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A $1.0\ \mathrm{g}$ sample containing $\ce{BaCl2*2H2O}$ was dissolved and an excess of $\ce{K2CrO4}$ solution added. After a suitable period, the $\ce{BaCrO4}$ was filtered, washed and redissolved in $\ce{HCl}$ to convert $\ce{CrO4^2-}$ to $\ce{Cr2O7^2-}$. An excess of $\ce{KI}$ was added, and the liberated iodine was titrated with $84.7\ \mathrm{mL}$ of $0.137\ \mathrm{M}$ sodium thiosulphate. Calculate the percent purity of $\ce{BaCl2*2H2O}$.

I first found the number of mili-equivalents of sodium thiosulphate and equated that to the number of moles of pure $\ce{BaCl2*2H2O}$ times the n-factor i.e. $2$.

From this equation I get the number of moles of pure $\ce{BaCl2*2H2O}$ and then multiply this by its molecular mass to get the mass of pure $\ce{BaCl2*2H2O}$. But it comes greater than $1\ \mathrm{g}$. How to solve this?

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Chromium(VI) is reduced to chromium(III) while oxidizing iodide to iodine. So one equivalent of chromate corresponds to 3 equivalents of iodine ($\ce{I}$, not $\ce{I2}$).

$$\ce{BaCl2 \cdot 2 H2O} \equiv \ce{BaCrO4} ~\mathrm{(chrome(VI))} \equiv \frac{3}{2}~\ce{I2} \equiv 3~\ce{S2O3^{2-}}$$

If you take this into account you should probably end up with 87.5% purity.

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  • $\begingroup$ K 2 Cr 2 O 7 + 7H 2 SO 4 + 6KI → 4K 2 SO 4 + Cr 2 (SO 4 ) 3 + 7H 2 O +3I 2 , Is this not the reaction? Also the what would be the product when I reacts with ‎Na2S2O3? Can you please elaborate a little ? $\endgroup$ – rony Dec 31 '16 at 17:54
  • $\begingroup$ Your reaction equation is correct. However, spectator ions are not very interesting for understanding what's going on. The essential thing is that 1 mole chromium(IV) oxidizes 3 moles iodide, and 1 mole $\ce{I2}$ will oxidize 2 moles thiosulfate, hence 1 mole iodine atoms oxidize 1 mole thiosulfate. Therefore 3 moles thiosulfate are equivalent to 1 mole chromium(VI) and hence to 1 mole barium chloride. $\endgroup$ – aventurin Dec 31 '16 at 18:39

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