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Oxyacids of Phosphorus

The equivalen...

Question

The equivalent weight of $H_{2} P O_{4}$ in the given reaction is:
$N a O H + H_{2} P O_{4} \to N a H_{2} P O_{4} + H_{2} O$ .

49

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98

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32.67

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24.5

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Solution

The correct option is B $98$
According to the problem:
For given reaction,
$N a O H + H_{3} P O_{4} \to N a H_{2} P O_{4} + H_{2} O$
Hence,
Equivalent mass of the acid= $\frac{M o l e c u l a r m a s s}{N u m b e r o f r e p l a c a b l e H^{+} (b a s i c i t y)}$
Therefore
Equivalent mass of $H_{3} P O_{4} = M o l e c u l a r m a s s o f H_{3} P O_{4}$
$∴$ Equivalent mass of $H_{3} P O_{4} = \frac{3 \times 1 + 1 \times 31 + 4 \times 16}{1} = 98$

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Similar questions

N a H_{2} P O_{4}

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

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_{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_{2} P O_{4}^{-}

(H_{3} P O_{4})

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

H_{3} P O_{4}

H_{3} P O_{4} + H_{2} O ⇋ H_{3} O^{+} + H_{2} P O_{4}^{-}

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

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

H_{2} P O_{4}^{-}

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