Question

One gram of $N{a}_{3}As{O}_4$ is boiled with an excess of solid 𝐾𝐼 in the presence of strong 𝐻𝐶𝑙. The iodine evolved is absorbed in 𝐾𝐼 solution and titrated against 0.2 𝑁 hypo solution. Assuming the reaction to be $As{O}_4^{3-}+2H{}^{+}+2I{}^{-}\to As{O}_3^{3-}+H_{2}O+I_2$, calculate the volume of hypo consumed. [Atomic weight of 𝐴𝑠 = 75]

• A. $48.1$ mL
• B. $30.3$ mL
• C. $38.4$ mL
  
• D. $24.7$ mL

Answer 1

The correct option is A $48.1$ mL

Calculating the moles of $I_2$ produced

$As{O}_4^{3-}+2H{}^{+}+2I{}^{-}\to As{O}_3^{3-}+H_{2}O+I_2$

Moles of $As{O}_4^{3-}$ reacted$=\frac{Mass}{Molarmass}$

$=\frac{1}{208}=0.0048$

Calculating the volume of hypo consumed in the chemical reaction involved

According to the reaction,

$I_2+2S_2{O}_3^{2-}\to S_4{O}_6^{2-}+2I{}^{-}$

1 𝑚𝑜𝑙 of $I_2$ reacts with 2 𝑚𝑜𝑙 of $S_2{O}_3^{2-}$

So, 0.0048 𝑚𝑜𝑙 of $I_2$ will react with
2 × 0.0048 𝑚𝑜𝑙 of $S_2{O}_3^{2-}$
$Normalityofhyposolution=\frac{Numberofmoles}{Volumeofthesolution\left(inL\right)}\times n-factor$
$Volumeof=\frac{Numberofmoles}{Normality}\times n-factor$

The 𝑛 − 𝑓𝑎𝑐𝑡𝑜𝑟 of hypo solution is 1.

$=\frac{0.0096}{0.2}\times 1=0.048L=48mL$

Thus, 48 𝑚𝐿 of 0.2 𝑁 hypo solution is consumed.

So, option (A) is the correct answer.