# Why can't a weaker acid protonate the conjugate base of a stronger acid?

So I've heard that, supposedly, a weaker acid cannot protonate the conjugate base of a stronger acid to give that stronger acid. For instance, adding ethanoic acid to sodium sulphate will not result in sulphuric acid and sodium ethanoate. Is this true? If so, why?

• This is not true. Acid strength depends on medium (the same acid can be weak in one solvent and strong in another), and of course you can chemically convert, say, $\ce{H2S}$ to $\ce{H2SO4}$. – andselisk Mar 10 at 9:22

Say you have a sodium chloride solution and you bubble in some carbon dioxide. The latter is only a weak acid, but it does generate some solvated hydrogen ions in solution and there are also chloride ions from the salt. These are all you need to claim that there is now hydrochloric acid in your solution! But there's a catch: you can claim only as much hydrochloric acid as the solvated hydrogen ions you generated, which is not a lot. You're overlooking the true major constituents if you emphasize this point of view.

Here's the moral of the story: You can make a strong acid from a weak one. But, the amount of strong acid you can capture at equilibrium will be limited. The amounts of the various acids you can identify must all be consistent with generating the same concentration of protons, and perforce that corresponds to less strong acid than weak acid.

I interpret the question to mean that given two hypothetical acids $$\ce{HWeak}$$ and $$\ce{HStrong}$$, why does the following reaction not occur: $$\ce{HWeak} \ + \ \ce{Strong-} \overset{?}{\rightarrow} \ce{Weak-} \ + \ \ce{HStrong}$$ ?

The answer lies in the definition of "strong and weak" acids. To recap: A Brønsted–Lowry acid is a substance capable of dissociating an $$\ce{H+}$$ ion, commonly called a proton. A strong acid is thus an acid that does so easily, or equivalently, whose conjugate anion has a very low affinity for protons. A weak acid has a conjugate anion that more readily accepts a proton.

The wording above already shows that acids are in principle always about equilibria: $$\ce{HAcid} \rightleftharpoons \ce{H+} \ + \ \ce{Acid-}$$ meaning that the reaction can occur in either direction. For an acid, the equilibrium lies on the right side of the equation and more strongly so for a strong acid.

This means that the equilibrium $$\ce{HWeak} \ + \ \ce{Strong-} \rightleftharpoons \ce{Weak-} \ + \ \ce{HStrong}$$ tends to the left side, but is of course subject to all factors affecting such an equilibrium. If you would continously remove the strong acid by some means or if you start with a strong acid concentration of zero, some reaction to the right will occur. However, if there is another base (i.e. acceptor of proton such as water), $$\ce{HStrong}$$ would probably pass the proton to it.

To summarize, the answer is: you can make some strong acid with a weak one, but not in large yields. Typically, the reaction will occur the other way around due to the way strong and weak acids are defined.