The equivalent weights of H_3PO_4 in the foll | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$98, 49, 32.67$

Vedclass · Answer

(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:
$H_3PO_4 + OH^{-} \rightarrow H_2PO_4^- + H_2O$
$H_3PO_4 + 2OH^{-} \rightarrow HPO_4^{2-} + 2H_2O$
$H_3PO_4 + 3OH^{-} \rightarrow PO_4^{3-} + 3H_2O$

Similar Questions

In a metal oxide,there is $20\%$ oxygen by weight. Its equivalent weight is

The equivalent mass of an element is $4$ and the vapor density of its chloride is $59.25$. Find the valency of the element.

In the reaction $2Na_2S_2O_3 + I_2 \to Na_2S_4O_6 + 2NaI$,the equivalent weight of $Na_2S_2O_3$ (molecular weight = $M$) is equal to:

The equivalent mass of a metal is $29.73$ and the vapour density of its chloride is $130.4$. Find out the atomic mass of the metal.

If the equivalent mass of a metal with a valency of $+2$ is $12$,then the molar mass of its oxide is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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