The conjugate base of $H_{2} P O_{4^{-}}$ is:

P O_{4}^{3 -}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P_{2} O_{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H_{3} P O_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H P O_{4}^{2 -}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C $H P O_{4}^{2 -}$
The concept of Conjugate Acid-Base pair was given by Lowry-Bronsted Theory.
According to the theory:
Acids are Proton $(H^{+})$ donors
Bases are Proton $(H^{+})$ acceptors
Hence, in order to get the conjugate base of a species, we should consider that substance as an acid and vice versa.
For example-
$H_{2} S O_{4} \to H S O_{4}^{-} + H^{+}$
Therefore,
Conjugate base of $H_{2} S O_{4}$ is $H S O_{4}^{-}$ .
Using this logic, for the given example:
$H_{2} P O_{4}^{-} \to H P O_{4}^{- 2} + H^{+}$
Hence,
Conjugate base of $H_{2} P O_{4}^{-}$ is $H P O_{4}^{- 2}$ .

Suggest Corrections

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

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}_{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 -}

H_{3} P O_{4}

H P O_{4}^{2 -}

Join BYJU'S Learning Program