As a example, the ionization of dibasic acid $\mathrm{H}_{2} \mathrm{X}$ in aqueous solution is represented in two step.

$(i)$ $\mathrm{H}_{2} \mathrm{X}_{\text {(aq) }}+\mathrm{aq}+\mathrm{H}_{\text {(aq) }}^{+}+\mathrm{HX}_{\text {(aq) }}^{-}$

$(ii)$ $\mathrm{HX}_{\text {(aq) }}^{-}+\mathrm{aq}+\mathrm{H}_{\text {(aq) }}^{+}+\mathrm{X}_{\text {(aq) }}^{2-}$

If equilibrium constant of $\mathrm{K}_{a}$ $(i)$ and $\mathrm{K}_{a}$ $(ii)$ of this both equilibrium $(i)$ and $(ii)$ then,

$\therefore \mathrm{K}_{a}$ $(i)$ $=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HX}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{X}\right]}, \mathrm{K}_{a}$ $(ii)$ $=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{X}^{2-}\right]}{\left[\mathrm{HX}^{-}\right]}$

So, $\mathrm{K}_{a}$ (i) $\times \mathrm{K}_{a}$ $(ii)$ $=\frac{\left[\mathrm{H}^{+}\right]^{2}\left[\mathrm{X}^{2-}\right]}{\left[\mathrm{H}_{2} \mathrm{X}\right]}$ but

Reaction $(i)$ + Reaction $(ii)$

$\mathrm{H}_{2} \mathrm{X}_{(\mathrm{aq})}+\mathrm{aq} \square 2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{aq})}^{2-}$

For this, equilibrium constant $\mathrm{K}_{a}$ $(iii)$ is,

$\mathrm{K}_{a}$ $(iii)$ $=\frac{\left[\mathrm{H}^{+}\right]^{2}\left[\mathrm{X}^{2-}\right]}{\left[\mathrm{H}_{2} \mathrm{X}\right]}$

So, For dibasic acid,

$\mathrm{K}_{a}$ $(iii)$ $=\mathrm{K}_{a}$ $(i)$ $\times \mathrm{K}_{a}$ $(ii$)…. ….(Eq.-$i$)

where, $\mathrm{K}_{a}$$ (i)$ = First ionization constant, $\mathrm{K}_{a}$ $(ii)$ is second ionization constant.

For any polybasic acid respectively $\mathrm{K}_{a}$ (i), $\mathrm{K}_{a}$ $(ii)$…. than

$\mathrm{K}_{a}=\mathrm{K}_{a}$ $(i)$ $\times \mathrm{K}_{a}$ $(ii)$ $\times \ldots . . \quad$….(Eq.-ii)

Generally $\mathrm{K}_{a}$ (i) $>\mathrm{K}_{a}$ $(ii)$ $>\mathrm{K}_{a}$ $(iii)$…. as the after formation ion the remove of proton is difficult.