Question

$12$ g of an impure sample of arsenious oxide was dissolved in water containing $7.5$ g of sodium bicarbonate and the resulting solution was diluted to $250$ mL. $25$ mL of this solution was completely oxidized by $22.4$ mL of solution of iodine. $25$ mL of this iodine solution reacted with same volume of a solution containing $24.8$ g of sodium was completely oxidized by $22.4mL$ of a solution thiosulphate $\left(N{a}_2{S}_2{O}_{3}5H_{2}O\right)$ in one litre. Calculate the percentage of arsenious oxide in the sample. (Atomic mass of $As=75$)

• A. $12.4\%$
• B. $3.2\%$
• C. $9.2\%$
• D. None of the above

Answer 1

The correct option is C $9.2\%$

Given, $A{s}_2{O}_3$ sample$=12.0$ g

It reacts with $NaHC{O}_3$ to give $N{a}_{3}As{O}_3$. Its reaction with $I_2$ shows the reactions :
$(A{s}^{3+}{)}_2⟶(A{s}^{5+}{)}_2+4e$
$I_2+2e⟶2I^{-}$

Meq. of $A{s}_2{O}_3$ in 25 mL = Meq. of $I_2$
$=22.4\times N$ ...(i)

Also, normality of $I_2$ can be evaluated as Meq. of $I_2=$ Meq. of hypo

$N\times 25=\frac{24.8}{248\times 1}\times 25$
$N=\frac{1}{10}$

The reactions here are :
$I_2+2e⟶2I^{-}$
$2S_2{O}_3^{2-}⟶S_4{O}_6^{2-}+2e$

$\therefore$ By Eq. (i), Meq. of $A{s}_2{O}_3$ in 25 mL$=22.4\times \frac{1}{10}=2.24$

$\therefore$ Meq. of $A{s}_2{O}_3$ in 50 mL $=2.24\times \frac{250}{25}=22.4$

or $\frac{w}{E_{A{s}_2{O}_3}}\times 1000=22.4$

or $\frac{w}{\left(\frac{198}{4}\right)}\times 1000=22.4$

$\therefore w_{A{s}_2{O}_3}=1.1088$ g

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