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
Answer: [H3O+] = 0.00656 M
