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Example 13.10: Determination of Kb from Equilibrium Concentrations

Caffeine, \(\ce{C8H10N4O2}\) is a weak base. What is the value of Kb for caffeine if a solution at equilibrium has \(c_{\mathrm{\ce{C8H10N4O2}}} = 0.050\) M; \(c_{\mathrm{\ce{C8H10N4O2H+}}} = 5.0\times 10^{-3}\) M; and \(c_{\mathrm{\ce{OH-}}} = 2.5\times 10^{-3}\) M?

Solution

\([\ce{C8H10N4O2}]_{\mathrm{eq}}\) \(= 0.050\)


\([\ce{C8H10N4O2H+}]_{\mathrm{eq}}\) \(= 5.0\times 10^{-3}\)


\([\ce{OH-}]_{\mathrm{eq}}\) \(= 2.5\times 10^{-3}\)


At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:


\(K_{\mathrm{b}}\) \(= \dfrac{[\ce{C8H10N4O2H+}]_{\mathrm{eq}} \cdot [\ce{OH-}]_{\mathrm{eq}}}{[\ce{C8H10N4O2}]_{\mathrm{eq}}}\)

\(\ \ \ =\dfrac{5.0\times 10^{-3} \cdot 2.5\times 10^{-3}}{0.050}\)

\(\ \ \ =\dfrac{1.250\times 10^{-5}}{0.050}\)

\(\ \ \ =2.5\times 10^{-4}\)