Question

Write the reaction and the corresponding K_b equilibrium expression for each of the following substances (acting as bases in water).

a. ${\mathbf{NH}}_{\mathbf{3}}$b.${\mathbf{CN}}^{\mathbf{-}}$ c. pyridine,${\mathbf{C}}_{\mathbf{5}}{\mathbf{H}}_{\mathbf{5}}\mathbf{N}$ d. aniline,data-custom-editor="chemistry" ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{NH}}_{\mathbf{2}}$

Short answer

e,

Answer

A base dissociates in water to give the corresponding cation (after accepting H⁺ ion) and hydroxyl ion. The base dissociation constant is the measure of the basic strength of a basic solution.

(a) $$\begin{gathered} {\mathrm{NH}}_{3_{\left(\mathrm{aq}\right)}}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}\rightleftharpoons {{\mathrm{NH}}_4^{+}}_{\left(\mathrm{aq}\right)}+{{\mathrm{OH}}^{-}}_{\left(\mathrm{aq}\right)} \\ {\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{NH}}_4^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{NH}}_3\right]} \end{gathered}$$

(b) $$\begin{gathered} \mathrm{C}{{\mathrm{N}}^{-}}_{\left(\mathrm{aq}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}\mathit{}\rightleftharpoons {\mathrm{HCN}}_{\left(\mathrm{aq}\right)}+{{\mathrm{OH}}^{-}}_{\left(\mathrm{aq}\right)} \\ {\mathrm{K}}_{\mathrm{b}}=\frac{\left[\mathrm{HCN}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{CN}}^{-}\right]} \end{gathered}$$

(c) $$\begin{gathered} {\mathrm{C}}_5{\mathrm{H}}_5{\mathrm{N}}_{\left(\mathrm{aq}\right)}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_6{{\mathrm{NH}}^{+}}_{\left(\mathrm{aq}\right)}+{{\mathrm{OH}}^{-}}_{\left(\mathrm{aq}\right)} \\ {\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}\right]} \end{gathered}$$

(d) $$\begin{gathered} {\mathrm{C}}_5{\mathrm{H}}_5{\mathrm{NH}}_{2\left(\mathrm{aq}\right)+}{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5{{\mathrm{NH}}_3^{+}}_{\left(\mathrm{aq}\right)}+{\mathrm{OH}}^{-} \\ {\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{C}}_6{\mathrm{H}}_5{{\mathrm{NH}}_3}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\right]} \end{gathered}$$

Step-by-step solution

NH3 in Water

NH₃ acts as a base and forms cation and hydroxyl ion.

$$\begin{gathered} {\mathrm{NH}}_{3_{\left(\mathrm{aq}\right)}}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}\rightleftharpoons {{\mathrm{NH}}_4^{+}}_{\left(\mathrm{aq}\right)}+{{\mathrm{OH}}^{-}}_{\left(\mathrm{aq}\right)} \\ {\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{NH}}_4^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{NH}}_3\right]} \end{gathered}$$

CN- in Water

CN^- acts as a base as it accepts a ion and produces and hydroxyl ion.

$$\begin{gathered} \mathrm{C}{{\mathrm{N}}^{-}}_{\left(\mathrm{aq}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}\mathit{}\rightleftharpoons {\mathrm{HCN}}_{\left(\mathrm{aq}\right)}+{{\mathrm{OH}}^{-}}_{\left(\mathrm{aq}\right)} \\ {\mathrm{K}}_{\mathrm{b}}=\frac{\left[\mathrm{HCN}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{CN}}^{-}\right]} \end{gathered}$$

Pyridine in Water

Pyridine acts as a Bronsted Lowry base as it accepts a ion and produces pyridinium ion and hydroxyl ion.

$$\begin{gathered} {\mathrm{C}}_5{\mathrm{H}}_5{\mathrm{N}}_{\left(\mathrm{aq}\right)}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_6{{\mathrm{NH}}^{+}}_{\left(\mathrm{aq}\right)}+{{\mathrm{OH}}^{-}}_{\left(\mathrm{aq}\right)} \\ {\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}\right]} \end{gathered}$$

Aniline in Water

Aniline acts as a Bronsted Lowry base as it accepts a ${\mathrm{H}}^{+}$ion(due to the presence of lone pair of electrons) and produces ${\mathrm{C}}_6{\mathrm{H}}_5{{\mathrm{NH}}_3}^{+}$(aniliniumion) and hydroxyl ion.

$$\begin{gathered} {\mathrm{C}}_5{\mathrm{H}}_5{\mathrm{NH}}_{2\left(\mathrm{aq}\right)+}{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{C}}_6{\mathrm{H}}_5{{\mathrm{NH}}_3^{+}}_{\left(\mathrm{aq}\right)}+{\mathrm{OH}}^{-} \\ {\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{C}}_6{\mathrm{H}}_5{{\mathrm{NH}}_3}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\right]} \end{gathered}$$