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Lewis Theory

Conjugate bas...

Question

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

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Solution

The correct option is A ${P O}_{4}^{3 -}$
According to Acid-base theory, Acids are electrolytes which donate $H^{+}$ ion and bases are which accept $H^{+}$ ions.
The conjugate base of $H P O_{4}^{2 -}$ is $P O_{4}^{3 -}$ as $H P O_{4}^{2 -}$ donates its $H^{+}$ ion to give its conjugate base.

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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^{+}]

[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^{+} + 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_{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}

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