Question

The equivalent weight of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ in the following reaction is:
${\text{2N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_3} + {{\text{I}}_{\text{2}}} \to {\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}} + {\text{2NaI}}$
A. ${\text{M}}$
B. ${\text{M/8}}$
C. ${\text{M/0}}{\text{.5}}$
D. ${\text{M/2}}$

Answer 1

Hint: Determine the equivalent weight of -${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ using the steps as follows:
-Determine the oxidation numbers of the species in the reaction.
-Determine the N-factor of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$
-Determine the equivalent weight of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$.

Complete step by step answer:
Step 1: Determine the oxidation numbers of the species in the reaction as follows:
The reaction is,
${\text{2N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_3} + {{\text{I}}_{\text{2}}} \to {\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}} + {\text{2NaI}}$
Determine the oxidation number of sulfur in ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ as follows:
The total charge on ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is zero. The oxidation number of ${\text{Na}}$ is ${\text{ + 1}}$ and the oxidation number of ${\text{O}}$ is $ - 2$. Let the oxidation number of sulfur be $x$.Thus,
${\text{(2 \times Oxidation number of Na)}} + {\text{(2 \times Oxidation number of S)}} + {\text{(3 \times Oxidation number of O)}} = {\text{Total charge on N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3}$
Thus,
\[(2 \times + 1) + (2 \times x) + (3 \times - 2) = 0\]
           \[( + 2) + (2x) + ( - 6) = 0\]
                                       \[2x = + 6 - 2\]
                                       \[2x = + 4\]
                                         \[x = + 2\]
Thus, the oxidation number of sulfur in ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is ${\text{ + 2}}$.
Determine the oxidation number of ${{\text{I}}_{\text{2}}}$ as follows:
The oxidation number of any molecule in its elemental form is zero.
Thus, the oxidation number of ${{\text{I}}_{\text{2}}}$ is zero.
Determine the oxidation number of sulfur in ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}}$ as follows:
The total charge on ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}}$ is zero. The oxidation number of ${\text{Na}}$ is ${\text{ + 1}}$ and the oxidation number of ${\text{O}}$ is $ - 2$. Let the oxidation number of sulfur be $x$.Thus,
${\text{(2 \times Oxidation number of Na)}} + {\text{(4 \times Oxidation number of S)}} + {\text{(6 \times Oxidation number of O)}} = {\text{Total charge on N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}}$
Thus,
\[(2 \times + 1) + (4 \times x) + (6 \times - 2) = 0\]
           \[( + 2) + (4x) + ( - 12) = 0\]
                                          \[4x = + 12 - 2\]
                                          \[4x = + 10\]
                                            \[x = + 2.5\]
Thus, the oxidation number of sulfur in ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}}$ is \[ + 2.5\].
Determine the oxidation number of iodine in ${\text{NaI}}$ as follows:
The total charge on ${\text{NaI}}$ is zero. The oxidation number of ${\text{Na}}$ is ${\text{ + 1}}$. Let the oxidation number of iodine be $x$.Thus,
${\text{(1 \times Oxidation number of Na)}} + {\text{(1 \times Oxidation number of I)}} = {\text{Total charge on NaI}}$
Thus,
\[(1 \times + 1) + (1 \times x) = 0\]
                        \[x = - 1\]
Thus, the oxidation number of iodine in ${\text{NaI}}$ is \[ - 1\].
Thus,
${\text{2N}}{{\text{a}}_{\text{2}}}\mathop {{{\text{S}}_{\text{2}}}}\limits^{ + 2} {{\text{O}}_3} + \mathop {{{\text{I}}_{\text{2}}}}\limits^0 \to {\text{N}}{{\text{a}}_{\text{2}}}\mathop {{{\text{S}}_{\text{4}}}}\limits^{ + 2.5} {{\text{O}}_{\text{6}}} + {\text{2Na}}\mathop {\text{I}}\limits^{ - 1} $
Step 2: Determine the N-factor of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ as follows:
The oxidation number of sulfur changes from ${\text{ + 2}}$ to \[ + 2.5\]. Thus,
${\text{N}} - {\text{factor of N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3} = {\text{Number of sulfur atoms}} \times {\text{Change in oxidation state}}$
Thus,
${\text{N}} - {\text{factor of N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3} = 2 \times (2.5 - 2)$
${\text{N}} - {\text{factor of N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3} = 2 \times (0.5)$
${\text{N}} - {\text{factor of N}}{{\text{a}}_2}{{\text{S}}_2}{{\text{O}}_3} = 1$
Thus, the N-factor of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is $1$.
Step 3: Determine the equivalent weight of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ as follows:
The equivalent weight is the ratio of the molecular weight to the N-factor.
Consider that the molecular weight of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is ${\text{M}}$. Thus,
${\text{Equivalent weight of N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}} = \dfrac{{\text{M}}}{1}$
${\text{Equivalent weight of N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}} = {\text{M}}$
Thus, the equivalent weight of ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is ${\text{M}}$.
So, the correct answer is “Option A”.

Note: The N-factor for the acids is the number of replaceable ${{\text{H}}^{\text{ + }}}$ ions while the N-factor for the bases is the number of replaceable ${\text{O}}{{\text{H}}^ - }$ ions.