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?


1 Answer 1


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.

  • $\begingroup$ 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, 2016 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, 2016 at 18:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.