Asolution 0.35 m of para- bromobenzoic acid, brc7h4cooh, has a percentage dissociation of 1.69%. determine the ka of para- bromobenzoic acid.
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Ответ:
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.
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