In the reaction Ca(OH)_2 + H_3PO_4 to CaHPO_4 | vedclass

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Question

(D) The equivalent weight of an acid is calculated as $\frac{	ext{Molecular weight}}{	ext{Basicity}}$.In the given reaction,$H_3PO_4$ reacts with $C

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(D) The equivalent weight of an acid is calculated as $\frac{	ext{Molecular weight}}{	ext{Basicity}}$.In the given reaction,$H_3PO_4$ reacts with $Ca(OH)_2$ to form $CaHPO_4$. Here,only $2$ hydrogen atoms of $H_3PO_4$ are replaced by $Ca^{2+}$ ions.Therefore,the n-factor (basicity) of $H_3PO_4$ in this reaction is $2$.Molecular weight of $H_3PO_4 = (3 	imes 1) + 31 + (4 	imes 16) = 3 + 31 + 64 = 98$.Equivalent weight $= \frac{98}{2} = 49$.

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

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