Question

What will be the valence factor or n-factor $\left(n_f\right)$ for $H_{2}S{O}_4$ when
(i) It reacts with $NaOH$ to form $NaHS{O}_4$
(ii) It reacts with $NaOH$ to form $N{a}_{2}S{O}_4$

• A. $n_f$ of $H_{2}S{O}_4$ is 2 for both the cases
• B. (i) $n_f$ is $1$ and (ii) $n_f$ is $2$
• C. $n_f$ of $H_{2}S{O}_4$ is 1 for both the cases
• D. None of the above

Answer 1

The correct option is B (i) $n_f$ is $1$ and (ii) $n_f$ is $2$

For an acid $n_f$ is the replaceable $H^{+}$ ions in an reaction:

$H_{2}S{O}_4\left(aq\right)+NaOH\left(aq\right)⟶NaHS{O}_4\left(aq\right)+H_{2}O\left(l\right)$
Since
For the reaction above one $H^{+}$ gets replaced to form $NaHS{O}_4$
(i) $n_f$ of $H_{2}S{O}_4$ is $1$
$H_{2}S{O}_4\left(aq\right)+NaOH\left(aq\right)⟶N{a}_{2}S{O}_4\left(aq\right)+2H_{2}O\left(l\right)$
For the reaction above two $H^{+}$ gets replaced to form $N{a}_{2}S{O}_4$
(ii) $n_f$ of $H_{2}S{O}_4$ is $2$