The pH of a \(0.10 - M\) solution of caffeine is \(11.16. \) Determine \({K_b}\) for caffeine from these data:

\[{C_8}{H_{10}}\;{N_4}{O_2}(aq) + {H_2}O(l)\rightleftharpoons{C_8}{H_{10}}\;{N_4}{O_2}{H^ + }(aq) + O{H^ - }(aq)\]

This problem is quite similar to the last one, the only difference is that we must calculate the \({{\bf{K}}_{\bf{b}}}\) from the pH of the solution rather than the \({{\bf{K}}_{\bf{a}}}\).

\({{\bf{K}}_{\bf{b}}}{\bf{ = }}\frac{{\left( {{{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{10}}}}{{\bf{N}}_{\bf{2}}}{{\bf{O}}_{\bf{4}}}{{\bf{H}}^{\bf{ + }}}} \right){\bf{ \times }}\left( {{\bf{O}}{{\bf{H}}^{\bf{ - }}}} \right)}}{{\left( {{{\bf{C}}_{\bf{8}}}{{\bf{H}}_{{\bf{10}}}}{{\bf{N}}_{\bf{2}}}{{\bf{O}}_{\bf{4}}}} \right)}}\)

We can see that in order to calculate the\({{\bf{K}}_{\bf{b}}}\), we need to calculate the concentration of the\({\bf{O}}{{\bf{H}}^{\bf{ - }}}\)rather than the\({{\bf{H}}^{\bf{ + }}}\). If we determine the pOH, we can calculate the\({\bf{O}}{{\bf{H}}^{\bf{ - }}}\)concentration. The rule that the sum of the pH and pOH is always 14 can be used.

We can easily calculate the \(pOH\) now that we know the \(pH\):

\(\begin{aligned}pOH &= 14 - pH\\\;\;\;\;\;\;\; &= 14 - 11.16\\\;\;\;\;\;\;\; &= 2.84.\end{aligned}\)

From the \(pOH,\) we can now compute the concentration of \(O{H^ - }\)ions:

\(\begin{aligned}\;\;\;pOH &= - log\left( {O{H^ - }} \right)\\\left( {O{H^ - }} \right) &= 1{0^{ - pOH}} M\\\;\;\;\;\;\;\;\;\;\; &= 1{0^{ - 2.84}} M\\\;\;\;\;\;\;\;\;\;\; &= 1.45 \times 1{0^{ - 3}} M.\end{aligned}\)

We can assume that the concentrations of \(O{H^ - }\) and \({C_8}{H_{10}}{N_2}{O_4}\) are equivalent. The caffeine molecule has an initial concentration of \(0.10 M\), while the ions have an initial concentration of \(0 M\). The concentration of the caffeine molecule lowers by \(x,\) while the concentrations of the ions rise by \(x\) until the equilibrium is established. We can simply determine the equilibrium concentration of the caffeine molecule because we know the ion concentrations:

\(\begin{aligned}\left( {{C_8}{H_{10}}{N_2}{O_4}} \right) &= 0.10 M - x\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= \left( {0.10 - 1.45 \times 1{0^{ - 3}}} \right) M\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= 0.10 M.\end{aligned}\)

We can compute the \({K_b}\) now that we have all of the equilibrium concentrations:

\(\begin{aligned}{K_b} &= \frac{{\left( {{C_8}{H_{10}}{N_2}{O_4}{H^ + }} \right) \times \left( {O{H^ - }} \right)}}{{\left( {{C_8}{H_{10}}{N_2}{O_4}} \right)}}\\ &= \frac{{1.45 \times 1{0^{ - 3}} \times 1.45 \times 1{0^{ - 3}}}}{{0.10}}\\ &= 2.10 \times 1{0^{ - 5}}.\end{aligned}\)

The \({K_b}\) value of caffeine equals \(2.10 \times 1{0^{ - 5}}.\)

\({K_b} = 2.10 \times 1{0^{ - 5}}.\)