Question

Chile salt petre is a natural source of $NaN{O}_3$ which also contains $NaI{O}_3$. The $NaI{O}_3$ can be used as a source of iodine produced in the following reactions:

$I{O}_3^{-}+3HS{O}_3^{-}⟶I^{-}+3H^{+}3S{O}_4^{2-}$ ... (i)

$5I^{-}+I{O}_3^{-}+6H^{+}⟶3I_2\left(s\right)+3H_{2}O$ ... (ii)

$1.00$ L of the starting solution which contains $5.80$ g $NaI{O}_3$ per litre is treated with stoichiometric quantity of $NaHS{O}_3$. Further, a quantity of the starting solution is added to the reaction mixture to bring about the second reaction. $[I=127,Na=23,S=32]$

How many equivalents of $I{O}_3^{-}$ are used in step (ii)?

• A. $6$
• B. $5$
• C. $0.06$
• D. $0.035$

Answer 1

The correct option is D $0.035$

Given, $5.8$ g $NaI{O}_3=\frac{5.8}{198}$ mol $NaI{O}_3$

$\therefore NaHS{O}_3=3\times \frac{5.8}{198}$ mol $=3\times \frac{5.8}{198}\times 104$ g $NaHS{O}_3$ $=9.14$ g

$I^{-}$ formed in step $I=\frac{5.8}{198}$ mol

$I{O}_3^{-}$ required in step $II=\frac{5.9}{198\times 5}$ mol $=\frac{5.8\times 198}{198\times 5}$ g $NaI{O}_3$ $=1.16$ g $NaI{O}_3$

$\therefore$ Volume $=\frac{1.16}{5.8}=0.2$ L

$\underset{+5}{I{O}_3^{-}}⟶\underset{-1}{I^{-}}$
Change in oxidation number $=6$ units
Thus, one mole $NaI{O}_3$ is $6$ equivalents.
$1.16$ g $NaI{O}_3=\frac{1.16}{198}$ mol

$=\frac{6.96}{198}$ equivalents (since $I{O}_3^{-}=6I$)

$=0.035$ equivalents

Hence, the correct option is $\left(D\right)$.