Â
HCOOH <-------------->Â Â HCOO- + H+
IÂ Â Â Â Â Â Â Â Â Â Â
0.010Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0.0050Â Â Â Â Â Â Â Â Â Â
0
CÂ Â Â Â Â Â Â Â Â Â Â Â Â
-x                           Â
0.0050+x        +x
EÂ Â Â Â Â Â Â Â Â Â
0.010 -
x                    Â
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.
