Dissociation equation
  HCOOH (aq) +  H2O (l)  Â
--- > HCOO-(aq) + H3O+(aq)
IÂ Â Â
0.10Â 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 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
C
-x                                                      Â
+x                   Â
+x
E
(0.10-x)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 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
x                       Â
x
Acidic species
H3O+]
Basic species
HCOO-
Ka expression
Ka = [HCOO-][H3O+] / [HCOOH]
1.8 x 10^-4 = x^2/ 0.10-x
x^2 = 1.8 E-4 * (010-x)
Lets solve this quadratic equation.
x = 0.004154
[H3O+]= 0.004154
pH = -log ([H3O+] = - log ( 0.004154)
= 2.38
pH of the solution would be 2.38
Cation is H3O+
anion is HCOO-
Hydrolysis equation of HCOO-
HCOO-(aq) + H2O (l) ---> HCOOH (aq) + OH-(aq)
