Question

What mass of \[HN{O_3}\] is needed to convert \[5g\] of iodine into iodic acid according to the reaction:
\[{I_2} + HN{O_3} \to HI{O_3} + N{O_2} + {H_2}O\]
A. \[12.4g\]
B. \[24.8g\]
C. \[0.248g\]
D. \[49.6g\]

Answer 1

Hint: The reaction taking place between iodine and nitric acid is not a simple reaction but a redox titration in which the oxidation states of iodine and nitrogen undergo a change as the reaction proceeds. Therefore the \[n - factor\] must be considered during the calculations of mass.

Complete answer:
The reaction given in the equation can be written as follows:
\[{I_2} + HN{O_3} \to HI{O_3} + N{O_2} + {H_2}O\]
The above reaction is a redox reaction in which iodine is the chemical species getting oxidized and nitrogen is getting reduced. The oxidation and the reduction half reactions can be written as follows:
Oxidation half
\[{I_2}(0) \to 2I( + 5) + 10{e^ - }\]
(The numbers inside round brackets indicate the oxidation states of elements)
Reduction half
\[HN{O_3}( + 5) + {e^ - } \to N{O_2}( + 4)\]
(The oxidation states are associated with the nitrogen atom)
The equivalent mass \[{M_{eq}}\] of both the iodine and nitric acid must remain equal during the redox reaction. The formula for calculating equivalent mass is given as follows:
\[{M_{eq}} = n - factor \times ({\text{number of moles)}}\]
And the number of moles can be calculated using the formula:
\[{\text{number of moles}} = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}\]
Therefore the number of moles of iodine is:
\[{\text{number of moles of iodine}} = \dfrac{{5g}}{{254gmo{l^{ - 1}}}}\]
\[{\text{number of moles of nitric acid}} = \dfrac{x}{{63gmo{l^{ - 1}}}}\] assuming that the mass of nitric acid is \[x\]
The \[n - factor\] for iodine involved in the oxidation half is \[10\].
The \[n - factor\] for reduction involved in the reduction half is \[1\].
Since the equivalents of iodine and nitric acid are equal, it can be expressed as follows:
\[{M_{eq}}({I_2}) = {M_{eq}}(HN{O_3})\]
\[10 \times \dfrac{{5g}}{{254gmo{l^{ - 1}}}} = 1 \times \dfrac{x}{{63gmo{l^{ - 1}}}}\]
Solving this equation for getting the value of \[x\] :
\[x = 12.4g\]

Hence the correct option is (A)

Note:
The \[n - factor\] is the measure of the number of electrons that are involved in a particular reaction per molecule. In an oxidation reaction the \[n - factor\] is the number of electrons lost and in a reduction reaction, it is the number of electrons gained.