We’re being asked **to calculate the equilibrium constant of an acid** if a **0.010 M solution of an acid is 8.0% ionized**.

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

$\overline{){\mathbf{\%}}{\mathbf{}}{\mathbf{ionization}}{\mathbf{=}}\frac{\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}}{\mathbf{[}\mathbf{HA}{\mathbf{]}}_{\mathbf{initial}}}{\mathbf{\times}}{\mathbf{100}}}$

Remember that ** weak acids** partially dissociate in water and that

The dissociation of the general acid HA is as follows:

HA(aq) + H_{2}O(l) ⇌ H_{3}O^{+}(aq) + A^{–}(aq)

A 0.010 M solution of an acid is 8.0% ionized. What is K_{a} for this acid?

A. 3.2 x 10^{-6}

B. 2.8 x 10^{-5}

C. 6.4 x 10^{-6}

D. 7.0 x 10^{-5}

E. 1.6 x 10^{-4}

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