H_2SeO_4 + NaOH to NaHSeO_4; E.W. of H_2SeO_4 | vedclass

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(B) The equivalent weight $(E.W.)$ of an acid is given by the formula: $E.W. = \frac{	ext{Molecular Weight (M)}}{	ext{Basicity (n-factor)}}$.In the

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$E_1 > E_2$

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(B) The equivalent weight $(E.W.)$ of an acid is given by the formula: $E.W. = \frac{	ext{Molecular Weight (M)}}{	ext{Basicity (n-factor)}}$.In the first reaction,$H_2SeO_4$ reacts with $1$ mole of $NaOH$ to form $NaHSeO_4$,so the $n$-factor is $1$. Thus,$E_1 = \frac{M}{1} = M$.In the second reaction,$H_2SeO_4$ reacts with $2$ moles of $NaOH$ to form $Na_2SeO_4$,so the $n$-factor is $2$. Thus,$E_2 = \frac{M}{2} = 0.5M$.Comparing the two,$E_1 = M$ and $E_2 = 0.5M$,therefore $E_1 > E_2$.

$H_2SeO_4 + NaOH \to NaHSeO_4$; $E.W.$ of $H_2SeO_4 = E_1$
$H_2SeO_4 + 2NaOH \to Na_2SeO_4$; $E.W.$ of $H_2SeO_4 = E_2$
What is the relation between $E_1$ and $E_2$?

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$H_2SeO_4 + NaOH \to NaHSeO_4$; $E.W.$ of $H_2SeO_4 = E_1$$H_2SeO_4 + 2NaOH \to Na_2SeO_4$; $E.W.$ of $H_2SeO_4 = E_2$What is the relation between $E_1$ and $E_2$?