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Compare the calculated percent ionization of acetic acid in the solution containing both 0.10 M acetic...

Compare the calculated percent ionization of acetic acid in the solution containing both 0.10 M acetic acid and 0.10 M sodium acetate to the calculated percent ionization for the solution containing 0.10 M acetic acid alone. In which solution is the percent ionization the lowest? Clearly and completely explain why this occurs.

Expert Solution

Alone acetic acid :

CH3COOH --------------> CH3COO- + H+

0.10 M 0 0

0.10 - x x x

Ka = x^2 / 0.10 - x

1.8 x 10^-5 = x^2 / 0.10 - x

x = 1.34 x 10^-3

% ionization = (1.34 x 10^-3) / 0.1 ) x 100

= 1.34 %

in solution of both

CH3COOH --------------> CH3COO- + H+

0.1 0.10 0

0.10 - x 0.10 + x x

Ka = (0.10 + x) x / (0.10 - x)

1.8 x 10^-5 = (0.10 + x) x / (0.10 - x)

x = 1.8 x 10^-5

% ionization = 1.8 x 10^-5 / 0.1 ) x 100

= 0.018 %

% ionization is lowest in solution containing both acetic acid and sodium acetate

queen_honey_blossom answered 2 years ago

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