The equivalent mass of $H_{3} P O_{4}$ in the reaction given below is :
$H_{3} P O_{4} + N a O H \to N a H_{2} P O_{4} + H_{2} O$

\frac{M}{1}

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\frac{M}{2}

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2 M

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\frac{M}{4}

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Solution

The correct option is B $\frac{M}{1}$
In this Reaction $2 - O H$ of $H_{3} P O_{4}$ is being substituted and changes to $N a H_{2} P O_{4}$

So , Equivalent Weight is $\frac{98}{2}$ = $49$

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Similar questions

H_{3} P O_{4}

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

(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

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_{3} P O_{3}

H_{3} P O_{3} \to H_{3} P O_{4} + P H_{3}

(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

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