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(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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kb value
The Kb value, or base ionization constant, measures the strength of a base in water. It represents the equilibrium constant for the ionization of a base into its conjugate acid and hydroxide ions. In other words, the higher the Kb, the stronger the base, indicating that the base dissociates more in water to produce hydroxide ions. For instance, in this exercise, ammonia (NH3) has a Kb value of \(1.8 \times 10^{-5}\), whereas hydroxylamine (H3NO) possesses a Kb of \(1.1 \times 10^{-8}\).
  • Ammonia is the stronger base because it has a higher Kb value.
  • Higher Kb reflects a base's greater tendency to attract protons, thus forming hydroxide ions more readily.

Understanding the Kb value is crucial in predicting reactions involving acids and bases, especially in comparing the relative strength of different substances in terms of their basicity.
Conjugate acid
In the context of acid-base chemistry, every base has a conjugate acid, which is the species formed when the base gains a proton. It is important to note that a strong base usually correlates with a weak conjugate acid and vice versa.
  • The conjugate acid of ammonia (\(NH_3\)) is the ammonium ion (\(NH_4^+\)).
  • The conjugate acid of hydroxylamine (\(H_3NO\)) is the hydroxylammonium ion (\(H_3NOH^+\)).

Because ammonia is a stronger base than hydroxylamine, its conjugate acid, \(NH_4^+\), is weaker in terms of acidity compared to \(H_3NOH^+\). This relationship helps us establish the link between the strength of a base and its corresponding conjugate acid.
Ion-product constant of water (Kw)
The ion-product constant of water (\(K_w\)) is a fundamental principle in acid-base chemistry. It relates the concentrations of hydrogen ions and hydroxide ions in water at equilibrium. At 25°C, \(K_w\) is \(1.0 \times 10^{-14}\). This constant is key to understanding the relationship between acidic and basic environments in water.
  • The equation \(K_w = [H^+][OH^-]\) describes the product of the concentrations of these ions.
  • Water can act as both an acid and a base - it self-ionizes slightly, which is why \(K_w\) is so small.

In the context of the original exercise, \(K_w\) is used to connect the base ionization constant (\(K_b\)) and the acid ionization constant (\(K_a\)) of the conjugate acid through \(K_w = K_a \times K_b\). This relationship allows for the calculation of \(K_a\) from a known \(K_b\) and vice versa.
Ka value
The \(K_a\) value, or acid ionization constant, measures the strength of an acid in solution. It represents the equilibrium constant for the ionization of an acid into its conjugate base and hydronium ions. A higher \(K_a\) value indicates a stronger acid, showing a greater tendency to donate protons. In this exercise, we calculate \(K_a\) using the relation \(K_a = \frac{K_w}{K_b}\):
  • For the ammonium ion (\(NH_4^+\)), \(K_a = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\).
  • For the hydroxylammonium ion (\(H_3NOH^+\)), \(K_a = \frac{1.0 \times 10^{-14}}{1.1 \times 10^{-8}} = 9.09 \times 10^{-7}\).

These \(K_a\) values highlight that \(H_3NOH^+\) is a stronger acid compared to \(NH_4^+\), as indicated by its larger \(K_a\) value.
Ammonia
Ammonia (NH3) is a common compound with significant importance in chemistry, particularly as both a base and a nucleophile. Its basic characteristics stem from its ability to accept a proton to form its conjugate acid, ammonium \(NH_4^+\).
The Kb value of ammonia, \(1.8 \times 10^{-5}\), reflects its strength as a base, which is stronger compared to many other bases like hydroxylamine. Ammonia forms a balanced equilibrium in water between its molecular form and its ionized form:
  • Ammonia serves various roles -- from industrial applications, such as fertilizers, to important reactions in biochemistry.
  • In aqueous solutions, it creates \eqref{NH_4^+} and \eqref{OH^-} ions which affect the pH.

Understanding ammonia's base behavior provides insights into how basicity impacts chemical reactions and equilibrium.
Hydroxylamine
Hydroxylamine \(\text{(H}_3\text{NO)}\), like ammonia, also acts as a base but has a considerably lower Kb value (\(1.1 \times 10^{-8}\)). This signifies it is a weaker base than ammonia.
Hydroxylamine's conjugate acid, hydroxylammonium \(\text{(H}_3\text{NOH}^+\)), is correspondingly stronger than the conjugate acid of ammonia due to its lower basicity. Despite being less basic, hydroxylamine has unique properties and uses:
  • Often used in the synthesis of chemical intermediates and pharmaceuticals.
  • It is involved in biological pathways, such as the biosynthesis of cyclic hydroxamic acids.

While hydroxylamine is weaker in basicity than ammonia, its role in chemical syntheses and reactions is diverse, highlighting the importance of knowing both its strengths and limitations as a base.

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Most popular questions from this chapter

Which member of each pair produces the more acidic aque- ous solution: (a) \(\mathrm{ZnBr}_{2}\) or \(\mathrm{CdCl}_{2},\) (b) \(\mathrm{CuCl}\) or \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2},(\mathrm{c})\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) or \(\mathrm{NiBr}_{2}\) ? Explain.

Determine the \(\mathrm{pH}\) of each of the following solutions \(\left(K_{a}\right.\) and \(K_{b}\) values are given in Appendix \(D\) ): (a) \(0.095 M\) hypochlorous acid, (b) \(0.0085 \mathrm{M}\) hydrazine, (c) \(0.165 \mathrm{M}\) hydroxylamine.

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)

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, \(\mathrm{CHO}_{2}^{-}\).

The amino acid glycine \(\left(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}\right)\) can participate in the following equilibria in water: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \\ \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \quad K_{a}=4.3 \times 10^{-3} \\ \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \\ +\mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{OH}^{-} \quad K_{b}=6.0 \times 10^{-5} \end{aligned} $$ (a) Use the values of \(K_{a}\) and \(K_{b}\) to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion: $$ \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH} \rightleftharpoons{ }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-} $$ What assumptions did you need to make? (b) What is the pH of a \(0.050 \mathrm{M}\) aqueous solution of glycine? (c) What would be the predominant form of glycine in a solution with \(\mathrm{pH} 13 ?\) With \(\mathrm{mH}^{1} ?\)

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