What is polyprotic acid ? Give examples of polyprotic acid and its ionization.
Polyprotic acid : The acids which have more than one ionizable proton per molecule of the acid. Such acids are known as polybasic or polyprotic acids.
Diprotic acid: The acid which have two ionizable proton per molecule of the acid. Such acids are known as Dibasic or Diprotic acid.
e.g., $\mathrm{H}_{2} \mathrm{X}+$ (aq) $+2 \mathrm{H}^{+}+\mathrm{X}^{2-} \quad \mathrm{K}_{a}$
The ionization of the reaction is in two steps is as under,
$(i)$ $\mathrm{H}_{2} \mathrm{X}+\mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HX}^{-} \quad \mathrm{K}_{a}$ $(i)$
$(ii)$ $\mathrm{HX}^{-}+\mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{X}^{2-} \quad \mathrm{K}_{a}$ $(ii)$
In it, $\mathrm{K}_{a}$ $(i)$ $>\mathrm{K}_{a}$ $(ii)$ and $\mathrm{K}_{a}$ (i) $\times \mathrm{K}_{a}$ $(ii)$ $=\mathrm{K}_{a}$
Examples of diprotic acid : Oxalic acid $\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)$, Sulphuric acid $\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)$, Carbonic acid $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)$, Sulphurous acid $\left(\mathrm{H}_{2} \mathrm{SO}_{3}\right)$, Ascorbic acid etc.
Examples of triprotic acid : Phosphoric acid $\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right)$, Cytaric acid.... etc.
In polyprotic acid solution constant mixture of acid like $\mathrm{H}_{2} \mathrm{~A}, \mathrm{HA}^{-}$and $\mathrm{A}^{2-}$.
$K _{ a_1,}, K _{ a_2 }$ and $K _{ a_3}$ are the respective ionization constants for the following reactions $(a), (b),$ and $(c)$.
$(a)$ $H _{2} C _{2} O _{4} \rightleftharpoons H ^{+}+ HC _{2} O _{4}^{-}$
$(b)$ $HC _{2} O _{4}^{-} \rightleftharpoons H ^{+}+ HC _{2} O _{4}^{2-}$
$(c)$ $H _{2} C _{2} O _{4} \rightleftharpoons 2 H ^{+}+ C _{2} O _{4}^{2-}$
The relationship between $K_{a_{1}}, K_{ a _{2}}$ and $K_{ a _{3}}$ is given as
Heat of neutralisation of weak acid and strong base is less than the heat of neutralisation of strong acid and strong base due to
For a concentrated solution of a weak electrolyte ( $K _{ eq }=$ equilibrium constant) $A _2 B _3$ of concentration ' $c$ ', the degree of dissociation " $\alpha$ ' is
The $pH$ of $0.1$ $M$ $HCN$ solution is $5.2$ calculate ${K_a}$ of this solution.
If $pK_a =\, -\,log K_a=4$ for a weak acid $HX$ and $K_a= C\alpha ^2$ then Van't Haff factor when $C = 0.01\,M$ is