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 $BaCr{O}_4$ was dissolved in dil $HCl$ to convert $BaCr{O}_4$ to $C{r}_2{O}_7^{2-}$ anion. The solution treated with excess sodium iodide and tri lodide produced was titrated with $21mL$ of $0.095M$ sodium thiosulphate. Calculate the $\%$ of $BaC{l}_2\cdot 2H_{2}O$ 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
$BaCr{O}_4+HCl\to C{r}_2{O}_7^{2-}+B{a}^{2+}$
$C{r}_2{O}_7^{2-}+I^{-}\to C{r}^{3+}+I_3^{-}$
$I_3^{-1}+S_2{O}_3^{2-}\to S_4{O}_6^{2-}+I^{-}$
Meqs of $N{a}_2{S}_2{O}_3=21\times 0.095\times 1=1.995Meqs$ of $I_3^{-}$
$\equiv$ Meqs of $C{r}_2{O}_7^{2-}$
$mM$ of $C{r}_2{O}_7^{2-}=\frac{1.995}{6}$
$mM$ of $Cr{O}_4^{2-}=\frac{1.995}{6}\times 2=0.665mM$ of $BaCr{O}_4$
Mass $\%$ of $BaC{l}_2\cdot 2H_{2}O=\frac{0.665\times 10^{-3}\times 244}{0.230}\times 100=70.5\%$