We’re being asked **to calculate the equilibrium constant of HA** if a **0.0538 M solution is 3.57% ionized**.

Recall that the ** percent ionization **is given by:

$\overline{){\mathbf{\%}}{\mathbf{}}{\mathbf{dissociation}}{\mathbf{}}{\mathbf{=}}\frac{\mathbf{[}\mathbf{A}{\mathbf{]}}_{\mathbf{initial}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{[}\mathbf{A}{\mathbf{]}}_{\mathbf{final}}}{\mathbf{[}\mathbf{A}{\mathbf{]}}_{\mathbf{initial}}}{\mathbf{\times}}{\mathbf{100}}}$

The dissociation is as follows:

N_{2}O_{4}(g) ⇌ 2NO_{2}(g)

From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table and K_{a} expression.

The ** K_{c} expression** for is:

$\overline{){{\mathbf{K}}}_{{\mathbf{c}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}{\mathbf{=}}\frac{{\mathbf{\left[}{\mathbf{NO}}_{\mathbf{2}}\mathbf{\right]}}^{\mathbf{2}}}{\mathbf{\left[}{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{\right]}}}$

Note that each concentration is raised by the stoichiometric coefficient: [NO_{2}] is raised to 2 and [N_{2}O_{4}] is raised to 1.

Exactly 1.0 mol N_{2}O_{4} is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation

N_{2}O_{4}(g) ⇌ 2NO_{2}(g).

If at equilibrium N_{2}O_{4} is 28.0% dissociated, what is the value of the equilibrium constant, K_{c}, for the reaction under these conditions?

A) 0.44

B) 2.3

C) 0.31

D) 0.78

E) 0.11

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