# Problem: What is the percent ionization of a 2 M solution of acetic acid at 25 °C? a) 0.6% b) 0.3% c) 0.0036% d) 0.0018%

###### FREE Expert Solution

Recall that the percent ionization is given by:

The dissociation of acetic acid (CH3COOH) is as follows:

CH3COOH(aq) + H2O(l)  H3O+(aq) + A(aq); Ka = 1.8 × 10–5

From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table and Ka expression.

The Ka expression for CH3COOH is:

$\overline{){{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}{\mathbf{=}}\frac{\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{COO}}^{\mathbf{-}}\mathbf{\right]}}{\mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}\mathbf{COOH}\mathbf{\right]}}}$

${\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{COO}}^{\mathbf{-}}\mathbf{\right]}}{\mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}\mathbf{COOH}\mathbf{\right]}}$

$\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{=}\frac{\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{2}\mathbf{-}\mathbf{x}}$

$\frac{\mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}\mathbf{COOH}{\mathbf{\right]}}_{\mathbf{initial}}}{{\mathbf{K}}_{\mathbf{a}}}\gg \mathbf{500}$

Since the ratio is greater than 500, we can remove (–x) from the equation.

The equation simplifies to:

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###### Problem Details

What is the percent ionization of a 2 M solution of acetic acid at 25 °C?

a) 0.6%

b) 0.3%

c) 0.0036%

d) 0.0018%