Question

Calculate the amount (in milligrams) of $Se{O}_3^{-2}$ in solution on the basic of following data $20mL$ of $M/40$ 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 $5.0mL$ of $M/15$ solution of $NaAs{O}_2$. The reactions are given below. (Atomic mass of $K=39,Br=80,As=75,Na=23,O=16,Se=79)$
(a) $Se{O}_3^{-2}+Br{O}_3{}^{-}+H^{+}⟶Se{O}_4{}^{-2}+B{r}_2+H_{2}O$.
(b) $Br{O}_3^{-}+As{O}_2^{-}+H_{2}O⟶B{r}^{-}+As{O}_4^{-3}+H^{+}$

Answer 1

Answer: 123

m.eq. of $Se{O}_3{}^{2-}=$ m.eq. of $Br{O}_3{}^{-}$
$2\times$ m.mole of $Se{O}_3{}^{2-}=$ m.eq. of $Br{O}_3{}^{-}\times 5$
$2\times$ m.mole of $As{O}_2^{-}=$m.eq. of $Br{O}_3{}^{-}\times 6$
$\frac{2}{5}\times$ m.mole of $Se{O}_3{}^{2-}+\frac{2}{6}$ m.mole of $As{O}_2{}^{-}$
$=m\cdot$ mole of $Br{O}_3^{-}$(Total)
$\frac{2}{5}\times$ m.mole of $Se{O}_3{}^{2-}+\frac{2}{6}\times 5\times \frac{1}{1.5}=20\times \frac{1}{40}$
m. mole of $Se{O}_3{}^{2-}=\left(\frac{1}{2}-\frac{1}{9}\right)\frac{5}{2}=\frac{35}{36}$
wt. of $Se{O}_3{}^{2-}$ in $mg=\frac{35}{36}\times 127$
$=123.47mg\approx 123mg$