Question

12. Calculate the pH of a 0.10 M solution of potassium formate, KHCOO. Ka of formic acid (HCOOH) = 1.8 x 10^-4. Write the dissociation equation and label each ion as acidic, basic, or neutral. Also write the hydrolysis equation.

Dissociation equation: _____________________________________ cation: __________ anion: __________

Hydrolysis equation: _________________________________________________________________________

Answer 1

Dissociation equation

   HCOOH (aq) +   H2O (l)    --- > HCOO-(aq) + H3O+(aq)

I    0.10                                                   0                       0

C -x                                                        +x                     +x

E (0.10-x)                                               x                         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)