Question

A solution 0.35 m of para- bromobenzoic acid, brc7h4cooh, has a percentage dissociation of 1.69%. determine the ka of para- bromobenzoic acid.

Answer 1

Dissociation of para-bromobenzoic acid can be represented as:

4-BrC₆H₄COOH----->4-BrC₆H₄COO⁻ + H⁺

Dissociation constant of this acid can be calculated as:

Ka={[BrC₆H₄COO⁻][H⁺]}/[4-BrC₆H₄COOH]

as[4-BrC₆H₄COOH=Concentration of para- bromobenzoic acid=0.35 M

And as 1.69% of this acid dissociates to form [BrC₆H₄COO⁻] and [H⁺], so amount of these ions formed will be:

[BrC₆H₄COO⁻]=[H⁺]=1.69×0.35=0.59

So Now Ka=(0.59×0.59)/0.35

=0.99.

Answer 2

Dissociation of para-bromobenzoic acid can be represented as:

4-BrC₆H₄COOH----->4-BrC₆H₄COO⁻ + H⁺

Dissociation constant of this acid can be calculated as:

Ka={[BrC₆H₄COO⁻][H⁺]}/[4-BrC₆H₄COOH]

as[4-BrC₆H₄COOH=Concentration of para- bromobenzoic acid=0.35 M

And as 1.69% of this acid dissociates to form [BrC₆H₄COO⁻] and [H⁺], so amount of these ions formed will be:

[BrC₆H₄COO⁻]=[H⁺]=1.69×0.35=0.59

So Now Ka=(0.59×0.59)/0.35

=0.99.