Question

Calculate the percent ionization of 1.50 M aqueous acetic acid solution. For acetic acid, Ka = 1.8 × 10−5 . (a) 2.71% (b) 3.55% (c) 1.78% (d) 0.35% (e) None of the above

Answer 1

Pertinent equation for ionization of acetic acid AcOH & corresponding ICE table is given as,

AcOH   AcO^- (aq) + H⁺ (aq)

Initial conc. 1.50 M 0 0

Change -X +X +X

Eqm. conc. (1.50-X) X X

Ionization constant expressed as

Ka = [AcO^-][H⁺] /  [AcOH] ............ (1)

Ka = (X)(X) / (1.5-X) = X² / (1.50-X)

As AcOH is weak acid it ionizes very lettle and hence we can say X <<< 1.50 and hence 1.50-X 1.50.

"Small concentration assumption" holds true.

Ka = X² / 1.50

Ka = 1.8 x 10^-5. ............. (Given)

X² / 1.50 =  1.8 x 10^-5.

  X² = 1.50 x 1.8 x 10^-5.

  X² = 27.0 x 10^-6.

  X = 5.2 x 10^-3.

By ICE table,

[H⁺] = [AcO^-] = X = 5.2 x 10^-3.

Then

% Ionization = (Fraction ionized / initial amount of AcOH) x 100

% Ionization = ( 5.2 x 10^-3/ 1.50) x 100

% Ionization = 0.355%

No figure of the given matches.

So Answer: (e) None of the above.

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