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Answer to Question #15627 in Physical Chemistry for Matthias

Question #15627
This is a question I am presently doing right now which I found in my homework. I am quite sure that in (a) I have to use the Hasselbach equation though I am not sure how to work it out. If anyone knows any of the answers, and how you got to them, please feel free to share. Any help is greatly appreciated :)

CH3COO- (aq) + H2O (l) <==> CH3COOH (aq) + OH- (aq)

A 0.01mol dm^-3 solution has a pH of 8.87

a. Calculate [H+] and [OH-] in the solution

b. Find [CH3COOH] in the solution.

c. Calculate the acid dissociation constant of CH3COOH.

Thanks! :)
Expert's answer
pH = -log[H+]

[H+] = 10-8.87 = 1.5*10-9

pH + pOH = pKw = 14

14 - 8.87 = 5.13

pOH = -log[OH-]

[OH-] = 4.4*10-6

[OH-] = [CH3COO-]

Kb for CH3COO- = Kw/Ka
Kw/Ka = (CH3COOH)(OH-)/(CH3COO-)
Ka = 4.4*10-6

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