The percentage deprotonation of benzoic acid in a 0.110 m solution is 2.4%. What is the pH of the solution and the Ka of benzoic acid?
To solve this problem first I found the concentration of [H3O+] which is (0.024*0.110)= 2.6x10^-3 mol/L and then the pH= -log (2.6x10^-3)= 2.58.
However, I am confused on how to find ka. I know ka = [C6H5COO-][H3O+]/[C6H5COOH] and that both [C6H5COO-] and [H3O+] equals 2.6x10^-3.
How come [C6H5COOH]= 0.110(1-0.024)?
Hw 6D.9
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Renee Grange 1I
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Re: Hw 6D.9
The Ka value should be
Ka = [C6H3Co2-] [H3O+] / [C6H5COOH]
Ka = (2.64x10^-3)^2/((.110- (2.64 x 10^-3))
Ka = 6.5 x 10^-3
Ka = [C6H3Co2-] [H3O+] / [C6H5COOH]
Ka = (2.64x10^-3)^2/((.110- (2.64 x 10^-3))
Ka = 6.5 x 10^-3
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