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When a 19.8 mL sample of a 0.460 M aqueous acetic acid solution is titrated with a 0.398 M aqueous sodium hydroxide solution,
A 17.5 mL sample of a 0.430 M aqueous hydrofluoric acid solution is titrated with a 0.381 M aqueous potassium hydroxide solut
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Answer #1

(a) Ka for acetic acid = 1.8 x 10-5 pKa=-log(1.8 x 10-5) = 4.74 At mid point, the number of moles of acid (CH3COOH) is equal

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(b) Before any addition of base : Conc. HE + H20 F 1 F + H30* Initial 0.430 0 Change - -x + Equilibrium * x.x wa -= 3.5x104 0

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