In the solution containing both 0.10 M acetic acid and 0.10 M sodium acetate, the acetic...

4.2 Ratings (666 Votes)

pka of aceticacid = -logka

Â Â Â Â Â Â Â Â Â Â Â Â = -log(1.75*10^âˆ’5)
Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â Â Â Â Â Â Â = 4.76

aceticacid + CH3COONa = buffer

pH of acidic buffer = pka + log(salt/acid)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 4.76 + log(0.1/0.1)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 4.76

Â Â Â Â Â Â Â Â Â Â Â Â pH = 4.76

Â Â Â Â Â Â Â Â Â Â CH3COOH (aq) <-----> CH3COO-(aq) + H^+(aq)

Â Â initial 0.1 MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.1 MÂ Â Â Â Â Â 0

Â Â changeÂ Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â x

equilibrium 0.1-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â 0.1+xÂ Â Â Â Â Â Â x

Â Â Â ka = [CH3COO-][H+]/[CH3COOH]

Â Â Â 1.75*10^-5 = ((0.1+X)X)/(0.1-X)

Â Â Â X = 1.75*10^-5 M

percentage dissociation of CH3COOH = dissociated / initial*100

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (1.75*10^-5)/0.1*100

Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.0175 %

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