Question

Chile saltpetre, a source of $NaN{O_3}$ , also contains $NaI{O_3}$ . The $NaI{O_3}$ is a source of ${I_2}$ produced as shown in the following equations:
Step (I) : $IO_3^ - + 3HSO_3^ - + {I^ - } + 3{H^ - } + 3SO_4^{2 - }$
Step (II) : $5{I^ - } + IO_3^ - + 6{H^ - } \to 3{I_2}\left( S \right) + 3{H_2}O$
One litre sample of chile saltpetre solution containing $6.6{\text{ }}g$ $NaI{O_3}$ is treated with $NaHS{O_3}$ . Now an additional amount of the same solution is added to the reaction mixture to bring about the second titration. Calculate the weight of $NaHS{O_3}$ required in step (I) and what additional volume of chile saltpetre must be added in step (II) to bring out complete conversion of ${I^ - }to{\text{ }}{I_2}$ .
A. $100mL$
B. $400mL$
C. $200mL$
D. None of these

Answer 1

Hint : The chemical compound with the chemical formula $NaN{O_3}$ in Chile is saltpetre, otherwise known as sodium nitrate. This alkali metal nitrate salt is commonly called Chile saltpetre, to distinguish it from normal saltpetre, potassium nitrate (because the large deposits were traditionally mined in Chile).

Complete Step By Step Answer:
Sodium nitrate is a very water-soluble white, delicate solid. It is a readily available source of the nitrate anion $\left( {NO_3^ - } \right)$ , useful to manufacture fertiliser, pyrotechnical and bombs, glass and pottery enamels, food preservatives (special meats) and solid rocket propellants, for several reactions on industrial scales. For these purposes it was extensively mined.
Molecular weight of $NaI{O_3}$ = $198g/mol$ $$
And of $NaHS{O_3}$ = $104g/mol$
$6{e^ - } + IO_3^ - \to {I^ - }\left( {n = 6} \right) \\\ 10{e^ - } + 2IO_3^ - \to {I_2}\left( {n = \dfrac{{10}}{2} = 5} \right) \\\ HSO_3^ - \to SO_4^{2 - } + 2{e^ - }\left( {n = \dfrac{2}{2} = 1} \right) \\\$
Milliequivalents of $NaHS{O_3}$ = Milliequivalents of $NaI{O_3}$
$= N \times V = \dfrac{{6.6 \times {{10}^3}}}{{\dfrac{{198}}{6}}} = 200$
Milliequivalent of $NaHS{O_3}$ = $200$
$\dfrac{{W \times {{10}^3}}}{{\dfrac{{104}}{2}}} = 200 \Rightarrow {W_{NaHS{O_3}}} = 10.4g$
Milliequivalents of ${I^{{ - ^{}}}}$ ​ formed in step (I) using $n - factor$ of $6 = 200$
$6{e^ - } + IO_3^ - \to {I^ - }$
In step (II), the valence factor or n−factor of ${I^ - }$ ​is $1$ .
$\left( {2{I^ - } \to {I_2} + 2{e^ - }} \right)$ and $n - factor$ of $IO_3^ -$ is $5$
$IO{e^ - } + 2IO_3^ - \to {I_2}\left( {n = \dfrac{{10}}{2} = 5} \right)$
Thus, milliequivalents of ${I^ - }$ formed using $n - factor$ of $1 = \dfrac{{200}}{6}$
$mEq$ of $NaI{O_3}$ used in step (II) $\dfrac{{200}}{6}$
$N \times V = \dfrac{{200}}{6}$
$\dfrac{{6.6}}{{\dfrac{{198}}{5}}} \times {V_{mL}} = \dfrac{{200}}{6}$
${V_{NaI{O_3}}} = \dfrac{{200 \times 198}}{{6 \times 5 \times 6.6}} = 200mL$ $$
Hence, the correct option is: (C) $200mL$ .

Note :
Sodium nitrate is used for heat storage and more recent heat transfer in solar power plants along with potassium nitrate and calcium nitrate. As energy storage material in prototype facilities, like the Andasol Solar Power Station and the Archimedes project, a mix of sodium nitrate, calcium nitrate and potassium nitrate is used. It is also used for optional microorganism breathing in the wastewater industry. Nitrosomonas, a microorganism, prefers oxygen to consume nitrates so that they grow faster in the wastewater.