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Bronsted Lowry Theory

The conjugate...

Question

The conjugate base of $H {P O}_{4}^{2 -}$ is

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

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H_{2} {P O}_{4}^{-}

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

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

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Solution

The correct option is B ${P O}_{4}^{3 -}$
$H {P O}_{4}^{2 -} + H_{2} O a c i d ⇌ H_{3} \oplus O + {P O}_{4}^{3 -} c o n j u g a t e b a s e$

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Similar questions

H_{2} {P O}_{4}^{-}

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

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

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

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

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

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

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

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

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

[H^{+}]

[H_{2} {P O}_{4}]

[H P O_{4}^{2 -}]

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

0.01 M

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

K_{1} = 7.225 \times 10^{- 3}

K_{2} = 6.8 \times 10^{- 8}

K_{3} = 4.5 \times 10^{- 13}

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

p K_{a_{1}} = 2.21

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

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