Determine the [H3O+] of a 0.210 M solution of formic acid.

Determine pH of this solution of formic acid.

formic acid is a week acid

Ka of the formic acid is 1.8 x 10^-4

we have to construct the ICE table

HCOOH + H2O (l) <----> H3O+ + HCOO-

I 0.21 0 0

C -x +x +x

E 0.21-x +x +x

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

1.8 x 10^-4 = [x] [x ] / [0.21-x]

x² + x 1.8 * 10^-4 - 3.78 x 10^-5 = 0

solve the quadratic equation

x = 0.00606 = [H3O+]

pH = -log[H3O+]

pH = -log[0.00606]

pH = 2.22