Question

Write the${\mathbf{K}}_{\mathbf{a}}$nd${\mathbf{K}}_{\mathbf{b}}$reactions of${\mathbf{NaHCO}}_{\mathbf{3}}$.

Short answer

The $K_a$and $K_b$reactions for${\mathrm{NaHCO}}_3$are:

role="math" localid="1663394686406" $\mathrm{HC}{\mathrm{O}}_3^{-}\stackrel{{\mathrm{K}}_{\mathrm{a}}}{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{co}}_3^{2-}\mathrm{HC}{{\mathrm{O}}_3^{-}}_{\rightleftharpoons }^{\mathrm{Ka}}{\mathrm{H}}^{+}+{\mathrm{co}}_3^{2-}$

${\mathrm{HCO}}_3^{-}+{\mathrm{H}}_2\mathrm{O}\stackrel{\mathrm{Kb}\hspace{0.17em}}{\rightleftharpoons }{\mathrm{H}}_{2{\mathrm{CO}}_3}+{\mathrm{OH}}^{-}{\mathrm{HCO}}_3^{-}+$ ${\mathrm{H}}_2{\mathrm{O}}_{\rightleftharpoons }^{\mathrm{Kb}}{\mathrm{H}}_2{\mathrm{CO}}_3+{\mathrm{OH}}^{-}$

Step-by-step solution

Acid dissociation constant 

A weak acid maintains equilibrium with its ions in an 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: $\mathrm{HA}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{A}}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}\mathrm{HA}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{A}}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}$

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

Step 2:Base dissociation constant

A weak base remains in equilibrium with its ions in an aqueous solution. $\mathrm{BOH}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{BH}}^{+}+{\mathrm{OH}}^{-}\mathrm{BOH}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{BH}}^{+}+{\mathrm{OH}}^{-}$ The equilibrium constant is known as the base dissociation constant, and it is given by the following relations.

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

 reaction for Kareaction forNaHCO3

The bicarbonate ion dissociates into one proton and one carbonate ion. An equilibrium is established. The reaction is given below.

${\mathrm{HCO}}_3^{-}\stackrel{\mathrm{Ka}}{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{CO}}_3^{2-}{\mathrm{HCO}}_3^{-}\stackrel{\mathrm{Ka}}{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{CO}}_3^{2-}$

Step 4:Kbreaction forNaHCO3

In this reaction, the bicarbonate ion reacts with water to form carbonic acid with the liberation of ${\mathrm{OH}}^{-}$ ions.

${\mathrm{HCO}}_3^{-}+{\mathrm{H}}_2\mathrm{O}\stackrel{\mathrm{Kb}}{={\mathrm{H}}_2}{\mathrm{co}}_3+{\mathrm{OH}}^{-}$

As a result, it's known as the base dissociation reaction.

Thus, the complete reaction between $Ka$and $Kb$reaction for${\mathrm{NaHCO}}_3$${\mathrm{HCO}}_3^{-}\stackrel{\mathrm{Ka}}{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{CO}}_3^{2-}$ ${\mathrm{HCO}}_3^{-}+{\mathrm{H}}_2\mathrm{O}\stackrel{\mathrm{Kb}}{\rightleftharpoons }{\mathrm{H}}_2{\mathrm{CO}}_3+{\mathrm{OH}}^{-}$ As a result, it's known as the base dissociation reaction