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# Find the equivalent weight of $H_3PO_4$ in the reaction. $Ca(OH)_2 +H_3 PO_4 \to CaHPO_4+2H_2O$

$(a)\;49 \\ (b)\;49.5 \\(c)\;48 \\(d)\;49.2$

$Ca(OH)_2+H_3PO_4 \to CaHPO_4 +2H_2O$
two hydrogen atoms of $H_3PO_4$ are replaced, therefore the equivalent weight of $H_3PO_4$ is half of its molecular weight.
Molecular weight of $H_3PO_4 =(1 \times 3)+31+(4 \times 16)=98$
Equivalent weight $=\large\frac{98}{2}$$=49$