Experimental measurement shows that 0.200 M arsenious acid, H2AsO3 (aq), is 0.0051% dissociated at 25 degrees Celsius....

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Chemistry

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

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4.3 Ratings (700 Votes)

Given:

[H2AsO3]=0.2 M

Percent dissociation is 0.0051 %

Solution

Reaction:

H₂AsO₃ (aq) + HÂ₂O (l) --- > HAsOÂÂ₃^- (aq) + H₃O⁺ (aq)

Acid dissociation constant expression:

ka = [HAsO₃^-] [ H_3ÂO⁺] / [H₂AsO₃]

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

H₂AsO₃ (aq) + HÂ₂O (l) --- > HAsOÂÂ₃^- (aq) + H₃O⁺ (aq)

IÂ Â Â Â Â Â Â Â Â Â 0.200Â 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 Â Â Â Â Â Â Â Â Â Â Â 0

CÂ Â Â Â Â Â Â Â -xÂ 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 +xÂ 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

EÂ Â Â Â Â Â Â Â Â (0.200-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 xÂ 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

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

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