Question

Write the ${\mathbf{K}}_{\mathbf{b}}$reaction of ${\mathbf{CN}}^{\mathbf{-}}$ Given that the ${\mathbf{K}}_{\mathbf{a}}$ value for $\mathbf{HCN}$is$\mathbf{6}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}$ calculate${\mathbf{K}}_{\mathbf{b}}$ for${\mathbf{CN}}^{\mathbf{-}}$.

Short answer

The reaction is given below.

${\mathrm{CN}}^{-}+{\mathrm{H}}_2\mathrm{O}\stackrel{{\mathrm{K}}_{\mathrm{b}}}{\leftleftarrows }\mathrm{HCN}+{\mathrm{OH}}^{-}{\mathrm{CN}}^{-}+{\mathrm{H}}_2\mathrm{O}\stackrel{{\mathrm{K}}_{\mathrm{b}}}{\leftleftarrows }\mathrm{HCN}+{\mathrm{OH}}^{-}$

The value of ${\mathrm{K}}_{\mathrm{b}}\mathrm{is}1.6\times {10}^{-5}$.

Step-by-step solution

Acid dissociation constant

A weak acid maintains equilibrium with its ions in aqueous solution. Since equilibrium is established, there must be an equilibrium constant for the reaction. The acid dissociation constant is given by the following relation:

$\mathrm{HA}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{A}}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}$

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{A}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[\mathrm{HA}\right]}$

Base dissociation constant

A weak base remains in equilibrium with its ions in an aqueous solution.

$\mathrm{BOH}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{BH}}^{+}+{\mathrm{OH}}^{-}$

The equilibrium constant is known as the base dissociation constant, and it is given by the following relations.

${\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{OH}}^{-}\right]\left[{\mathrm{BH}}^{+}\right]}{\left[\mathrm{BOH}\right]}$

Step 3: KB reaction for CN-

is the conjugate base of HCN. The acid-base reaction is given below.

${\mathbf{CN}}^{\mathbf{-}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\stackrel{{\mathbf{K}}_{\mathbf{b}}}{\mathbf{\leftleftarrows }}\mathbf{HCN}\mathbf{+}{\mathbf{OH}}^{\mathbf{-}}$

Calculation of Kb

The relation between $K_a$and $K_b$is:

$$\begin{gathered} K_w=K_a\times K_b \\ {K}_b=\frac{K_w}{K_a} \\ =\frac{1.0\times {10}^{-14}}{6.2\times {10}^{-10}} \\ =1.6\times {10}^{-5} \end{gathered}$$

The value of ${\mathrm{K}}_{\mathrm{b}}\mathrm{is}1.6\times {10}^{-5}$.