The equivalent weight of phosphoric acid (H_3 | vedclass

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(D) The molecular weight of $H_3PO_4$ is $3 	imes 1 + 1 	imes 31 + 4 	imes 16 = 98 \ g/mol$. In the given reaction,$NaOH + H_3PO_4 	o NaH_2PO_4 +

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$98$

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(D) The molecular weight of $H_3PO_4$ is $3 	imes 1 + 1 	imes 31 + 4 	imes 16 = 98 \ g/mol$. In the given reaction,$NaOH + H_3PO_4 	o NaH_2PO_4 + H_2O$,one molecule of $H_3PO_4$ reacts with one molecule of $NaOH$ to replace one $H^+$ ion. Therefore,the n-factor (basicity) of $H_3PO_4$ in this reaction is $1$. The equivalent weight is calculated as: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{	ext{n-factor}} = \frac{98}{1} = 98$.

The equivalent weight of phosphoric acid $(H_3PO_4)$ in the reaction $NaOH + H_3PO_4 \to NaH_2PO_4 + H_2O$ is

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