What is polyprotic acid ? Give examples of polyprotic acid and its ionization.
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 $\mathrm{A}^{2-}$.
Similar Questions
$0.1$ $mol$ 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.......$\%$
$0.1$ $mol$ 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.......$\%$
The $pH$ of the solution obtained on neutralisation of $40\, mL\, 0.1\, M\, NaOH$ with $40\, mL\, 0.1\, M\, CH_3COOH$ is
The $pH$ of the solution obtained on neutralisation of $40\, mL\, 0.1\, M\, NaOH$ with $40\, mL\, 0.1\, M\, CH_3COOH$ is
- [AIIMS 2007]
${K_a} = 1.4 \times {10^{ - 5}}$ of propanoic acid. Calculate its $pH$ of $0.1$ $M$ solution.
${K_a} = 1.4 \times {10^{ - 5}}$ of propanoic acid. Calculate its $pH$ of $0.1$ $M$ solution.
A $0.1\,N $ solution of an acid at room temperature has a degree of ionisation $ 0.1$ . The concentration of $O{H^ - }$ would be
A $0.1\,N $ solution of an acid at room temperature has a degree of ionisation $ 0.1$ . The concentration of $O{H^ - }$ would be
A certain amount of $H_2CO_3$ & $HCl$ are dissolved to form $1$ litre solution. At equilibrium it is found that concentration of $H_2CO_3$ & $CO_3^{-\,-}$ are $0.1\,M$ & $0.01\,M$ respectively. Calculate the $pH$ of solution. Given that for $H_2CO_3$ $K_{a_1} =10^{-5}$ & $K_{a_2} =10^{-8}$
A certain amount of $H_2CO_3$ & $HCl$ are dissolved to form $1$ litre solution. At equilibrium it is found that concentration of $H_2CO_3$ & $CO_3^{-\,-}$ are $0.1\,M$ & $0.01\,M$ respectively. Calculate the $pH$ of solution. Given that for $H_2CO_3$ $K_{a_1} =10^{-5}$ & $K_{a_2} =10^{-8}$