If the equilibrium of [H+] for methanoic acid is 3.4x10^-3 , what is the molarity of the acid?

Answer 1

What is the Ka for methanoic acid (also known as formic acid)?

HCOOH ==> H⁺ + COO^-

Ka = [H+][COO-]/[HCOOH] = (3.4x10^-3)² / [HCOOH] - 0.0034

Look up the Ka for methanoic acid, plug it in and solve for [HCOOH]