# How do I know that an acid and conjugate base are present in equimolar amounts?

Acetylsalicylic acid is a molecule in several popular drugs such as Magnyl and aspirin. It has a $$\mathrm{p}K_\mathrm{a}$$ of 3.5.

I'm about to calculate the $$\mathrm{pH}$$ of a $$\pu{0.010 M}$$ solution of acetylsalicylic acid.

I've read up on how to do this and in one video the author talks about "equimolar amounts". He says that now that we have equimolar amounts, we can write: $$\ce{\frac{[H+][A-]}{[HA]}} = \ce{\frac{[H+]^2}{[HA]}}$$

Which makes sense as there are equally many $$\ce{H}$$ and $$\ce{A}$$ ions. However when can I assume that an acid disassociated equally? Also, I thought that when it did this then it was called a strong acid? Why would I be using this formula at all for a strong acid?

If I can assume that acetylsalicylic acid disassociates into equal acid and base ions, then I can calculate my problem, but I'm skeptical as to assuming this is true.

• Think about the dissociation reaction. How could there be differing amounts of proton and conjugate base? With respect to strong vs. weak acid, a strong acid is one that dissociates completely. Acetylsalicylic acid is a weak acid, but that doesn't change the reaction stoichiometry, just where the equilibrium lies. – jerepierre Feb 23 '15 at 17:48
• Do you mean that we can always say that there are equally many protons and conjugate bases? Why would the formula exist as [H][A]/[HA] at all and not just as [H]^2/[HA]? – Paze Feb 23 '15 at 17:49
• Because you could add [A] from a different source, i.e. a salt. If you're considering that [H] and [A] can only come from the acid, then they must be equal. – jerepierre Feb 23 '15 at 17:51
• Oh, that makes sense. You should write this as an answer so I can choose best answer. – Paze Feb 23 '15 at 17:53

I'm really confused here!

No, you are not! You only (temporarily) mixed up two things, both of with you have already understood separately.

$$\ce{HA + H2O <=> H3O+ + A-}$$

1. Position of the equilibrium

This is described by the equilibrium constant $K$, which is the quotient of the rate constants for the forward and the back reaction: $K = \frac{k_{forward}}{k_{back}}$. For strong acids, of which you know that they completely dissociate, the equilibrium is on the product side.

2. Equimolarity

There's nothing out of nothing! If one molecule of $\ce{HA}$ dissociates is water, there exactly one molecule $\ce{H3O+}$ and exactly one molecule of $\ce{A-}$ generated. Not more, not less.

Q: How could another molecule of $\ce{A-}$ appear out of nothing?

A: It can't and it doesn't happen - not even in the LEGO universe. It is just a proton transfer reaction from $\ce{HA}$ to $\ce{H2O}$; the overall number of atoms does not change!

If is however possible that some $\ce{A-}$ may come from a different source, just by adding a salt $\ce{MA}$ of the conjugate base (think sodium acetylsalicylate) to the solution of your acid. jerepierre has mentioned this in his comment! But for the moment, this is another story.