Question

Write the conjugate base of \[\text{ }{{\text{H}}_{2}}\text{PO}_{4}^{-}\] ?
(A) \[\text{ }{{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}\]
(B) \[\text{ HPO}_{4}^{2-}\]
(C) \[\text{ PO}_{4}^{3-}\]
(D) \[\text{ }{{\text{H}}_{2}}\text{PO}_{3}^{-}\]

Answer 1

Hint: According to the Bronsted-Lowry acid-base concept, the conjugate base formed is the base member i.e.${{\text{X}}^{\text{-}}}$ of the pair of compounds that transforms by gaining or losing a proton. The conjugate base can gain a proton. But conjugate acid is more susceptible to losing a proton. Here the diphosphate ion acts as an acid. Thus its conjugate base form will be one proton fewer species.

Complete step by step solution:
The Bronsted –Lowry of the acid-base concept is based on the proton. According to the Bronsted –Lowry concept, the acid is the species that dissociates into its ions in water and it loses its proton. The species which is formed after losing the proton is called as the acid’s conjugate base.
Similarly, the species which accepts the proton in the chemical reaction is called the Bronsted-Lowry base and after gaining the proton it forms the conjugate acid of the base.
Let us consider a general acid-base reaction as follows:
$\text{Acid + Base }\rightleftharpoons \text{ Conjugate base + Conjugate Acid }$
We can say that according to the Bronsted-Lowry acid-base concept the acid $\text{Aci}{{\text{d}}_{\text{1}}}$ reacts with the base $\text{Bas}{{\text{e}}_{2}}$ to form $\text{Aci}{{\text{d}}_{2}}$and $\text{Bas}{{\text{e}}_{1}}$.where $\text{Aci}{{\text{d}}_{\text{1}}}$ loses a proton to form conjugate base $\text{Bas}{{\text{e}}_{1}}$and base $\text{Bas}{{\text{e}}_{2}}$ accept a proton to form conjugate acid $\text{Aci}{{\text{d}}_{2}}$
$\begin{align}
  & \\
 & \begin{matrix}
   \text{Aci}{{\text{d}}_{\text{1}}} & \text{+} & \text{Bas}{{\text{e}}_{\text{2}}} & \rightleftharpoons & \text{Bas}{{\text{e}}_{\text{1}}} & \text{+} & \text{Aci}{{\text{d}}_{\text{2}}} \\
   \text{(Acid)} & {} & \text{(Base)} & {} & \text{(Conjugate Base)} & {} & \text{(Conjugate Acid)} \\
\end{matrix} \\
\end{align}$
Let's have look on the question:
We have to find the conjugate base
$\text{ }{{\text{H}}_{\text{2}}}\text{PO}_{\text{4}}^{\text{-}}\text{ + Base }\rightleftharpoons \text{HPO}_{\text{4}}^{\text{2-}}\text{ + Base-H}$ $\text{ }{{\text{H}}_{\text{2}}}\text{PO}_{\text{4}}^{\text{-}}\text{ }$
 From the Bronsted-Lowry acid-base concept we know that the conjugate base is obtained when the species loses its proton.
Thus when $\text{ }{{\text{H}}_{\text{2}}}\text{PO}_{\text{4}}^{\text{-}}\text{ }$an acid and it is loses its proton as shown below:
${{\text{H}}_{\text{2}}}\text{PO}_{\text{4}}^{\text{-}}\text{ + Base }\rightleftharpoons \text{ HPO}_{\text{4}}^{\text{2-}}\text{+ Base-H}$
If the base is water then the reaction can be written as:
${{\text{H}}_{\text{2}}}\text{PO}_{\text{4}}^{\text{-}}\text{ + }{{\text{H}}_{\text{2}}}\text{O }\rightleftharpoons \text{ HPO}_{\text{4}}^{\text{2-}}\text{+ }{{\text{H}}_{3}}{{\text{O}}^{\text{+}}}$
Thus, the conjugate base of dihydrogen phosphate $\text{ (}{{\text{H}}_{\text{2}}}\text{PO}_{\text{4}}^{\text{-}}\text{ )}$ is phosphate ion$\text{ (HPO}_{\text{4}}^{\text{2-}})$

Hence, (B) is the correct option.

Note: The acid can generate conjugate base only in presence of a base. If the base is not present the reaction does not proceed. In other words, we can say that there must be a species that can accept the proton. Water is amphoteric. It can act as a base or acid depending on the reaction. It receives proto to form hydronium ion ${{\text{H}}_{3}}{{\text{O}}^{\text{+}}}$ conjugate acid.