We’re being asked **to calculate the percent ionization** of a **0.10 M and 0.010 M aqueous solution of HA**.

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. 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} = 9.5 × 10^{–7}**

A certain weak acid, HA, has a Ka value of 9.5×10^{–7}.

Calculate the percent dissociation of HA in a 0.10 M solution?

Calculate the percent dissociation of HA in a 0.010 M solution?

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