Question

Question:Write the dissociation reaction and the corresponding K_a equilibrium expression for each of the following acids in water.

a. $\mathbf{HCN}$

b. ${\mathbf{HOC}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}$

c. ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{NH}}_{\mathbf{3}}$

Short answer

Acids dissociate in water to give the corresponding anion, and hydronium ions and acid dissociation constant is the measure of the acid strength of a solution.

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

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

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

Step-by-step solution

Dissociation of HCN in Water.

HCN is a weak acid and dissociates in anion and hydronium ions.

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

Dissociation of HOC6H5 in Water.

Phenol act as a weak acid in water and gives phenoxide ion.

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

Dissociation of  in C6H5NH3+Water.

Amines on accepting proton give ammonium ions that are acidic in nature and give amine to donate H⁺ ion.

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