Question

One gram of $N{a}_{2}As{O}_4$ is boiled with excess of solid $KI$ in presence of strong $HCl$. The iodine evolved is absorbed in $KI$ solution and litrated against $0,2N$ hypersolution. Assuming the reaction to be
$AS{O}_4^{3-}+2H^{+}+2I^{-}\to As{O}_3^{2-}+H_{2}O+I_2$.
Calculate the volume of thiosulphate hypo consumed. [Atomic weight of $As=75$].

• A. $48.1mL$
• B. $58.4mL$
• C. $24.7mL$
• D. $30.3mL$

Answer 1

The correct option is A $48.1mL$

Given,

Mass of $N{a}_{3}As{O}_4=1gm$

Molar mass of $N{a}_{3}As{O}_4=208gm$

Normality of hyper solution=$0.2N$

The reaction involved is

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

Here oxidation state Here oxidation state

of $As=+5$ of $As=+3$

$\therefore C^{-}$ transfer=$2$

$\therefore$ n factor of $N{a}_{3}As{O}_4=2$

$\Rightarrow$ Normality of $N{a}_{3}As{O}_4=\frac{1}{\frac{208}{2}}=\frac{1}{104}N$

Now, at equivalence

No. of gm equivalents of $N{a}_{3}As{O}_4$= No. of gm equivalents of thiosulphate

$\Rightarrow \underset{N{a}_{3}As{O}_4}{N_1{V}_1}=\underset{thiosulphate}{N_2{V}_2}$

$\Rightarrow \frac{1}{104}\times 1=0.2\times V_2$ $[\because$ Volume of $N{a}_{3}As{O}_4=1L]$

$\Rightarrow V_2=0.04807L$

$=48.1ml$