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Bronsted Lowry Theory

Which of the ...

Question

Which of the following is not a conjugate acid-base pair?

H {P O}_{3}^{2 -}, {P O}_{3}^{3 -}

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

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H_{2} {P O}_{4}^{-}, H_{3} {P O}_{4}

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H_{2} {P O}_{4}^{-}, {P O}_{3}^{3 -}

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Solution

The correct option is D $H_{2} {P O}_{4}^{-}, {P O}_{3}^{3 -}$
$H_{2} {P O}_{4}^{-}, {P O}_{3}^{3 -}$ is not a conjugate acid-base pair as it differs by two protons.
A conjugate acid-base pair differs by one proton only. For example, $H C l$ and $C l^{-}$ ion differ by one proton only. Hence, they are the conjugate acid-base pair.
Thus,
(A) $H {P O}_{3}^{2 -}, {P O}_{3}^{3 -}$ , (B) $H_{2} {P O}_{4}^{-} {, H} {P O}_{4}^{2 -}$ and (C) $H_{2} {P O}_{4}^{-}, H_{3} {P O}_{4}$ are conjugate acid base pairs as they differ by one proton only.

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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^{+}]

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_{2} {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 -}

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