Question

In a 0.1000M solution of methanoic acid, [H+] = 4.2 x 10-3M. Calculate the Ka of this acid.

Answer 1

Answer:

Ka = 1.764 × 10^-4

Explanation:

The equation for the dissociation of methanoic acid which is a weak acid.

HCOOH + H2O  ⇄  H3O+ + HCO2-

But; [H3O+] = [H+] = 4.2 x 10-3M

Ka = [H3O+] [HCO2-]/ [HCOOH]

In this case;

[H3O+] = [HCO2-] = 4.2 x 10^-3M

[ HCOOH] = 0.100 M

Therefore;

Ka = [ 4.2 x 10^-3M] [4.2 x 10^-3M]/ [0.1000M]

    = 1.764 × 10^-4