Question

A mixture containing $A{s}_2{O}_3$ and $A{s}_2{O}_5$ required $20.10$ mL of $0.05N$ iodine for titration. The resulting solution is then acidified and excess of $KI$ was added. The liberated iodine required $1.1113$ g hypo $\left(N{a}_2{S}_2{O}_3.5H_{2}O\right)$ for complete reaction. Calculate the mass of mixture. The reactions are as follows:
$A{s}_2{O}_3+2I_2+2H_{2}O⟶A{s}_2{O}_5+4H^{+}+4I^{-}$
$A{s}_2{O}_5+4H^{+}+4I^{-}⟶A{s}_2{O}_3+2I_2+2H_{2}O$

• A. $0.2497$ g
• B. $0.485$ g
• C. $0.1257$ g
• D. $0.0846$ g

Answer 1

Answer: (A)

$0.2497$ g

Meq. of $I_2$ used $=20.10\times 0.05=1.005$
Let meq. of $A{s}_2{O}_3$ and $A{s}_2{O}_5$ in mixture be $a$ and $b$ respectively.

Addition of $I_2$ to mixture converts $A{s}_2^{3+}$ into $A{s}_2^{5+}$.
$\therefore$ Meq. of $A{s}_2{O}_3=$ Meq. of $I_2$ used $=$Meq. of $A{s}_2^{5+}$ formed
$\therefore a=1.005$ ...(i)
After oxidation, the mixture contains all the arsenic in $+5$ oxidation state, which is then reduced to $A{s}_2^{3+}$ by using $KI$ and hypo.
$\therefore$ Meq. of $A{s}_2{O}_3$ as $A{s}^{5+}+$ Meq. of $A{s}_2{O}_3$
$=$ Meq. of $I_2$ liberated $=$ Meq. of hypo used
$a+b=\left(\frac{1.113}{248}\right)\times 1000$
or $a+b=4.481$ ...(ii)
By Eqs. (i) and (ii), we get $b=4.481-1.005=3.476$
$\therefore$ Mass of $A{s}_2{O}_3=(\text{Meq.}\times \frac{\text{Eq. mass}}{1000})$ $(E_{A{s}_2{O}_3}=\frac{198}{4})$
$=\frac{1.005\times 198}{4\times 1000}=0.498$ g
Mass of $A{s}_2{O}_5=\frac{3.476\times 230}{4\times 1000}(E_{A{s}_2{O}_5}=\frac{230}{4})$
$=0.1999$ g
$\therefore$ Mass of mixture $=0.0498+0.1999=0.2497$ g