The fourth option on Edgen, "two alcohol functional groups". You're welcome :)
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
