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.
The dissociation constant of a substituted benzoic acid at $25^{\circ} \mathrm{C}$ is $1.0 \times 10^{-4}$. The $\mathrm{pH}$ of a $0.01 \ \mathrm{M}$ solution of its sodium salt is
Degree of dissociation of $0.1\,N\,\,C{H_3}COOH$ is (Dissociation constant $ = 1 \times {10^{ - 5}}$)
Ionisation constant of $CH_3COOH$ is $1.7 \times 10^{-5}$ and concentration of $H^+$ ions is $3.4 \times 10^{-4}$. Then find out initial concentration of $CH_3COOH$ Molecules
Derive ${K_a} \times {K_b} = {K_w}$ equation.
Accumulation of lactic acid $(HC_3H_5O_3),$ a monobasic acid in tissues leads to pain and a feeling of fatigue. In a $0.10\, M$ aqueous solution, lactic acid is $3.7\%$ dissociates. The value of dissociation constant, $K_a,$ for this acid will be