8. Write the reaction of acetic acid in water and calculate the pH of a 0.100...

Question

Question

science chemistry

Answer 1

Answer #1

The ionization reaction of acetic acid in water is

CH3COOH + H2O <=> CH3COO^- + H3O⁺

Initial Conc: 0.1 0 0

At Equilibrium: 0.1 - x x x

We know Ka= 1.8 x 10 ^ -5 for acetic acid

so

Ka= [CH3COO^-]*[H3O⁺]/[CH3COOH]

1.8 x 10 ^ -5 = x^2/(0.1-x)

On solving this equation we get

x = 0.0013 => [H+]

so pH = -log([H+]) = 2.88

Answer 2

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