Ionization of phosphoric acid gives:

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

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H P O_{4}^{- 2}

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P O_{4}^{- 3}

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all of these

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Solution

The correct option is D all of these
Ionisation of phosphoric acid is as shown below:
$1) H_{2} O + H_{3} {P O}_{4} \to H_{3} O^{+} + H_{2} {P O}_{4}^{-} 2) H_{2} O + H_{2} {P O}_{4}^{-} \to H_{3} O^{+} + H {P O}_{4}^{2 -} 3) H_{2} O + H {P O}_{4}^{2 -} \to H_{3} O^{+} + P O_{4}^{3 -}$
Hence, option D is the correct answer.

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Similar questions

H_{3} {P O}_{4} ⇌ H_{3} {P O}_{4}^{-} + H^{+}

p K_{a_{1}} = 2.21

H_{3} {P O}_{4}

(H_{3} {P O}_{4})

H_{3} {P O}_{4} ⇌ H^{+} + H_{2} {P O}_{4}^{-}; K_{a_{1}} = 10^{- 3}

H_{2} {P O}_{4}^{-} ⇌ H^{+} + H {P O}_{4}^{- 2}; K_{a_{2}} = 10^{- 7}

H P O_{4}^{- 2} ⇌ H^{+} + {P O}_{4}^{- 3}; K_{a_{3}} = 10^{- 13}

0.1 M

H_{3} {P O}_{4}

{P O}_{4}^{3 -}

N a H_{2} P O_{4}

N a_{2} H P O_{4}

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

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