In which of the following reactions, $1 g$ equivalent of $H_{3} {P O}_{4}$ reacts with $3 g$ equivalents of $N a O H$ ?

H_{3} {P O}_{4} + N a O H \to N a H_{2} {P O}_{4}

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H_{3} {P O}_{4} + 2 N a O H \to {N a}_{2} H {P O}_{4}

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H_{3} {P O}_{4} + 3 N a O H \to {N a}_{3} {P O}_{4}

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Solution

The correct option is D $H_{3} {P O}_{4} + 3 N a O H \to {N a}_{3} {P O}_{4}$
The reaction is a double-displacement reaction. In this reaction, two elements displace each other. Not only that, but it is also a neutralization reaction. A neutralization reaction involves the combination of an acid and base to form salt and water.
The reaction is :
$H_{3} P O_{4} + 3 N a O H ⟶ N a_{3} P O_{4} + 3 H_{2} O$

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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}

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

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

N a_{3} {P O}_{4}

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

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

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

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