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Question: What is the equivalent mass \({{I}}{{{O}}_{{4}}}^{{ - }}\) when it is converted into \({{{I}}_{{2}}}...

What is the equivalent mass IO4{{I}}{{{O}}_{{4}}}^{{ - }} when it is converted into I2{{{I}}_{{2}}}in acidic medium?
A. M6\dfrac{{{M}}}{{{6}}}
B. M7\dfrac{{{M}}}{{{7}}}
C. M5\dfrac{{{M}}}{{{5}}}
D. M4\dfrac{{{M}}}{{{4}}}

Explanation

Solution

Equivalent weight for a given mole of a substance is generally defined as molar mass divided by n-factor. If there are x equivalents, its weight would be x×Mnfactor\dfrac{{{{x \times M}}}}{{{{n - factor}}}}. N-factor is the capability of accepting or releasing electrons, which is reflected as change in oxidation number.

Complete step by step answer:
Consider the conversion of IO4{{I}}{{{O}}_{{4}}}^{{ - }}into I2{{{I}}_{{2}}}in acidic medium
2IO4+16H++14eI2+8H2O{{2I}}{{{O}}_{{4}}}^{{ - }}{{ + 16}}{{{H}}^{{ + }}}{{ + 14}}{{{e}}^{{ - }}}\xrightarrow{{}}{{{I}}_{{2}}}{{ + 8}}{{{H}}_{{2}}}{{O}}
Oxidation number of I=X×4(2)=1{{I = X \times 4}}\left( {{{ - 2}}} \right){{ = - 1}}
I=+7{{I = + 7}}
Here in LHS the oxidation state of I{{I}} is +7{{ + 7}} and in RHS the oxidation state of I2{{{I}}_{{2}}}is 0. This clearly indicates that Iodine is been reduced to 0 oxidation state in the reaction
Equivalent weight is given as
As 14 electrons are involved in the reaction and 2 moles of IO4{{I}}{{{O}}_{{4}}}^{{ - }} is been used thus
Equivalent  Weight=2×M14{{Equivalent\; Weight = }}\dfrac{{{{2 \times M}}}}{{{{14}}}}
Equivalent  Weight=M7{{Equivalent \;Weight = }}\dfrac{{{M}}}{{{7}}}

So, the correct answer is Option B.

Additional information:
Equivalent weight has got units of mass and dimensions where in case of atomic weight it is dimensionless. Equivalent weights were before calculated experimentally but now it is calculated by dividing molecular mass by the number of negative or positive electrical charges that is resulted by the dissolution of the compound. The number of gram equivalents of the solute present in 1litre of solution is nothing but normality. It means solution which contains one gram equivalent of the solute in one litre of the solution. It is given as 1N. Normality could be used to express the concentration of OH{{O}}{{{H}}^{{ - }}} and H3O+{{{H}}_{{3}}}{{{O}}^{{ + }}} in acid base titrations.

Note: In case of acid base titrations the equivalent weight of acid or base is the mass that reacts or supplies one mole of H+{{{H}}^{{ + }}} ions and in case of redox reactions the equivalent weight of each reactant reacts or supplies one mole of electrons. In precipitation reactions the number of ions which will precipitate in the given reaction is given by the equivalence factor.