The equivalent weight of H_3PO_4 in the react | vedclass

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(B) The reaction is $Ca(OH)_2 + H_3PO_4 \rightarrow CaHPO_4 + 2H_2O$. In this reaction,$H_3PO_4$ loses $2$ $H^+$ ions to form $HPO_4^{2-}$. Thus,the n

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$49$

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(B) The reaction is $Ca(OH)_2 + H_3PO_4 ightarrow CaHPO_4 + 2H_2O$. In this reaction,$H_3PO_4$ loses $2$ $H^+$ ions to form $HPO_4^{2-}$. Thus,the n-factor (valency factor) of $H_3PO_4$ is $2$. The molecular weight of $H_3PO_4 = (3 	imes 1) + 31 + (4 	imes 16) = 98 \ g/mol$. Equivalent weight $= \frac{	ext{Molecular Weight}}{	ext{n-factor}} = \frac{98}{2} = 49$.

The equivalent weight of $H_3PO_4$ in the reaction $Ca(OH)_2 + H_3PO_4 \rightarrow CaHPO_4 + 2H_2O$ is

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