# How do the hydrogen halides dissolve in water, and how does such dissolution lead to acidity? How do we express these concentrations?

Jun 2, 2017

You will have to restate this question.......

#### Explanation:

$\text{Concentration"="Moles of solute"/"Volume of solution}$.

Common hydrogen halides are $H C l \left(g\right)$ or $H I \left(g\right)$; these dissolve in water to give $\text{hydronium or acidium ions..........}$

$H C l \left(g\right) + {H}_{2} O \left(l\right) \rightarrow {H}_{3} {O}^{+} + C {l}^{-} \left(g\right)$

The ${H}_{3} {O}^{+}$ representation is more a conception than an actuality. As far as anyone knows it is cluster of water molecules with an extra ${H}^{+}$ associated with the cluster.

The following is taken from an answer to a related question.

$H C l \left(g\right)$ is a source of hydronium ion, ${H}_{3} {O}^{+}$ in aqueous solution.........

We may take a tank of $H C l \left(g\right)$, and we can bleed it in to water to give an AQUEOUS solution that we could represent as $H C l \left(a q\right)$ OR ${H}_{3} {O}^{+}$ and Cl^−. In each case this is a REPRESENTATION of what occurs in solution.

As far as anyone knows, the actual acidium ion in solution is
${H}_{5} {O}_{2}^{+}$ or ${H}_{7} {O}_{3}^{+}$, i.e. a cluster of 3 or 4 water molecules with an EXTRA ${H}^{+}$ tacked on.

Note that the ${H}^{+}$ is quite mobile, and passes, tunnels if you like, the extra ${H}^{+}$ from cluster to cluster. For this reason both ${H}^{+}$ and $H {O}^{-}$ have substantial mobility in aqueous solution.

${H}^{+} + {e}^{-} \rightarrow \frac{1}{2} {H}_{2} \left(g\right)$

Depending at which level you are at (and I don't know!, which is part of the problem in answering questions on this site), you might not have to know the details at this level of sophistication. The level I have addressed here is probably 1st/2nd year undergrad.........