The conjugate acid of $H P O_{4}^{2 -}$ is:

H_{3} P O_{3}

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H_{3} P O_{4}

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

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P O_{4}^{3 -}

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Solution

The correct option is C $H_{2} P O_{4}^{-}$
The conjugate acid of $H P O_{4}^{2 -}$ is $H_{2} P O_{4}^{-}$
$H_{2} P O_{4}^{-} \to H^{+} + H P O_{4}^{2 -}$
The base and its conjugate acid differ by one proton only.

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Similar questions

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})

H_{3} P O_{4} ⇌ H^{\oplus} + H_{2} P O_{4}^{⊝}; K_{a_{1}} :

H_{3} P O_{4}^{⊝} ⇌ H^{\oplus} + H P O_{4}^{2 -}; K_{a_{2}} :

H P O_{4}^{2 -} ⇌ H^{\oplus} + P O_{4}^{3 -}; K_{a_{3}} :

H_{3} P O_{4}

H P O_{4}^{2 -}

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

H_{3} P O_{4}_{1} H^{+} + H_{2} P O_{4}^{-}; K_{1}

H_{2} P O_{4}^{-}_{2} H^{+} + H P O_{4}^{2 -}; K_{2}

H P O_{4}^{2 -}_{3} H^{+} + P O_{4}^{3 -}; K_{3}

H_{3} P O_{4} ⇌ 3 H^{+} + P O_{4}^{3 -}

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