Question

Formic acid (HCO2H) has a dissociation constant of 1.8 x 10^-4 M. The acid dissociates 1:1. What is the [H+] if 0.35 mole of formic acid is dissolved in 1.0 liter of water?

Answer 1

Initial [ HCO2H] = moles * volume 
                           =0.35 moles * 1 L = 0.35 M

by using ICE table:

              HCO2H ↔ H+  + HCO2-
initial      0.35 M        0          0
change  - X             +X          +X
Equ       (0.35 - X)       X          X

∴ Ka = [H+][HCO2-] / [HCO2H]

by substitution:

1.8 x 10^-4 =  X^2 / (0.35-X) by solving for X

∴ X = 0.0079 or 7.9 x 10^-3

∴ [H+] = X = 7.9 x 10^-3 M