Question

${{K}_{{{a}_{1}}}},{{K}_{{{a}_{2}}}}$ and ${{K}_{{{a}_{3}}}}$ values for ${{H}_{3}}P{{O}_{4}}$ are ${{10}^{-3}},{{10}^{-8}}$ and ${{10}^{-12}}$ respectively, then:
(i) What is the dissociation constant of $HPO_{4}^{2-}$?
(ii) What is ${{K}_{b}}$ of $HPO_{4}^{2-}$?
(iii) What is ${{K}_{b}}$ of${{H}_{2}}PO_{4}^{-}$?
(iv) What is ${{K}_{b}}$ of $PO_{4}^{3-}$?

Answer 1

Hint: According to bronsted-lowry theory of acid-base, a compound that accepts hydrogen ion in a solution is a base and that gives off hydrogen ion in a solution is an acid. The species that forms after a compound gives off hydrogen ion will be its conjugate base and vice versa. For example if water gives off hydrogen ion then it will be an acid and hydroxide ion will be its conjugate base and similarly if water accepts hydrogen ion then it will be base and hydronium ion formed will be its conjugate acid.

Complete step by step solution:
${{H}_{3}}P{{O}_{4}}\to {{H}^{+}}+{{H}_{2}}PO_{4}^{-}$
${{H}_{2}}PO_{4}^{-}\to {{H}^{+}}+HPO_{4}^{2-}$
$HPO_{4}^{2-}\to {{H}^{+}}+PO_{4}^{3-}$
Answer (i) dissociation constant of $HPO_{4}^{2-}$ will be ${{K}_{{{a}_{3}}}}$ and ${{K}_{{{a}_{3}}}}$ is equal to${{10}^{-12}}$.
Answer (ii) product of dissociation constants of conjugate acid-base pair will always be equal to the ionic product of water.
Conjugate acid of $HPO_{4}^{2-}$ is ${{H}_{2}}PO_{4}^{-}$, therefore
${{K}_{{{a}_{2}}}}\times {{K}_{{{b}_{2}}}}={{K}_{w}}$
${{K}_{{{b}_{2}}}}=\dfrac{{{K}_{w}}}{{{K}_{{{a}_{2}}}}}$
${{K}_{{{b}_{2}}}}=\dfrac{{{10}^{-14}}}{{{10}^{-8}}}$
${{K}_{{{b}_{2}}}}={{10}^{-6}}$
So ${{K}_{b}}$ of $HPO_{4}^{2-}$ is ${{10}^{-6}}$
Answer(iii) product of dissociation constants of conjugate acid-base pair will always be equal to the ionic product of water.
Conjugate acid of ${{H}_{2}}PO_{4}^{-}$ is ${{H}_{3}}PO_{4}^{{}}$, therefore
${{K}_{{{a}_{1}}}}\times {{K}_{{{b}_{1}}}}={{K}_{w}}$
${{K}_{{{b}_{1}}}}=\dfrac{{{K}_{w}}}{{{K}_{{{a}_{1}}}}}$
${{K}_{{{b}_{1}}}}=\dfrac{{{10}^{-14}}}{{{10}^{-3}}}$
${{K}_{{{b}_{1}}}}={{10}^{-11}}$
So ${{K}_{b}}$ of ${{H}_{2}}PO_{4}^{-}$ is ${{10}^{-11}}$
Answer (iv)
Higher the value of dissociation constant of acid, stronger the acid will be. That clears that ${{H}_{3}}P{{O}_{4}}$ is stronger acid than ${{H}_{2}}PO_{4}^{-}$ and ${{H}_{2}}PO_{4}^{-}$ is stronger acid than $HPO_{4}^{2-}$
Stronger the acid, weaker the conjugate base will be and vice versa.
That suggests that $PO_{4}^{3-}$ is stronger base than $HPO_{4}^{2-}$ and $HPO_{4}^{2-}$ is the stronger base than ${{H}_{2}}PO_{4}^{-}$ and their ${{K}_{b}}$ values will be in the order of their basic strengths.
Therefore, ${{K}_{{{b}_{3}}}}>{{K}_{{{b}_{2}}}}>{{K}_{{{b}_{1}}}}$

Additional information:
Orthophosphoric acid is poly protic acid that gives more than one hydrogen ion when dissolved into the water.

Note: Let’s understand why the product of dissociation constant of an acid and its conjugate base is equal to the ionic product of water.
This is dissociation reaction of orthophosphoric acid in water
\[{{H}_{3}}P{{O}_{4}}(aq)\to {{H}_{2}}PO_{4}^{-}+{{H}^{+}}\]
Its dissociation constant will be
${{K}_{a}}=\dfrac{[{{H}_{2}}PO_{4}^{-}][{{H}^{+}}]}{[{{H}_{3}}P{{O}_{4}}]}$
Reaction of dissociation of conjugate base in water
${{H}_{2}}PO_{4}^{-}(aq)+{{H}_{2}}O(l)\to {{H}_{3}}P{{O}_{4}}(aq)+O{{H}^{-}}(aq)$
Dissociation constant of conjugate base will be
${{K}_{b}}=\dfrac{[{{H}_{3}}P{{O}_{4}}][O{{H}^{-}}]}{[{{H}_{2}}PO_{4}^{-}]}$