When the Pka for formic acid = 3.77 and Pka = -㏒ Ka 3.77 = -㏒ Ka ∴Ka = 1.7x10^-4
when Ka = [H+][HCOO-}/[HCOOH]
when we have Ka = 1.7x10^-4 &[HCOOH] = 0.21 m so by substitution: by using ICE table value 1.7x10^-4 = X*X / (0.21-X) (1.7x10^-4)*(0.21-X) = X^2 by solving this equation for X