Determine the [h3o+] of a 0.250 m solution of formic acid

Formic acid is a weak acid so it ionizes partailly and you need the ionization constant.

This constat is Ka = 1.77 * 10 ^ - 4

Now write the equilibrium reaction

HCOOH(aq) + H2O ⇄ H3O+(aq) + HCOO-(aq)

0.250M - x                       x                  x

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

1.77x10^-4 = x^2 / (0.250-x)

Solve for x and you get: x =

 x = 0.00656

[H3O+] = 0.00656 M

Olá. porque a reação na temperatura é muito rapida a do copo : )