Question

How does a Brønsted-Lowry acid form its conjugate base when dissolved in water? How is the water involved in this process?

Answer 1

The Brønsted-Lowry acid-base theory designates `H_3O^+` as the acidium species in solution.........and thus water, `OH_2`, is the proton carrier...........

Explanation:

And this `H_3O^+`, the hydronium ion, is simply a protonated water molecule.........i.e. `H^+ + H_2O rarr H_3O^+` (mass and charge are conserved as always!)

`HCl(g)` (for instance) is a source of hydronium ion, `H_3O^+` in aqueous solution.........

We may take a tank of `HCl(g)`, and we can bleed it in to water to give an AQUEOUS solution that we could represent as `HCl(aq)` OR `H_3O^+` and `Cl^−`.

`HCl(g) stackrel(H_2O)rarrunderbrace(H_3O^(+))_("hydronium ion") +Cl^-`

In each case this is a REPRESENTATION of what occurs in solution. If we bleed enuff gas in, we achieve saturation at a concentration of approx. `10.6*mol*L^-1` with respect to hydrochloric acid.

As far as anyone knows, the actual acidium ion in solution is
`H_5O_2^+` or `H_7O_3^+`, i.e. a cluster of 2 or 3 or 4 water molecules with an EXTRA `H^+` tacked on. We represent it in solution (without loss of generality) as `H_3O^+`, the `"hydronium ion"`, which is clearly the conjugate acid of `H_2O`.

Note that the `H^+` is quite mobile, and passes, tunnels if you like, the extra `H^+` from cluster to cluster. If you have ever played rugby, I have always liked to compare this to when the forwards form a maul, and can pass the pill from hand to hand to the back of the maul while the maul is still formed.

Of course, tunnelling, proton transfer, is more likely in a cluster of water molecules, so the analogy might not be particularly apt in that there is definite transfer of a ball in a maul, but a charge in a water cluster is conceivably tunnelled. The same applies to the transfer of an hydroxide ion. For this reason both `H^+` and `HO^-` have substantial mobility in aqueous solution.

And we know that hydroxide/hydronium ions obey the following equilibrium in aqueous solution under standard conditions of temperature and pressure........

`H_3O^+ + HO^(-) rarr2H_2O`; `K_w=[H_3O^+][HO^-]=10^(-14)`.

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.........