Question

A mixture of salts was treated to determine the barium content. A 0.230g sample was dissolved and treated with excess potassium chromate. The precipitate of BaCrO₄ was dissolved in dil HCl to convert BaCrO₄ to $\text{Cr}_{2} \text{O}_{7}^{2 -}$ anion. The solution treated with excess sodium iodide and tri Iodide produced was titrated with 21 mL of 0.095 M sodium thiosulphate. Calculate the % of $\text{BaCl}_{2} \cdot 2 \text{H}_{\text{2}} \text{O}$ in the sample. [given Molar Mass of BaCl2.2H2O = 244]

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

Answer 1

Answer: (A)

The skeletal reactions are
$\text{BaCrO}_{4}+\text{HCl} & \rightarrow & \text{Cr}_{2}\text{O}_{7}^{2 -}+\text{Ba}^{2 +} \\ \text{Cr}_{2}\text{O}_{7}^{2 -}+\text{I}^{-} & \rightarrow & \text{Cr}^{3 +}+\text{I}_{3}^{-} \\ \text{I}_{3}^{- 1}+\text{S}_{2}\text{O}_{3}^{2 -} & \rightarrow & \text{S}_{4}\text{O}_{6}^{2 -}+\text{I}^{-}$
Meqs of Na₂S₂O₃ = 21 x 0.095 x 1 = 1.995 Meqs of $\text{I}_{3}^{-}$
$≡$ Meqs of $\text{Cr}_{2} \text{O}_{7}^{2 -}$
mM of $\text{Cr}_{2} \text{O}_{7}^{2 -} = \frac{1 \text{.} 9 9 5}{6}$
mM of $\text{Cr} \text{O}_{4}^{2 -} = \frac{1 \text{.} 9 9 5}{6} \times 2 = 0 \text{.} 6 6 5$ mM of BaCrO₄
Mass % of BaCl₂ . 2H₂O $= \frac{0 \text{.} 6 6 5 \times 1 0^{- 3} \times 2 4 4}{0 \text{.} 2 3 0} \times 1 0 0 = 7 0 \text{.} 5 \text{\%}$