Question

Calculate the mass of $Se{O}_3^{2-}$ in solution on the basis of following data. $20$ mL of $\frac{M}{60}$ solution of $KBr{O}_3$ was added to a definite volume of $Se{O}_3^{2-}$ solution. The bromine evolved was removed by boiling and excess of $KBr{O}_3$ was back titrated with $5.1$ mL of $\frac{M}{25}$ solution of $NaAs{O}_2$. The reactions are given below:

(a) $Se{O}_3^{2-}+Br{O}_3^{-}+H^{+}⟶Se{O}_4^{2-}+B{r}_2+H_{2}O$
(b) $Br{O}_3^{-}+As{O}_2^{-}+H_{2}O⟶B{r}^{-}+As{O}_4^{3-}+H^{+}$

• A. $0.064$ g
• B. $0.084$ g
• C. $0.042$ g
• D. None of these

Answer 1

The correct option is B $0.084$ g

The reactions are as follows:
In (a) $S{e}^{4+}⟶S{e}^{6+}+2e$
$10e+2B{r}^{5+}⟶B{r}_2^0$
$\therefore$ Eq. mass $KBr{O}_3=\frac{M}{5}$ (Valence factor=5)
In (b) $6e+B{r}^{5+}⟶B{r}^{-}$
$\therefore$ Eq. mass $KBr{O}_3=\frac{M}{6}$ (Valence factor=6)
$A{s}^{3+}⟶A{s}^{5+}+2e$
Left Meq. of $Br{O}_3^{-}$ of valence factor 6 $=$ Meq. of $As{O}_2^{-}=5.1\times \frac{1}{25}\times 2=0.0408$
Left Meq. of $Br{O}_3^{-}$ of valence factor (5) $=\frac{0.408\times 5}{6}=0.34$
Meq. of $Br{O}_3^{-}$ of valence factor (5) added $=20\times \frac{1}{60}\times 5$ $=\frac{5}{3}=1.67$
$\therefore$ Meq. of $Br{O}_3^{-}$ used for $Se{O}_3^{2-}=1.67-0.34=1.33$
Meq. of $Se{O}_3^{2-}=1.33$
or $\frac{w}{\frac{127}{2}}\times 1000=1.33$
$w_{Se{O}_3^{2-}}=0.084$ g
Hence, the mass of $Se{O}_3^{2-}$ is $0.084$ g.