Answer to Question #15627 in Physical Chemistry for Matthias
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.
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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