Question

Chile salt petre, 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: $I{O}_3^{⊝}+3HS{O}_3^{⊝}\to I^{⊝}+3H^{⊝}+3S{O}_4^{2-}$
Step II: $5I^{⊝}+I{O}_3^{⊝}+6H^{⊝}\to 3I_2\left(s\right)+3H_{2}O$
One litre sample of chile salt petre solution containing $6.6$ g $NaI{O}_3$ is treated with $NaHS{O}_3$. Now an additional amount of 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 salt petre must be added in step II to bring out complete conversion of $I^{⊝}$ to $I_2$.

• A. $100$ ml
• B. $400$ ml
• C. $200$ ml
• D. None of these

Answer 1

The correct option is C $200$ ml

Molecular weight of $NaI{O}_3=198$ g/mol and $NaHS{O}_3=104$ g/mol.

$6e^{-}+I{O}_3^{⊝}\to I^{⊝}(n=6)$
$10e^{-}+2I{O}_3^{⊝}\to I_2(n=\frac{10}{2}=5)$
$HS{O}_3^{⊝}\to S{O}_4^{2-}+2e^{-}(n=2)$
$2I^{⊝}\to I_2+2e^{-}(n=\frac{2}{2}=1)$

Milliequivalents of $NaHS{O}_3=$ Milliequivalents of $NaI{O}_3$

$=N\times V=\frac{6.6\times {10}^3}{\frac{198}{6}}=200$

Milliequivalents of $NaHS{O}_3=200$
$\frac{W\times {10}^3}{\frac{104}{2}}=200$ $⟹W_{NaHS{O}_3}=10.4g$

Milliequivalents of $I^{⊝}$ formed in step I using $n-$factor of $6=200$

$6e^{-}+I{O}_3^{⊝}\to I^{⊝}$

In step II, the valence factor or $n-$factor of $I^{⊝}$ is $1$.

$(2I^{⊝}\to I_2+2e^{-})$ and $n-$factor of $I{O}_3^{⊝}$ is $5$.

$IO{e}^{-}+2I{O}_3^{⊝}\to I_2(n=\frac{10}{2}=5)$
Thus, milliequivalents of $I^{⊝}$ formed using $n-$factor of $1=\frac{200}{6}$

$mEq$ of $NaI{O}_3$ used in step II $=\frac{200}{6}$
$N\times V=\frac{200}{6}$

$\frac{6.6}{\frac{198}{5}}\times V_{mL}=\frac{200}{6}$

$V_{NaI{O}_3}=\frac{200\times 198}{6\times 5\times 6.6}=200$ mL