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 hydrogen ion concentration in weak acid of dissociation constant ${K_a}$ and concentration $c$ is nearly equal to
$0.1$ $mol$ of $H_2S(g)$ is kept in a $0.4$ litre vessel at $1000\,K$. For the reaction -
$2{H_2}S(g)\,\rightleftharpoons\,2{H_2}(g)\, + \,{S_2}(g)\,;\,{K_c} = {10^{ - 6}}\% $ dissociation of $H_2S$ is.......$\%$
If the $pKa$ of lactic acid is $5$,then the $pH$ of $0.005$ $M$ calcium lactate solution at $25^{\circ}\,C$ is $........\times 10^{-1}$ (Nearest integer)
The ionisation constant of acetic acid is $1.8 \times 10^{-5}$. The concentration at which it will be dissociated to $2\%$, is
The percentage of pyridine $(C_5H_5N)$ that forms pyridinium ion $(C_5H_5N^+H)$ in a $0.10\, M$ aqueous pyridine solution ($K_b$ for $C_5H_5N = 1.7 \times 10^{-9}$) is