$P O_{4}^{- 3}$ is a _____ valent anion

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Solution

It is will trivalent anion. as it has - 3 charge so we can say that ,its a trivalent and its anion as it has negative charge .

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[F e (C N)_{6}]^{- 4}, P O_{4}^{- 3}, S O_{4}^{- 2}, C l^{-}

(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}

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

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

Is a 100% co-valent compound possible?

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

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

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p H = p K_{2} + {log}_{10} \frac{[H {P O}_{4}^{2 -}]}{[H_{2} {P O}_{4}^{-}]}

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

P O_{4}^{- 3}

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