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The acid dissociation constant $\mathbf{\left(}{\mathbf{K}}_{\mathbf{a}}\mathbf{\right)}$is a numerical representation of an acid's strength in solution. The dissociation constant${\mathbf{\text{K}}}_{\mathbf{\text{a}}}\mathbf{\text{=}}\frac{{\mathbf{\text{[A}}}^{\mathbf{\text{-}}}{\mathbf{\text{][H}}}^{\mathbf{\text{+}}}\mathbf{\text{]}}}{\mathbf{\text{[HA]}}}$is commonly stated as a quotient of the equilibrium concentrations (in$\mathit{m}\mathit{o}\mathit{l}\mathbf{/}\mathit{L}$).

To write the$K_a$ expression of${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}$ in water, we should write the balanced reaction first –

${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}{\text{(aq) +H}}_{\text{2}}\text{O(l)}⇌{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) + HPO}}_{\text{4}}^{\text{2 -}}\text{(aq)}$

Then write the$K_a$ expression using the formula below. Note that it only includes aqueous species.

$$\begin{gathered} {\text{K}}_{\text{a}}\text{=}\frac{\text{[product]}}{\text{[reactant]}} \\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][HPO}}_{\text{4}}^{\text{2 -}}\text{]}}{{\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\text{]}} \end{gathered}$$

Therefore, the expression is obtained as ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][HPO}}_{\text{4}}^{\text{2 -}}\text{]}}{{\text{[H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\text{]}}$.

To write$K_a$ the expression of${\text{H}}_{\text{3}}{\text{PO}}_{\text{2}}$ in water, we should write the balanced reaction first –

${\text{H}}_{\text{3}}{\text{PO}}_{\text{2}}{\text{(aq) +H}}_{\text{2}}\text{O(l)}⇌{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) +H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{-}}\text{(aq)}$

Then write the$K_a$ expression using the formula below. Note that it only includes aqueous species.

$$\begin{gathered} {\text{K}}_{\text{a}}\text{=}\frac{\text{[product]}}{\text{[reactant]}} \\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[H}}_{\text{3}}{\text{PO}}_{\text{2}}\text{]}} \end{gathered}$$

Therefore, the expression is obtained as${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][H}}_{\text{2}}{\text{PO}}_{\text{2}}^{\text{-}}\text{]}}{{\text{[H}}_{\text{3}}{\text{PO}}_{\text{2}}\text{]}}$ .

To write the$K_a$ expression of${\text{HSO}}_{\text{4}}^{\text{-}}$ in water, we should write the balanced reaction first –

${\text{HSO}}_{\text{4}}^{\text{-}}{\text{(aq) +H}}_{\text{2}}\text{O(l)}⇌{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{(aq) + SO}}_{\text{4}}^{\text{2 -}}\text{(aq)}$

Then write the$K_a$ expression using the formula below. Note that it only includes aqueous species.

$$\begin{gathered} {\text{K}}_{\text{a}}\text{=}\frac{\text{[product]}}{\text{[reactant]}} \\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][SO}}_{\text{4}}^{\text{2 -}}\text{]}}{{\text{[HSO}}_{\text{4}}^{\text{-}}\text{]}} \end{gathered}$$

Therefore, the expression is obtained as${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][SO}}_{\text{4}}^{\text{2 -}}\text{]}}{{\text{[HSO}}_{\text{4}}^{\text{-}}\text{]}}$ .