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

The dissociation of a weak acid, HA is as follows:

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

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

Let X = initial concentration of HA

The ** K_{a} expression** for HA is:

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

Note that each concentration is raised by the stoichiometric coefficient: [HA], [H_{3}O^{+}] and [A^{–}] are raised to 1.

**[H _{3}O^{+}] = [A^{–}] = x = 1.8 x 10^{-3}**.

We can then plug this into the K_{a} expression:

In a 0.050 M solution of a weak monoprotic acid, [H^{+}] = 1.8 x 10^{-3}. What is its K_{a}?

A. 3.6 x 10^{-2}

B. 9.0 x 10^{-5}

C. 6.7 x 10^{-5}

D. 1.6 x 10^{-7}

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