What must be the molarity of an aqueous solution of trimethylamine, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N},\) if it has a \(\mathrm{pH}=11.12 ?\) $$\begin{aligned} \left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+} &+\mathrm{OH}^{-} \\ K_{\mathrm{b}}=6.3 \times 10^{-5} \end{aligned}$$

First you need to calculate the concentration of \( OH^- \) ions using the given pH value. The formula connecting pH, pOH and \( OH^- \) is \( pOH = 14 - pH \), and concentration of \( OH^- \) ions is given by \( [OH^-] = 10^{-pOH} \). From the given pH=11.12, we have pOH = 14 - 11.12 = 2.88, so \( [OH^-] = 10^{-2.88} = 1.32 \times 10^{-3} \) M.

Then, you need to calculate the concentration of \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+} \) using the base dissociation constant \( K_{\mathrm{b}} \) and the calculated \( OH^- \) concentration. From \( K_{\mathrm{b}} = \frac{[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}][OH^-]}{[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}]}\) and assuming at equilibrium, \( [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}] = [OH^-] \) and \( [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}] = [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}] - [OH^-] \), we substitute and obtain \( [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}] = \sqrt{K_{\mathrm{b}} / [OH^-]} = \sqrt{6.3 \times 10^{-5} / 1.32 \times 10^{-3}} = 0.00735 \) M.

Finally, you calculate the initial molarity of the trimethylamine solution before any dissociation took place. As \( [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}] = [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}] - [OH^-] \), we have initial concentration of trimethylamine \( [\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}]_0 = 0.00735 - 1.32 \times 10^{-3} = 0.00603 \) M.