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Chapter 16: Problem 80

(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 \(\mathrm{H}_{3} \mathrm{NOH}^{+}\)

Short Answer

Expert verified
(a) Ammonia is the stronger base as it has a higher \(K_{b}\) value (\(1.8 \times 10^{-5}\)). (b) The hydroxylammonium ion is the stronger acid since ammonia is the stronger base and its conjugate acid, the ammonium ion, is weaker. (c) The \(K_{a}\) values are: Ammonium ion: \(K_{a(\mathrm{NH}_{4}^{+})} = 5.56 \times 10^{-10}\) Hydroxylammonium ion: \(K_{a(\mathrm{H}_{3}\mathrm{NOH}^{+})} = 9.09 \times 10^{-7}\)

Step by step solution

01

Compare Kb values of ammonia and hydroxylamine

The given Kb values for ammonia and hydroxylamine are: Ammonia (NH3): \(K_{b} = 1.8 \times 10^{-5}\) Hydroxylamine (H3NO): \(K_{b} = 1.1 \times 10^{-8}\) A higher Kb value means a stronger base. Since ammonia has a higher Kb value than hydroxylamine, ammonia is the stronger base.
02

Determine the stronger acid

The conjugate acid of ammonia is the ammonium ion (NH4+), and the conjugate acid of hydroxylamine is the hydroxylammonium ion (H3NOH+). In general, the stronger a base is, the weaker its conjugate acid will be. Since ammonia is the stronger base, its conjugate acid, the ammonium ion, will be the weaker acid. Therefore, the hydroxylammonium ion is the stronger acid.
03

Apply the ion-product constant of water

Recall that the ion-product constant of water (\(K_{w}\)) relates the acidity constant (\(K_{a}\)) of a conjugate acid to its base's basicity constant (\(K_{b}\)) through the following equation: \(K_{a} \times K_{b} = K_{w}\) The ion-product constant of water is \(K_w = 1.0 \times 10^{-14}\) at 25°C.
04

Calculate Ka for NH4+

Using the relationship between Ka, Kb, and Kw, we can calculate Ka for ammonium ion (NH4+) as follows: \(K_{a} \times K_{b} = K_{w}\) \(K_{a} = \frac{K_{w}}{K_{b}}\) \(K_{a(\mathrm{NH}_{4}^{+})} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\)
05

Calculate Ka for H3NOH+

Similarly, we can calculate Ka for hydroxylammonium ion (H3NOH+) as follows: \(K_{a(\mathrm{H}_{3}\mathrm{NOH}^{+})} = \frac{K_{w}}{K_{b}} = \frac{1.0 \times 10^{-14}}{1.1 \times 10^{-8}} = 9.09 \times 10^{-7}\) Now, we have calculated the Ka values for both the ammonium ion and the hydroxylammonium ion.

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Most popular questions from this chapter

Designate the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each equation, and also designate the conjugate acid and conjugate base of each on the right side. (a) \(\mathrm{HBrO}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{BrO}^{-}(a q)\) (b) \(\mathrm{HSO}_{4}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\) (c) \(\mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) dimethylamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} ;\) (b) carbonate ion, \(\mathrm{CO}_{3}^{2-} ;(\mathbf{c})\) formate ion, \(\mathrm{CHO}_{2}^{-}\).

Explain the following observations: (a) HCl is a stronger acid than \(\mathrm{H}_{2} \mathrm{~S} ;\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is a stronger acid than \(\mathrm{H}_{3} \mathrm{AsO}_{4} ;\) (c) \(\mathrm{HBrO}_{3}\) is a stronger acid than \(\mathrm{HBrO}_{2}\); (d) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is a stronger acid than \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-} ;(\mathrm{e})\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) is a stronger acid than phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\).

Calculate the concentration of an aqueous solution of \(\mathrm{NaOH}\) that has a pH of \(11.50 .\)

Identify the Lewis acid and Lewis base among the reactants in each of the following reactions: (a) \(\mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{3}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) $$ \begin{array}{l} \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}(a q)+3 \mathrm{ClO}_{4}^{-}(a q) \\ \text { (b) } \mathrm{CN}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{OH}^{-}(a q) \end{array} $$ (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(g)+\mathrm{BF}_{3}(g) \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NBF}_{3}(s)\) $$ \text { (d) } \mathrm{HIO}(l q)+\mathrm{NH}_{2}^{-}(l q) \rightleftharpoons \mathrm{NH}_{3}(l q)+\mathrm{IO}^{-}(l q) $$ (lq denotes liquid ammonia as solvent)
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