From the equilibrium concentrations given, calculate \(K_{\mathrm{a}}\) for each
of the weak acids and \(K_{\mathrm{b}}\) for each of the weak bases.
(a) \(\mathrm{NH}_{3}:\left[\mathrm{OH}^{-}\right]=3.1 \times 10^{-3}
\mathrm{M}\)
\(\left[\mathrm{NH}_{4}+\right]=3.1 \times 10^{-3} \mathrm{M}\)
\(\left[\mathrm{NH}_{3}\right]=0.533 \mathrm{M}\).
(b) \(\mathrm{HNO}_{2}:\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.011
\mathrm{M}\)
\(\left[\mathrm{NO}_{2}^{-}\right]=0.0438 \mathrm{M}\)
\(\left[\mathrm{HNO}_{2}\right]=1.07 \mathrm{M}\).
(c) \(\left(\mathrm{CH}_{3}\right)_{3}
\mathrm{N}:\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}\right]=0.25
\mathrm{M}\)
\(\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\right]=4.3 \times
10^{-3} \mathrm{M}\)
\(\left[\mathrm{OH}^{-}\right]=3.7 \times 10^{-3} \mathrm{M}\).
(d) \(\mathrm{NH}_{4}^{+}: \quad\left[\mathrm{NH}_{4}^{+}\right]=0.100
\mathrm{M}\)
\(\left[\mathrm{NH}_{3}\right]=7.5 \times 10^{-6} \mathrm{M}\)
\(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=7.5 \times 10^{-6} \mathrm{M}\).