Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral molecules, but they are frequently much more soluble as their acid salts. Assuming that \(\mathrm{pH}\) in the stomach is 2.5 , indicate whether each of the following compounds would be present in the stomach as the neutral base or in the protonated form: nicotine, $K_{b}=7 \times 10^{-7}\(; caffeine, \)K_{b}=4 \times 10^{-14} ;\( strychnine, \)K_{b}=1 \times 10^{-6} ;\( quinine, \)K_{b}=1.1 \times 10^{-6} .$

We are given the \(K_{b}\) values and the relationship \(K_{w} = K_{a} K_{b}\), where \(K_{w}\) (the ion product constant for water) is equal to \(1 \times 10^{-14}\). We can calculate the \(K_{a}\) values for each compound as follows: Nicotine: \(K_{b} = 7 \times 10^{-7}\) \[K_{a} = \frac{K_{w}}{K_{b}} = \frac{1 \times 10^{-14}}{7 \times 10^{-7}} = 1.43 \times 10^{-8}\] Caffeine: \(K_{b} = 4 \times 10^{-14}\) \[K_{a} = \frac{K_{w}}{K_{b}} = \frac{1 \times 10^{-14}}{4 \times 10^{-14}} = 2.5 \times 10^{-1}\] Strychnine: \(K_{b} = 1 \times 10^{-6}\) \[K_{a} = \frac{K_{w}}{K_{b}} = \frac{1 \times 10^{-14}}{1 \times 10^{-6}} = 1 \times 10^{-8}\] Quinine: \(K_{b} = 1.1 \times 10^{-6}\) \[K_{a} = \frac{K_{w}}{K_{b}} = \frac{1 \times 10^{-14}}{1.1 \times 10^{-6}} = 9.09 \times 10^{-9}\]

Using the calculated \(K_{a}\) values, we can find the \(\mathrm{pK}_{a}\) values for each compound: Nicotine: \[ \mathrm{pK}_{a} = - \log_{10} (1.43 \times 10^{-8}) = 7.84 \] Caffeine: \[ \mathrm{pK}_{a} = - \log_{10} (2.5 \times 10^{-1}) = 0.60 \] Strychnine: \[ \mathrm{pK}_{a} = - \log_{10} (1 \times 10^{-8}) = 8.0 \] Quinine: \[ \mathrm{pK}_{a} = - \log_{10} (9.09 \times 10^{-9}) = 8.04 \]

The \(\mathrm{pH}\) in the stomach is 2.5, so we can now compare the \(\mathrm{pK}_{a}\) values to determine the form of each compound: - Nicotine: \(\mathrm{pH} < \mathrm{pK}_{a}\) (2.5 < 7.84), so it will be present in its protonated form. - Caffeine: \(\mathrm{pH} > \mathrm{pK}_{a}\) (2.5 > 0.60), so it will be present in its neutral base form. - Strychnine: \(\mathrm{pH} < \mathrm{pK}_{a}\) (2.5 < 8.0), so it will be present in its protonated form. - Quinine: \(\mathrm{pH} < \mathrm{pK}_{a}\) (2.5 < 8.04), so it will be present in its protonated form. Therefore, in the stomach with a \(\mathrm{pH}\) of 2.5, nicotine, strychnine, and quinine will be present in their protonated forms, while caffeine will be present in its neutral base form.