilibrium constant expression, we get:

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

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

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

`= (10^(-3.7) "M")^2/(0.02 "M" - 10^(-3.7) "M")`

`K_a = 2.0105xx10^(-6)` `"M"`

Thus:

`color(blue)("pKa") = -log("K"_a)`

`= -log(2.01059xx10^-6)`

`~~ color(blue)(5.70)`

where the logarithm of any number is unitless.