4
$\begingroup$

$\ce{KBr}$ reacts with concentrated $\ce{H3PO4}$ to give $\ce{HBr}$ and ($\ce{KH2PO4}$ or $\ce{K3PO4}$) (not sure which one, if someone knows it, please tell).

Why isn't bromine gas liberated?

$\endgroup$
6
  • $\begingroup$ Think about the oxidation state of bromine. $\endgroup$
    – J. LS
    May 4, 2015 at 18:08
  • $\begingroup$ @J.LS So it can't change from -1 to 0? $\endgroup$
    – Larry
    May 4, 2015 at 18:30
  • 6
    $\begingroup$ That would require a pretty strong oxidizing agent. Phosphoric acid is an oxidizing agent, but not strong enough. You could probably produce significant amounts of bromine if you used sulfuric acid. $\endgroup$
    – J. LS
    May 4, 2015 at 19:02
  • 1
    $\begingroup$ Also, bromine exists as a liquid at room temperature. $\endgroup$
    – ringo
    May 4, 2015 at 20:19
  • 1
    $\begingroup$ If you are just after generating bromine, you can pass chlorine gas through a bed of KBr. This instantly precipitates pure bromine leaving KCl. The heat of the reaction vaporizes the bromine, so the reaction is best performed in a flask fitted with a condenser to liquify the bromine vapor. $\endgroup$
    – docscience
    May 7, 2015 at 13:55

1 Answer 1

4
$\begingroup$

Firstly, $\ce{H3PO4}$ is not a strong enough oxidizing agent to remove the electrons from the $\ce{Br-}$ ions in order for them to then form $\ce{Br2}$, which exists as a liquid at room temperature. Assuming we are working at room temperature, using a strong enough oxidizing agent such as $\ce{H2SO4}$ or a more reactive halogen like $\ce{Cl2_{(g)}}$ or $\ce{F2_{(g)}}$ would be enough to oxidize the $\ce{Br-}$ to $\ce{Br2_{(l)}}$.

If we desired to evolve $\ce{Br2_{(g)}}$, we would have to heat $\ce{Br2_{(l)}}$ to its boiling point of $58.8\mathrm{^oC}$ and then continue to supply it with enough heat to vaporize it. As docscience stated, however, some oxidizing reactions might be exothermic enough to vaporize $\ce{Br2_{(l)}}$ itself. In solution this would depend on $[\ce{HBr}]$ as to whether or not the heat evolved is enough to vaporize the $\ce{Br2_{(l)}}$, but in the solid form $\ce{Br2_{(l)}}$ is vaporized instantly in the extremely exothermic reaction of $\ce{KBr}$ with either $\ce{F2_{(g)}}$ or $\ce{Cl2_{(g)}}$.

$\endgroup$

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.