The correct order for the increasing extent of hydrolysis is:

P O_{4}^{3 -} < H P O_{4}^{2 -} < H_{2} P O_{4}^{-}

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H_{2} P O_{4}^{-} < H P O_{4}^{2 -} < P O_{4}^{3 -}

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H P O_{4}^{2 -} < P O_{4}^{3 -} < H_{2} P O_{4}^{-}

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P O_{4}^{3 -} < H_{3} P O_{4}^{-} < H P O_{4}^{2 -}

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Solution

The correct option is B $H_{2} P O_{4}^{-} < H P O_{4}^{2 -} < P O_{4}^{3 -}$
The extent of hydrolysis depends upon the charge on any anions present, more charge means more hydrolysis.

So, order of hydrolysis is $H_{2} P O_{4}^{-} < H P O_{4}^{2 -} < P O_{4}^{3 -}$

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

H_{3} P O_{4}

H_{3} P O_{4} + O H^{-} \to H_{2} P O_{4}^{-} + H_{2} O

H_{3} P O_{4} + 2 O H^{-} \to H P O_{4}^{2 -} + 2 H_{2} O

H_{3} P O_{4} + 3 O H^{-} \to P O_{4}^{3 -} + 3 H_{2} O

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