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Factors Affecting Alpha

In a 0.1 M ...

Question

In a $0.1 M$ solution of $H_{3} P O_{4}$ the ionic species present are (besides $H_{3} O^{+}$ and $O H^{-}$ ):

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

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{H_{2} P O_{2}}^{3 -}, {H_{2} P O_{4}}^{-}, {P O_{4}}^{3 -}

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

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

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Solution

The correct option is A ${H_{2} P O_{4}}^{-}, {H P O_{4}}^{2 -}, {P O_{3}}^{3 -}$
$H_{3} P O_{4}$ $=$ $H^{+}$ $+$ $H_{2} P O_{4}^{-}$
$H_{2} P O_{4}^{-}$ $=$ $H^{+}$ $+$ $H P O_{4}^{- 2}$
$H P O_{4}^{- 2}$ $=$ $H^{+}$ $+$ $P O_{4}^{- 3}$
Thus $H_{3} P O_{4}$ will dissociate into the ions $H_{2} P O_{4}^{-}$ , $H P O_{4}^{- 2}$ , $P O_{4}^{- 3}$ .
Option A is the correct answer.

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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} 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

0.01 M N a H_{2} P O_{4}

H_{3} P O_{4}

H_{3} P O_{4} + H_{2} O ⇋ H_{3} O^{+} + H_{2} P O_{4}^{-}

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

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

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

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