Consider various species generated when $H_{3} P O_{4}$ is dissolved in water. According to the Bronsted Lowry’s Concept among these, the conjugate acid of $H P O_{4}^{2 -}$ is :

H P O_{4}^{2 -}

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H_{2} P O_{4}^{-}

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P O_{4}^{3 -}

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H_{3} O^{+}

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Solution

The correct option is B $H_{2} P O_{4}^{-}$
When a Bronsted-lowry acid is in equilibrium, it will donate a proton (as acids do) in the form of a $H^{+}$ , and therefore become a base (willing to accept a proton). This resulting base is called the conjugate base. Again when in equilibrium, it will accept a proton, and become a conjugate acid.
Here is the equilibrium expression:
$H P O_{4}^{2 -} + H_{2} O ⇌ H_{2} P O_{4}^{-} (conjugate acid) + O H^{-}$ .
Simply the conjugate acid of any species is that species plus a proton; so it's $H_{2} P O_{4}^{-}$ .

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Similar questions

H_{3} P O_{4}

H P O_{4}^{2 -}

H_{2} O + H_{3} P O_{4} + H_{3} O^{+} + H_{2} P O_{4}^{-}, p^{k_{1}} = 2.15

H_{2} O + H_{2} P O_{4}^{-} + H_{3} O^{+} + H P O_{4}^{2 -}, p^{k_{2}} = 7.20

p^{H}

N a H_{2} P O_{4}

H_{3} P O_{4}, H_{3} P O_{4} ⇋ H_{2} P O_{4}^{-} + H^{+} (K_{1})

H_{2} P O_{4}^{-} ⇋ H P O_{4}^{2 -} + H^{+} (K_{2})

H P O_{4}^{2 -} ⇋ P O_{4}^{3 -} + H^{+} (K_{3})

Identify Bronsted acids in the reaction : $H_{2} C O_{3} + H P O_{4}^{2 -} ⇌ H_{2} P O_{4}^{-} + H C O_{3}^{-}$

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