A 0.250 m ascorbic acid solution is 1.79% ionized at 25 degrees. what is the pka...

A 0.250 m ascorbic acid solution is 1.79% ionized at 25 degrees. what is the pka for the ascorbic acid at this temp??

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Expert Solution

Let ascorbic acid be represented by HA . Acid dissociation is given by

1.79 % of 0.25 = (1.7/100) x 0.25 = 0.004475

HA (aq) <------------> H+ (aq) + A-(aq)

Initially 0.25 0 0

equilibrium 0.25 - 0.004475 0.004475 0.004475

Ka = [H+] [A-] /[HA]

= ( 0.004475)( 0.004475) / ( 0.25-0.004475)

= 8.16 x 10^ - 5

pka = -log Ka = -log ( 8.16 x 10^-5) = 4.09

queen_honey_blossom answered 1 year ago

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