(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 or the hydroxylammonium ion? (c) Calculate \(K_{a}\) values for \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{H}_{3} \mathrm{NOH}^{+}.\)

As we are given the Kb values for ammonia and hydroxylamine, we can compare them directly: \(K_{b,\,NH_3} = 1.8 \times 10^{-5}\) \(K_{b,\,H_2NOH} = 1.1 \times 10^{-8}\) Since \(K_{b,\,NH_3} > K_{b,\,H_2NOH}\), ammonia is the stronger base.

Stronger bases have weaker conjugate acids. Therefore, since ammonia is the stronger base, its conjugate acid, the ammonium ion (NH4+), will be weaker than the hydroxylammonium ion (H3NOH+). This means that the hydroxylammonium ion is the stronger acid.

To calculate the Ka values for the ammonium and hydroxylammonium ions, we can use the relationship between Ka, Kb, and Kw. The ion-product constant of water, Kw, is given by: \(K_w = K_a \cdot K_b\) \(K_w = 1.0 \times 10^{-14}\) at 25°C For ammonium (NH4+), we have: \(K_{a,\,NH_4^+} = \frac{K_w}{K_{b,\,NH_3}}\) \(K_{a,\,NH_4^+} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}}\) \(K_{a,\,NH_4^+} = 5.56 \times 10^{-10}\) For hydroxylammonium (H3NOH+), we have: \(K_{a,\,H_3NOH^+} = \frac{K_w}{K_{b,\,H_2NOH}}\) \(K_{a,\,H_3NOH^+} = \frac{1.0 \times 10^{-14}}{1.1 \times 10^{-8}}\) \(K_{a,\,H_3NOH^+} = 9.09 \times 10^{-7}\) So, the Ka values for ammonium and hydroxylammonium ions are approximately \(5.56 \times 10^{-10}\) and \(9.09 \times 10^{-7}\), respectively.