The equivalent weight of H_3PO_4 in the react | vedclass

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(A) The molecular weight $(Mw)$ of $H_3PO_4$ is calculated as: $(3 	imes 1) + 31 + (4 	imes 16) = 3 + 31 + 64 = 98 \ g/mol$.In the reaction $Ca(OH)_

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$49$

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(A) The molecular weight $(Mw)$ of $H_3PO_4$ is calculated as: $(3 	imes 1) + 31 + (4 	imes 16) = 3 + 31 + 64 = 98 \ g/mol$.In the reaction $Ca(OH)_2 + H_3PO_4 	o CaHPO_4 + 2H_2O$,two hydrogen atoms of $H_3PO_4$ are replaced by one $Ca^{2+}$ ion.Therefore,the $n$-factor for $H_3PO_4$ is $2$.The equivalent weight $(Ew)$ is given by the formula: $Ew = \frac{Mw}{n-factor} = \frac{98}{2} = 49$.

The equivalent weight of $H_3PO_4$ in the reaction is $Ca(OH)_2 + H_3PO_4 \to CaHPO_4 + 2H_2O$
$(Ca = 40, P = 31, O = 16)$

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The equivalent weight of $H_3PO_4$ in the reaction is $Ca(OH)_2 + H_3PO_4 \to CaHPO_4 + 2H_2O$$(Ca = 40, P = 31, O = 16)$