Question

A mixture of salts was treated to determine the barium content. A $0.230 g$ sample was dissolved and treated with excess potassium chromate. The precipitate of $BaCrO _{4}$ was dissolved in dil $HCl$ to convert $BaCrO _{4}$ to $Cr _{2} O _{7}^{2-}$ anion. The solution treated with excess sodium iodide and tri lodide produced was titrated with $21 mL$ of $0.095\, M$ sodium thiosulphate. Calculate the $\%$ of $BaCl _{2} \cdot 2 H _{2} O$ in the sample. [given Molar Mass of $BaCl 2.2 H 2 O =244]$

• A. 70.5 %
• B. 30.25 %
• C. 15.12 %
• D. 110 %

Answer 1

Answer: (A)

The skeletal reactions are
$BaCrO _{4}+ HCl \rightarrow Cr _{2} O _{7}^{2-}+ Ba ^{2+}$
$Cr _{2} O _{7}^{2-}+ I ^{-} \rightarrow Cr ^{3+}+ I _{3}^{-}$
$I _{3}^{-1}+ S _{2} O _{3}^{2-} \rightarrow S _{4} O _{6}^{2-}+ I ^{-}$
Meqs of $Na _{2} S _{2} O _{3}=21 \times 0.095 \times 1=1.995 Meqs$ of $I _{3}^{-}$
$\equiv$ Meqs of $Cr _{2} O _{7}^{2-}$
$mM$ of $Cr _{2} O _{7}^{2-}=\frac{1.995}{6}$
$mM$ of $CrO _{4}^{2-}=\frac{1.995}{6} \times 2=0.665 mM$ of $BaCrO _{4}$
Mass $\%$ of $BaCl _{2} \cdot 2 H _{2} O =\frac{0.665 \times 10^{-3} \times 244}{0.230} \times 100=70.5\%$