Question

$12.0$ of an impure sample of arsenious oxide $\left(A{s}_2{O}_3\right)$ was dissolved in water containing $7.5$ g of sodium bicarbonate $\left(NaHC{O}_3\right)$ and the resulting solution was diluted to $250$ mL. $25$ mL 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 $24.8$ g of $N{a}_2{S}_2{O}_3\cdot 5H_{2}O$ in $1L$. The percentage of $A{s}_2{O}_3$ in the sample is :

• A. $9.24$%
• B. $8.24$%
• C. $9.74$%
• D. none of the above

Answer 1

The correct option is A $9.24$%

Given, $A{s}_2{O}_3=12.0g$. It reacts with $NaHC{O}_3$ to give $N{a}_{3}As{O}_3$. Its reaction with $I_2$ shows the following redox changes.
$a$. $\stackrel{+3}{\underset{\underset{2x=2}{2x-6=0}}{A{s}_2{O}_3}}\to \stackrel{+5}{\underset{\underset{2x=10}{2x-10=0}}{A{s}_2{O}_5}}+4e^{-}(n=4)$
$b$. $I_2+2e^{-}\to 2I^{⊝}(n=2)$
$c$. $2S_2{O_3}^{2-}\to S_4{O_6}^{2-}+2e^{-}$ $(n=\frac{2}{2}=1)$
$mEq$ of $A{s}_2{O}_3$ in $25$ mL $=mEq$ 0f $I_2=22.4\times N$
Also, $mEq$ of $I_2=mEq$ of hypo $\left(S_2{O_3}^{2-}\right)$ $=N\times V$
$N\times 25=\frac{24.8}{\frac{248}{1}}\times 25$ (Given same volume of hypo reacts with iodine) $[EwofN{a}_2{S}_2{O}_3.5H_{2}O=\frac{248}{1}(n=1\left)\right]$
$\therefore N_{I_2}=\frac{1}{10}$
$mEq$ of $A{s}_2{O}_3$ in $25$ mL $=22.4\times \frac{1}{10}=2.24$
or, $mEq$ of $A{s}_2{O}_3$ in $250$ mL$=\frac{2.24\times 250}{25}=22.4$
$\frac{W}{Ew}\times {10}^3=22.4$ $[EwofA{s}_2{O}_3=\frac{198}{4}(n=4\left)\right]$

$\frac{W}{\frac{198}{4}}\times {10}^3=22.4$
$W_{A{s}_2{O}_3}=\frac{22.4\times 198}{4\times {10}^3}=1.1088$

% of $A{s}_2{O}_3=\frac{1.1088}{12}\times 100=9.24$%