a.
pH = 8.87
8.87 = -log(H^+)
Solve for (H^+), then
pH + pOH = pKw = 14
You know pH and pKw, solve for pOH and
pOH = -log(OH^-) which lets you solve for OH.
b.
This is a hydrolysis problem.
You know OH^- from part a. That = (CH3COOH).
c.
Kb for CH3COO^- = (Kw/Ka for CH3COOH)
(Kw/Ka) = (CH3COOH)(OH^-)/(CH3COO^-)
You know Kw, CH3COOH, OH, and CH3COO^-(hat's 0.01 in the problem), solve for Ka.
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! :)
1 answer
1 answer