# Why is no bromine liberated in the reaction of potassium bromide and concentrated phosphoric acid?

$\ce{KBr}$ 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?

• Think about the oxidation state of bromine. May 4 '15 at 18:08
• @J.LS So it can't change from -1 to 0? May 4 '15 at 18:30
• 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. May 4 '15 at 19:02
• Also, bromine exists as a liquid at room temperature. May 4 '15 at 20:19
• 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. May 7 '15 at 13:55

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 $\ce{Cl2_{(g)}}$.