Question

12.0 of an impure sample of arsenious oxide $({As}_{2}{O}_{3})$ was dissolved in water containing 7.5g of sodium bicarbonate $(NaHC{O}_{3})$ and the resulting solution was diluted to 250mL. 25mL of this solution was completely oxidised by 22.4 mL of a solution of ${I}_{2}$. 25 mL of this solution reacted with the same volume of a solution containing 24.8g of ${Na}_{2}{S}_{2}{O}_{3}.5{H}_{2}O$ in 1L. The percentages of $({As}_{2}{O}_{3})$ in the sample is:
a.) 9.24%
b.) 8.24%
c.) 9.74%
d.) none of the above

Answer 1

Hint: This question involves the concept of milliequivalents (Meq.). It is the relation between the normality and the volume of a substance.

Complete step by step answer:
In this question, 12g of ${As}_{2}{O}_{3}$ sample reacts with 7.5g of $NaHC{O}_{3}$ and gives ${Na}_{3}As{O}_{3}$. The redox reaction that takes place is as follows:

${ As }_{ 2 }^{ 3+ }\quad \longrightarrow \quad { As }_{ 2 }^{ 5+ }\quad +\quad { 2e }^{ - }$
${ I }_{ 2 }\quad +\quad { 2e }^{ - }\quad \longrightarrow \quad { 2I }^{ - }$

In the above reaction, we can see that arsenic gets oxidized from +3 oxidation state to +5 oxidation state and iodine gets reduced from 0 oxidation state to -1 oxidation state.
Therefore, from this we can say that the Meq of ${As}_{2}{O}_{3}$ in 25 mL of solution is equal to the Meq of ${I}_{2}$.
Therefore, the Meq of ${As}_{2}{O}_{3}$ in 25 mL is equal to the Meq of ${I}_{2}$
$\implies$ Meq of ${As}_{2}{O}_{3}$ in 25 mL = volume of ${I}_{2}$ x normality
$\implies$ Meq of ${As}_{2}{O}_{3}$ = 22.4 x N
Now, we know that the normality (N) is calculated by using the law of equivalence.
The reaction involved in this process are:

$ { I }_{ 2 }\quad +\quad { 2e }^{ - }\quad \longrightarrow \quad { 2I }^{ - }$
$ 2{ S }_{ 2 }{ O }_{ 3 }^{ 2- }\quad \longrightarrow \quad { S }_{ 4 }{ 0 }_{ 6 }^{ 2- }\quad +{ 2e }^{ - }$

From the reaction we can say that the n-factor of S is 1. Now, the Meq of iodine is equal to the Meq of hypo thiosulfate.
Thus, Meq of ${I}_{2}$ = Meq of ${Na}_{2}{S}_{2}{O}_{3}.5{H}_{2}O$
$\implies$ Meq of ${I}_{2}$ = Volume x normality
$ V\quad \times \quad { N }_{ { I }_{ 2 } }\quad =\quad \cfrac { Weight }{ Equivalent\quad weight } \quad \times \quad Volume$
$ 25\quad \times \quad { N }_{ { I }_{ 2 } }\quad =\quad \cfrac { 24.8 }{ 248 } \quad \times \quad 25$
$ { N }_{ { I }_{ 2 } }\quad =\quad \cfrac { 1 }{ 10 }$

Now, we have to calculate the Meq of ${As}_{2}{O}_{3}$ in the solution
Meq of ${As}_{2}{O}_{3}$ in 25 mL = $22.4 \times \cfrac {1}{10} = 2.24$
Meq of ${As}_{2}{O}_{3}$ in 250 mL = $2.24 \times \cfrac {250}{25} = 22.4$

Now, we know that the Meq of ${As}_{2}{O}_{3}$ = $\cfrac { Weight }{ Equivalent\quad weight } \quad \times \quad 1000$
$\implies 22.4\quad =\quad \cfrac { Weight }{ \cfrac { 198 }{ 4 } } \quad \times \quad 1000$
Here, equivalent weight of ${As}_{2}{O}_{3}$ = $\cfrac {M}{4}$
$\implies Weight\quad =\quad \cfrac { 22.4\quad \times \quad 198 }{ 4\quad \times \quad 1000 } \quad =\quad 1.1088\quad g$

Now, the percentage of ${As}_{2}{O}_{3}$ present can be calculated as
%$\quad of\quad { As }_{ 2 }{ O }_{ 3 }\quad =\quad \cfrac { Weight }{ Total\quad weight } \quad \times \quad 100$
Substituting the values in the above equation we get,
%$\quad of\quad { As }_{ 2 }{ O }_{ 3 }\quad =\quad \cfrac { 1.1088 }{ 12 } \quad \times \quad 100\quad =\quad 9.24%$
Therefore, % of ${As}_{2}{O}_{3}$ present in the sample is 9.24%.
So, the correct answer is “Option A”.

Note: While solving these types of questions, we need to keep in mind a few things, like the difference between the equivalent weight and the molecular weight. The first one is defined as molecular weight divided by the valence factor and the later is the sum of atomic weights of the individual atoms in a molecule.