Posted by **Matthias** on Saturday, September 29, 2012 at 5:03am.

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! :)

- Chemistry -
**DrBob222**, Saturday, September 29, 2012 at 2:48pm
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.

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