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If $K{a_1},\,K{a_2}\,{\text{and}}\,K{a_3}\,$ for ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$are ${10^{ - 3}}$, ${10^{ - 8}}$ and ${10^{ - 12}}$ respectively. If ${K_w}\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right) = {10^{ - 14}}$,
(i) What is the dissociation constant for ${\text{HP}}{{\text{O}}_{\text{4}}}^{2 - }$
(ii) What is ${K_b}$ of ${\text{HP}}{{\text{O}}_{\text{4}}}^{2 - }$
(iii) What is ${K_b}$ of ${{\text{H}}_2}{\text{P}}{{\text{O}}_{\text{4}}}^ - $
(iv) What is order of ${K_b}$ of ${\text{P}}{{\text{O}}_{\text{4}}}^{3 - }$ $\left( {K{b_3}} \right)$
, ${\text{HP}}{{\text{O}}_{\text{4}}}^{2-}$ $\left({K{b_2}}\right)$ and ${{\text{H}}_2}{\text{P}}{{\text{O}}_{\text{4}}}^ - $$\left( {K{b_1}} \right)$

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Last updated date: 03rd Mar 2024
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IVSAT 2024
Answer
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Hint: We know that some acids such as oxalic acid, sulphuric acid etc. have more than one ionizable proton per molecule of the acid. These acids are known as polybasic acids. These acids will ionize or dissociate in more than one step. Phosphoric acid $\left( {{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}} \right)$ is a polybasic acid.

Complete step by step answer:
(i) Given that first $\left( {K{a_1}} \right)$, second $\left( {\,K{a_2}\,} \right)$ and third ionization constant of ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ are ${10^{ - 3}}$ , ${10^{ - 8}}$ and ${10^{ - 12}}$ respectively.
First dissociation reaction is,
\[{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}} \rightleftharpoons {{\text{H}}^ + } + {\left( {{{\text{H}}_{\text{2}}}{\text{P}}{{\text{O}}_{\text{4}}}} \right)^ - }\,,\,K{a_1} = {10^{ - 3}}\]…… (1)
Second dissociation reaction is,
${\left( {{{\text{H}}_{\text{2}}}{\text{P}}{{\text{O}}_{\text{4}}}}\right)^-}\rightleftharpoons {{\text{H}}^ + } + {\left( {{\text{HP}}{{\text{O}}_{\text{4}}}} \right)^{2 - }},K{a_2} = {10^{ - 8}}$….. (2)
Third dissociation constant is,
${\left( {{\text{HP}}{{\text{O}}_{\text{4}}}} \right)^{- 2}} \rightleftharpoons {{\text{H}}^ + } + {\left( {{\text{P}}{{\text{O}}_{\text{4}}}} \right)^{ - 3}},K{a_3} = {10^{ - 12}}$…… (3)
From equation (2), we get to know that dissociation constant of ${\text{HP}}{{\text{O}}_{\text{4}}}^{2 - }$ is ${10^{ - 8}}$.
(ii) For a reaction, $Ka$ is calculated for conjugate acid and ${K_b}$ is calculated for conjugate base. To calculate ${K_b}$, we have to use the below formula.
${K_w} = Ka \times {K_b}$
Here, ${K_w}$ is ionization constant of water, $Ka$ is ionization constant and ${K_b}$ is basic constant.
Given that $Ka$ of ${\text{HP}}{{\text{O}}_{\text{4}}}^{2 - }$ is ${10^{ - 8}}$ and ${K_w} = {10^{ - 14}}$
So, $ \Rightarrow {K_b} = \dfrac{{{K_w}}}{{Ka}} = \dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 8}}}} = {10^{ - 6}}$
Therefore, ${K_b}$ of ${\text{HP}}{{\text{O}}_{\text{4}}}^{2 - }$ is ${10^{ - 6}}$.
(iii) Now we have to calculate the ${K_b}$ of ${{\text{H}}_2}{\text{P}}{{\text{O}}_{\text{4}}}^ - $
as in part (ii).
Given that $Ka$ of ${{\text{H}}_2}{\text{P}}{{\text{O}}_{\text{4}}}^ - $ is ${10^{ - 3}}$ and ${K_w} = {10^{ - 14}}$
So, $ \Rightarrow {K_b} = \dfrac{{{K_w}}}{{Ka}} = \dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 3}}}} = {10^{ - 11}}$
Therefore, ${K_b}$ of ${{\text{H}}_2}{\text{P}}{{\text{O}}_{\text{4}}}^ - $ is ${10^{ - 11}}$.
(iv) So, from the above calculation we calculated that,
${K_b}$ of ${{\text{H}}_2}{\text{P}}{{\text{O}}_{\text{4}}}^ - $ $\left( {{K_b}_{_1}} \right)$
is ${10^{ - 11}}$
${K_b}$ of ${\text{HP}}{{\text{O}}_{\text{4}}}^{2 - }$$\left( {{K_b}_{_2}} \right)$ is ${10^{ - 6}}$
Now, we have calculate the ${K_b}$ of ${\text{P}}{{\text{O}}_{\text{4}}}^{3 - }$ $\left( {{K_b}_{_3}} \right)$.
Given that $Ka$ of ${\text{P}}{{\text{O}}_{\text{4}}}^{ - 3}$ is ${10^{ - 12}}$
and ${K_w} = {10^{ - 14}}$
$ \Rightarrow {K_b} = \dfrac{{{K_w}}}{{Ka}} = \dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 12}}}} = {10^{ - 2}}$
So, the value of $K{b_3} = {10^{ - 2}}$
${K_b}_{_3} > {k_b}_{_2} > {k_b}_{_1}$

Note: It is to be noted that it is always difficult to remove a positively charged species (proton) from a negatively charged ions as compared to a neutral molecule because of the strong electrostatic force of attraction between proton and the anion with which it is present. As a result, the removal of proton from a monovalent anion is quite difficult and is even more difficult to remove the same from a divalent anion. Thus, the order of dissociation constants can be justified.
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