We’re being asked **to calculate the percent ionization** of a **0.250 M solution of HC_{2}H_{3}O_{2}. **

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

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

We know the initial concentration of HA, **0.10 M**. We just need to find the concentration of H^{+} at equilibrium.

Since HA has a low K_{a} value, it’s a weak acid.

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

The dissociation of HA is as follows:

HA(aq) + H_{2}O(l) ⇌ H_{3}O^{+}(aq) + A^{–}(aq); **K _{a} = 8.4 × 10^{–7}**

Find the percent ionization of a 0.250 *M* solution of HC_{2}H_{3}O_{2}. (Note: *K** _{a}* = 1.8×10

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