Can I get help on this?

Question

At equilibrium, an aqueous solution of formic acid (HCHO2) has these concentrations: [H+]=0.0021 M; [CHO2−]=0.0021 M; and [HCHO2]=0.0245 M.

HCHO2(aq)↽−−⇀H+(aq)+CHO−2(aq)

What is the value of K of this equilibrium?

𝐾=

Answer 1

HCHO₂ ==> H⁺ + CHO₂^- ... ionization of formic acid

Ka = [H⁺][CHO₂^-] / [HCHO₂]

Ka = (0.0021)(0.0021) / (0.0245)

Ka = 0.00018 = 1.8x10^-4