(a) Given that \(K_{b}\) for ammonia is \(1.8 \times 10^{-5}\) and that for hydroxylamine is \(1.1 \times 10^{-8}\), which is the stronger base? (b) Which is the stronger acid, the ammonium ion or the hydroxylammonium ion? (c) Calculate \(K_{a}\) values for \(\mathrm{NH}_{4}^{+}\) and $\mathrm{H}_{3} \mathrm{NOH}^{+}$.

The \(K_b\) values are given for ammonia and hydroxylamine. A higher \(K_b\) value indicates a stronger base. - Ammonia, \(K_b = 1.8 \times 10^{-5}\) - Hydroxylamine, \(K_b = 1.1 \times 10^{-8}\) Since the \(K_b\) value of ammonia is greater than that of hydroxylamine, ammonia is the stronger base.

The conjugate acids of the bases in question are the ammonium ion (\(\mathrm{NH}_{4}^{+}\)) and the hydroxylammonium ion (\(\mathrm{H}_{3}\mathrm{NOH}^{+}\)). The stronger base has a weaker conjugate acid. Since ammonia is the stronger base, its conjugate acid, the ammonium ion, will be the weaker acid. That means the hydroxylammonium ion is the stronger acid.

To obtain the \(K_a\) values, we will first need the equilibrium constant for water (\(K_w\)), which is \(1.0 \times 10^{-14}\) at 25°C. We can use the formula $$K_a \times K_b = K_w$$ to calculate \(K_a\) values for each ion. For the ammonium ion, \(K_b\) = 1.8 x 10^{-5}: $$K_a \times (1.8 \times 10^{-5}) = 1.0 \times 10^{-14}$$ $$K_a = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}}$$ $$K_a = 5.56 \times 10^{-10}$$ For the hydroxylammonium ion, \(K_b\) = 1.1 x 10^{-8}: $$K_a \times (1.1 \times 10^{-8}) = 1.0 \times 10^{-14}$$ $$K_a = \frac{1.0 \times 10^{-14}}{1.1 \times 10^{-8}}$$ $$K_a = 9.09 \times 10^{-7}$$ So, the \(K_a\) values for \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{H}_{3}\mathrm{NOH}^{+}\) are: - \(K_a (\mathrm{NH}_4^+) = 5.56 \times 10^{-10}\) - \(K_a (\mathrm{H}_{3}\mathrm{NOH}^{+}) = 9.09 \times 10^{-7}\)