Question

Calculate the amount (in milligrams) of $SeO _{3}^{-2}$ in solution on the basic of following data $20 \,mL$ of $M / 40$ solution of $KBrO _{3}$ was added to a definite volume of $SeO _{3}{ }^{-2}$ solution. The bromine evolved was removed by boiling and excess of $KBrO _{3}$ was back titrated with $5.0 \,mL$ of $M / 15$ solution of $NaAsO _{2}$. The reactions are given below. (Atomic mass of $K=39, Br =80, As =75, Na = 23, O =16, Se =79)$
(a) $SeO _{3}^{-2}+ BrO _{3}{ }^{-}+ H ^{+} \longrightarrow SeO _{4}{ }^{-2}+ Br _{2}+ H _{2} O$.
(b) $BrO _{3}^{-}+ AsO _{2}^{-}+ H _{2} O \longrightarrow Br ^{-}+ AsO _{4}^{-3}+ H ^{+}$

Answer 1

m.eq. of $SeO _{3}{ }^{2-}=$ m.eq. of $BrO _{3}{ }^{-}$
$2 \times$ m.mole of $SeO _{3}{ }^{2-}=$ m.eq. of $BrO _{3}{ }^{-} \times 5$
$2 \times$ m.mole of $AsO _{2}^{-}=$m.eq. of $BrO _{3}{ }^{-} \times 6$
$\frac{2}{5} \times$ m.mole of $SeO _{3}{ }^{2-}+\frac{2}{6}$ m.mole of $AsO _{2}{ }^{-}$
$= m \cdot$ mole of $BrO _{3}^{-}$(Total)
$\frac{2}{5} \times$ m.mole of $SeO _{3}{ }^{2-}+\frac{2}{6} \times 5 \times \frac{1}{1.5}=20 \times \frac{1}{40}$
m. mole of $SeO _{3}{ }^{2-}=\left(\frac{1}{2}-\frac{1}{9}\right) \frac{5}{2}=\frac{35}{36}$
wt. of $SeO _{3}{ }^{2-}$ in $mg =\frac{35}{36} \times 127$
$=123.47 \,mg \approx 123\, mg$