Question

Formic acid (HCO2H) is an organic acid secreted by ants and stinging nettles.

Calculate the pH in 0.21M HCO2H (Ka=1.8

Answer 1

HCOOH + H₂O -------------> H₃O⁺ + HCOO^-

    0.21                                    0             0

   0.21 - x                                x              x

Ka = [ H₃O⁺][HCOO^-]/[HCOOH]

1.8 x 10^-4 = x X x/0.21 - x

   x² + 1.8 x 10^-4x - 0.378 x 10^-4 = 0

x = - 1.8 x 10^-4+ sqrt [(1.8 x 10^-4 )²+ 4 x 1 x 0.378 x 10^-4]/2

x = 0.6 x 10^-2

concentration of species

[HCOOH] = 0.21 - x

                 = 0.21 - 0.006

                 = 0.204M

[HCOO^-] = 0.006M

[H₃O⁺]   = 0.006M

[OH^-] = Kw/[H₃O⁺] = 10^-14/0.006

          = 0.166 x 10^-11M

P^H = -log[H₃O⁺]

     = -log[6 x 10^-3]

P^H   = 2.22