Answer to Question #15627 in Physical Chemistry for Matthias
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.
a. 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
b. [OH-] = [CH3COO-]
c. Kb for CH3COO- = Kw/Ka Kw/Ka = (CH3COOH)(OH-)/(CH3COO-) Ka = 4.4*10-6