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Conjugate bas...

Question

Conjugate base of $H_{3} {P O}_{4}$ ?

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

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

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H_{4} {P O}_{4}^{+}

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

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Solution

The correct option is A $H_{2} {P O}_{4}^{-}$
${H_{2} P O_{4}}^{-}$ is conjugate base of $H_{3} P O_{4}$ .

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Similar questions

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_{2} {P O}_{4}^{2 -}; p K_{2} = 7.20

0.01 M

N a H_{2} {P O}_{4}

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^{+}

p K_{a_{1}} = 2.21

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

[H^{+}]

H_{3} P O_{4}

H_{3} {P O}_{4} ⇌ H^{+} + H_{2} {P O}_{4}^{-} {K a}_{1} = 10^{- 4}

H_{2} {P O}_{4}^{-} ⇌ H^{+} + H {P O}_{4}^{- 2} {K a}_{2} = 10^{- 9}

H {P O}_{4}^{- 2} ⇌ H^{+} + {P O}_{4}^{3 -} {K a}_{3} = 10^{- 13}

(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 -}

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