# For the following chemical reaction: #2HBr(aq) + Ba(OH)_2 -> 2H_2O(l) + BaBr_2(aq), how do you write the net ionic equation, including the phases?

Jan 26, 2017

${H}^{+} + H {O}^{-} \rightarrow {H}_{2} O \left(l\right)$
${H}^{+}$ is a convenience. Another representation is as ${H}_{3} {O}^{+}$, that is $\text{hydronium ion}$. As far as anyone knows, the actual species is a cluster of 4-5 water molecules, with an EXTRA ${H}^{+}$, i.e. to give ${H}_{9} {O}_{4}^{+}$ or ${H}_{11} {O}_{5}^{+}$. The ${H}^{+}$ is conceived to pass rapidly from cluster to cluster, to tunnel if you like. If you have ever played rugby, think of a maul, where the forwards can pass the pill from hand to hand while moving forward. Perhaps this is not the best analogy, because of the tunnelling ability of the protium ion, whereas a rugby ball has to be passed from hand to hand.
In this given reaction, the barium ions and bromide ions are along for the ride; they remain is solution as their water solvates. $B {a}^{2 +}$ is arguably the aquated ion, ${\left[B a {\left(O {H}_{2}\right)}_{6}\right]}^{2 +}$, and the bromide ion is arguably ${\left[B r {\left({H}_{2} O\right)}_{4 - 6}\right]}^{-}$, we write $B {a}^{2 +}$ etc. as a shorthand. Bond formation occurs between the hydroxide ion, and the protium ion to give the water molecule.