Question

Calculate the percent ionization of formic acid in a solution that is 0.010 M HCOOH and 0.005 M HCOONa. K_a (HCOOH) = 1.7x10^{-4 ? 3.4% is correct answer}

Answer 1

           HCOOH <-------------->   HCOO- + H+

I            0.010                           0.0050           0

C              -x                             0.0050+x         +x

E           0.010 - x                      0.0050+x         +x

   Ka = [HCOO-][H+] / [HCOOH]

1.7 x 10^-4 = x(0.0050 +x) / 0.010 -x

1.7 x 10^-6 - (1.7 x 10^-4)x = 0.0050x + x^2

x^2 + (0.00517)x - (1.7 x 10^-6) = 0

Solving the quadratic equation, we get the roots,

x =3.38 x 10^-4

[H+] = x = 3.38 x 10^-4

[HCOO-] = 3.38 x 10^-4

% ionization = [HCOO-]final / [HCOOH]initial x 100

= (3.38 x 10^-4 / 0.010) x 100

= 3.38 %

=3.4%

the addition of formate suppresses the ionization of formic acid by shifting the ionization equilibrium towards the un

-ionized acid.