Write the ${\mathit{K}}_{\mathbf{a}\mathbf{2}}$reaction of phosphoric acid $\left(H_3{\mathrm{PO}}_4\right)$and the ${\mathbf{K}}_{\mathbf{b}\mathbf{2}}$reaction of disodium oxalate$\mathbf{\left(}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{\right)}$.

A weak acid or base does not completely dissociate in an aqueous solution. The degree of dissociation of a weak acid or a weak base is very low. When dissolved in water, a weak acid or a weak base establishes an equilibrium condition between the ions and the molecules.

For example, if HA is a weak acid and BOH is a weak base, then the ionization reactions are:

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

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

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

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

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

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.

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

The formula of phosphoric acid is ${\mathrm{H}}_3{\mathrm{PO}}_4$. Thus, there are three hydrogens that can be removed. The ionization reactions are:

${\mathrm{H}}_3{\mathrm{PO}}_4\stackrel{{\mathrm{K}}_{\mathrm{a}1}}{\rightleftharpoons }{\mathrm{HPO}}_4^{-}+{\mathrm{H}}^{+}$

${\mathrm{H}}_2{\mathrm{PO}}_4^{-}\stackrel{{\mathrm{K}}_{\mathrm{a}2}}{\rightleftharpoons }{\mathrm{HPO}}_4^{2-}+{\mathrm{H}}^{+}$

Sodium oxalate can take up two protons. The reactions are:

$$\begin{gathered} {\mathrm{C}}_2{\mathrm{O}}_4^{-2}+{\mathrm{H}}_2\mathrm{O}\stackrel{{\mathrm{K}}_{\mathrm{b}1}}{\rightleftharpoons }{\mathrm{C}}_2{\mathrm{O}}_4{\mathrm{H}}^{-}+{\mathrm{OH}}^{-} \\ {\mathrm{HC}}_2{\mathrm{O}}_4^{-}+{\mathrm{H}}_2\mathrm{O}\stackrel{{\mathrm{K}}_{\mathrm{b}2}}{\rightleftharpoons }{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4+{\mathrm{OH}}^{-} \end{gathered}$$