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

1 Answer
Jul 6, 2017

Answer:

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.........