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Example 13.9: Determination of \(K_{\mathrm{a}}\) from Equilibrium Concentrations

Acetic acid is the principal ingredient in vinegar ( Figure 11); that's why it tastes sour. At equilibrium, a solution contains \(c_{\mathrm{\ce{CH3CO2H}}} = 0.0787\ \mathrm{M}\) and \(c_{\mathrm{\ce{H3O+}}}\)\( = \)\(c_{\mathrm{\ce{CH3CO2-}}}\)\( = 1.18\times 10^{-3}\ \mathrm{M}\) What is the value of \(K_{\mathrm{a}}\) for acetic acid?.
Figure 11Vinegar is a solution of acetic acid, a weak acid. (credit: modification of work by “HomeSpot HQ”/Flickr)

Solution

\([\ce{CH3CO2H}]_{\mathrm{eq}}\) \(= 0.0787\)


\([\ce{H3O+}]_{\mathrm{eq}}\) \(= 1.18\times 10^{-3}\)


\([\ce{CH3CO2-}]_{\mathrm{eq}}\) \(= 1.18\times 10^{-3}\)


\(K_{\mathrm{a}}\) = ?


We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:



\(K_{\mathrm{a}}\) \(= \dfrac{[\ce{H3O+}]_{\mathrm{eq}} \cdot [\ce{CH3CO2-}]_{\mathrm{eq}}}{[\ce{CH3CO2H}]_{\mathrm{eq}}}\)

\(\ \ \ =\dfrac{1.18\times 10^{-3} \cdot 1.18\times 10^{-3}}{0.0787}\)

\(\ \ \ =\dfrac{1.392\times 10^{-6}}{0.0787}\)

\(\ \ \ =1.77\times 10^{-5}\)