The equivalent weights of H_3PO_4 in the foll | vedclass

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(A) The equivalent weight of an acid is calculated as: $	ext{Equivalent Weight} = \frac{	ext{Molar Mass}}{	ext{Basicity (n-factor)}}$. The molar ma

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$98, 49, 32.67$

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(A) The equivalent weight of an acid is calculated as: $	ext{Equivalent Weight} = \frac{	ext{Molar Mass}}{	ext{Basicity (n-factor)}}$. The molar mass of $H_3PO_4$ is $98 \ g/mol$. $1$. In the first reaction,$H_3PO_4 + OH^{-} ightarrow H_2PO_4^- + H_2O$,the $n$-factor is $1$. So,$	ext{Eq. Wt.} = \frac{98}{1} = 98$. $2$. In the second reaction,$H_3PO_4 + 2OH^{-} ightarrow HPO_4^{2-} + 2H_2O$,the $n$-factor is $2$. So,$	ext{Eq. Wt.} = \frac{98}{2} = 49$. $3$. In the third reaction,$H_3PO_4 + 3OH^{-} ightarrow PO_4^{3-} + 3H_2O$,the $n$-factor is $3$. So,$	ext{Eq. Wt.} = \frac{98}{3} = 32.67$. Thus,the values are $98, 49, 32.67$.

The equivalent weights of $H_3PO_4$ in the following reactions are respectively:
$H_3PO_4 + OH^{-} \rightarrow H_2PO_4^- + H_2O$
$H_3PO_4 + 2OH^{-} \rightarrow HPO_4^{2-} + 2H_2O$
$H_3PO_4 + 3OH^{-} \rightarrow PO_4^{3-} + 3H_2O$

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The equivalent weights of $H_3PO_4$ in the following reactions are respectively: $H_3PO_4 + OH^{-} \rightarrow H_2PO_4^- + H_2O$ $H_3PO_4 + 2OH^{-} \rightarrow HPO_4^{2-} + 2H_2O$ $H_3PO_4 + 3OH^{-} \rightarrow PO_4^{3-} + 3H_2O$