The equivalent weight of H_3PO_4 in the follo | vedclass

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(B) The given reaction is: $H_3PO_4 + Ca(OH)_2 	o CaHPO_4 + 2H_2O$ In this reaction,$H_3PO_4$ reacts with $Ca(OH)_2$ to form $CaHPO_4$. The dissociat

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$49$

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(B) The given reaction is: $H_3PO_4 + Ca(OH)_2 	o CaHPO_4 + 2H_2O$ In this reaction,$H_3PO_4$ reacts with $Ca(OH)_2$ to form $CaHPO_4$. The dissociation of $H_3PO_4$ in this process is: $H_3PO_4 	o HPO_4^{2-} + 2H^{+}$ Since $H_3PO_4$ loses $2$ hydrogen ions $(H^{+})$,its valency factor ($n$-factor) is $2$. The molecular weight of $H_3PO_4 = (3 	imes 1) + 31 + (4 	imes 16) = 3 + 31 + 64 = 98 \ g/mol$. Equivalent weight $= \frac{	ext{Molecular Weight}}{n	ext{-factor}} = \frac{98}{2} = 49$.

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

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