Acid-base equilibria
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Why do only some ethanoic acid molecules ionise in water, when nearly all molecules of hydrochloric acid do?
Igloo says
This is an excellent and thoughtful question, and to answer it you need to consider the two processes:
[1] HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq)
[2] CH3COOH(aq) + H2O(l) <=> H3O+(aq) + CH3COO-(aq)
In each case I have shown the proton transfer taking place between the acid molecule and the water molecule.
As you know, since process [1] is considered to be virtually complete, we use an irreversible arrow for the reaction, whereas a reversible arrow is more appropriate for the equilibrium which is established in process [2].
In both processes, the water molecule is acting as a Bronsted-Lowry base and is being converted into a hydroxonium ion. The difference you ask about must lie in the conversion of the acid molecules into the respective anions, Cl-(aq) and CH3COO-(aq). In the case of process [1], a hydrogen-chlorine bond is being broken, whereas in process [2] a hydrogen-oxygen bond is broken. I would imagine that these are of comparable strength, so bond energies are unlikely to be responsible for the difference in the extent to which each process takes place. However, in process [1] the creation of Cl-(aq) is likely to be considerably more exothermic than that of CH3COO-(aq), largely due to the fact that it is much, much smaller and therefore has a considerably higher charge density than that of its organic counterpart. Appropriate data supports this view: the enthalpy of hydration of the gaseous chloride ion is high at -378 kJ mol-1, and although I have been unable to track down that of the corresponding ethanoate ion, I am sure that it will be less negative (less exothermic).
As you will know from your work on entropy changes, the two processes are likely to have negligible entropy changes for the system, since in both cases two moles of aqueous reactants form two moles of aqueous products, yet because process [1] is much more exothermic than process [2], this will give a much higher positive value for the entropy change in the surroundings. In other words, the total entropy change for process [1] is very positive, i.e. the process is very spontaneous, with a correspondingly high equilibrium constant, whereas that of process [2] will be much lower, giving rise to an equilibrium mixture.
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updated: 25 January 2006
