Determine the [h3o+] of a 0.210 m solution of formic acid.

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

∴X = 0.0059
∴[H+] = 0.0059
∴PH= -㏒ [H+]
       = -㏒ 0.0059
       = 2.23