Question

(a) Determine the pH and the percent ionization of a 0.205 M formic acid solution.

(b) Also determine the pH and percent ionization of formic acid in a solution 0.205 M in formic acid, and 0.295 M in sodium formate, NaOOCH.

Formic acid, HCOOH, has an equilibrium constant, Ka = 1.77 x 10-4 M.

Answer 1

(A) Let us write the reaction taking place

          HCCOH --------> HCOO^- +    H⁺

        0.205              -         -

        0.205-x                x    x

Ka = 1.77*10-4 M

Ka expression will be,

Substituting the values of Ka and concentrations in terms of x in above equation

   1.77*10^-4 (0.205-x) = x²   

   1.77*10^-4 *0.205- 1.77*10^-4 x =x²   

x² +1.77*10^-4 x- 1.77*10^-4 *0.205=0

On solving the above quadratic equation, we get x=5.9358 *10^-3 , -6.1128*10^-3

Since the value of concentration cannot be negative, we will discard the negative root

hence , x= [H+] = 5.9358 *10^-3 M

pH can be calculated as follows,

pH = - Log([H+]) = - Log( 5.9358 *10^-3) = 2.2265

% ionization = equilibrium/ initial * 100 =(5.9358 *10^-3 /0.205 )*100 = 2.8955%

b) The given solution is actually a buffer solution.

        Ka = 1.77*10^-4

         [HCOO-] =0.295 M

        [ HCOOH]=0.205

For a buffer solution we can use the handerson-hasselbalch equation ,

handerson-hasselbalch equation is given by,

Now we will convert Ka to pKa for further calculation

pKa = -Log(Ka) = -Log(1.77*10^-4) = 3.75

Now let us substitute the values of pKa and weak acid and conjugate base concentration in the Handerson equation

    pH = 3.75 + 0.15806 =3.908

Let us calculate H+ concentration

pH = - Log([H+])

3.9008 = - Log([H+])

From above equation, [H+] = 1.2357*10^-4 M

Percent ionization = [H+] / [HCCOH] *100= (1.2357*10^-4 / 0.205)*100 = 0.06028 %