What is the normality of $0.3 M H_{3} P O_{4}$ when it undergoes the reaction as?
$H_{3} P O_{4} + 2 O H^{-} \to H P O_{3}^{2 -} + 2 H_{2} O$ .

0.3 N

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0.15 N

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0.60 N

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0.90 N

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Solution

The correct option is C $0.60 N$

$H_{3} P O 4 + 2 O H^{-} \to H P O_{3}^{- 2} + 2 H_{2} O$

Normality $= n f a c t o r \times m o l a r i t y$

N factor=Number of $H^{+}$ ions replaced.Here n factor is 2 ,since 2 $H^{+}$ ions are replaced.

Normality $= 2 \times 0.3 = 0.6$ N
Hence option C is coorrect.

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Similar questions

H_{3} P O_{4}

H_{3} P O_{4} + O H^{-} \to H_{2} P O_{4}^{-} + H_{2} O

H_{3} P O_{4} + 2 O H^{-} \to H P O_{4}^{2 -} + 2 H_{2} O

H_{3} P O_{4} + 3 O H^{-} \to P O_{4}^{3 -} + 3 H_{2} O

H_{2} O_{2}

H_{2} O_{2} + 2 H^{+} + 2 e^{-} \to 2 H_{2} O

H_{2} O_{2} - 2 e^{-} \to O_{2} + 2 H^{+}

H_{2} O_{2} + 2 e^{-} \to 2 O H^{-}

H_{2} O_{2} + 2 O H^{-} 2 e^{-} \to O_{2} + 2 H_{2} O

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

{H_{3} P O}_{4} + C a {(O H)}_{2} ⟶ {C a H P O}_{4} + 2 H_{2} O

C a (O H)_{2} + H_{3} P O_{4} \to C a H P O_{4} + 2 H_{2} O

H_{3} P O_{4}

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