One K_{a} → monoprotic acid

let **HA** → barbituric acid

**Equilibrium reaction:** **HA**_{(aq)} + H_{2}O_{(l)} **⇌ A ^{-}_{(aq)} + H_{3}O^{+}_{(aq)} **

**Step 1:**** **Construct an ICE chart for the reaction.

**Step 2:** Write the K_{a} expression.

${\mathbf{K}}_{{\mathbf{a}}}\mathbf{=}\frac{\mathbf{products}}{\mathbf{reactants}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{A}}^{\mathbf{-}}\mathbf{\right]}\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}}{\left[\mathrm{HA}\right]}$

*Solids and liquids are not included in the expression*

**Step 3:**** **Calculate the equilibrium concentrations.

${\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\left[\mathbf{x}\right]\left[\mathbf{x}\right]}{\left[\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{-}\mathbf{x}\right]}\phantom{\rule{0ex}{0ex}}\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{-}\mathbf{x}}$

$\frac{{\mathbf{\left[}\mathbf{HA}\mathbf{\right]}}_{\mathbf{initial}}}{{\mathbf{K}}_{\mathbf{a}}}\mathbf{>}\mathbf{500}\mathbf{;}\mathbf{}\mathbf{ignore}\mathbf{}\mathbf{-}\mathbf{x}$

Calculate the pH at the following point in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid (K_{a} = 9.8 × 10^{−5}) with 0.100 M KOH.

(a) no KOH added

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