Question

$36mL$ of $0.5M$ $B{r}_2$ solution when made alkaline undergoes complete disproportionation into $B{r}^{⊝}$ and $Br{O}_3^{⊝}$. The resulting solution requires $45mL$ of $As\left(III\right)$ solution to reduce $Br{O}_3^{⊝}$ to $B{r}^{⊝}$. Given that $As\left(III\right)$ is oxidised to $As\left(V\right)$, what is the molarity of $As\left(III\right)$ solution?

• A. $0.2$
• B. $0.1$
• C. $0.4$
• D. $0.5$

Answer 1

The correct option is C $0.4$

$\underset{36mL/A{s}^{3+}}{6B{r}_2+12\stackrel{⊝}{O}H}\to 10B{r}^{⊝}+\underset{\underset{45mL/A{s}^{3+}}{\downarrow }}{2Br{O}_3^{⊝}}+6H_{2}O$
$6$ mmol $B{r}_2\equiv 2$ mmol $Br{O}_3^{⊝}$
$(36\times 0.5)=\frac{1}{3}\times 36\times 0.5$ mmol $Br{O}_3^{⊝}$
mmol $Br{O}_3^{⊝}=6.0$
$Br{O}_3^{⊝}+3A{s}^{3+}+6H^{⨁}\to B{r}^{⊝}+3A{s}^{5+}+3H_{2}O$
$1$ mmol $Br{O}_3^{⊝}\equiv 3$ mmol $A{s}^{3+}$
$6.0$ mmol $Br{O}_3^{⊝}\equiv$ $18$ mmol $A{s}^{3+}$
$(MA{s}^{3+}=\frac{18}{45}=0.4M)$