Question

Experimental measurement shows that 0.200 M arsenious acid, H₂AsO₃ (aq), is 0.0051% dissociated at 25 degrees Celsius. Determine the value of the acid-dissociation constant of H₂AsO₃ (aq) at 25 degrees Celsius.

Answer 1

Given:

[H2AsO3]=0.2 M

Percent dissociation is 0.0051 %

Solution

Reaction:

H₂AsO₃ (aq) + H­₂O (l) --- > HAsO­­₃^- (aq) + H₃O⁺ (aq)

Acid dissociation constant expression:

ka = [HAsO₃^-] [ H_3­O⁺] / [H₂AsO₃]

In order to get the ka value we must know the equilibrium concentrations of these all the species involved.
ICE

H₂AsO₃ (aq) + H­₂O (l) --- > HAsO­­₃^- (aq) + H₃O⁺ (aq)

I           0.200                                       0                                  0

C         -x                                             +x                                +x

E          (0.200-x)                                 x                                  x

Ka = x²/ (0.200-x)

Percent dissociation

0.00051 = (x/0.200) * 100

x = 1.02 E-5 lets plug this value of x in ka expression

ka = ( 1.02 E-5)²/ (0.200-1.02 E-5)

ka = 5.20 x 10^-10