Explain ionization and ionization constant in di and polyprotic acid.
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.
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