0.1M solution of methanoic acid is found to have a ph of 2.4 at 25 degree celcius. Calculate the dissociation constant k at this temperature.

Answer 1

Methanoic acid, CH3COOH, I will call HAc.
HAc ==> H^+ + Ac^-
Set up an ICE chart and substitute into Ka. Ka = (H^+)(Ac^-)/(HAc)