I have a question that asks me to describe why adding distilled water to my vinegar sample would not affect my calculated Molarity (determining amount of acid in the vinegar be titration with a base).

At first thought, it wouldn't, because the same amount of acid is still there, we'll still have the same amount of base reacting. So you'll still get the same amount by measuring how much base was needed.

But that got me thinking. There is a lot of diluting acids with water, but isn't water amphiprotic? Why doesn't it react with the acid, or react with it enough to matter at least? What information am I missing here?


1 Answer 1


First off, your thinking about the titration volume of a dilute solution vs the concentrate from which it was made is correct.

Why we don't need to worry about water acting as an acid or a base in this context: Consider the reaction that occurs between a weak acid, HA, and water (a very weak base.) $\ce{HA + H2O <-> A- + H3O+}$

The products of the reaction are a weak base and a strong acid (hydronium, the strongest acid that can exist in aqueous solution.) If I add a base to this mixture, it doesn't matter whether that base reacts with an undissociated HA molecule or with a hydronium ion, in either case it has removed a particle of one acidic species from the solution, leaving behind A-, the conjugate base of HA. The total number of such events that has to occur before the titration is done is a fixed value, regardless of how many of the events involved hydronium or HA.

  • $\begingroup$ That's interesting. Glad I thought to ask. Thanks! $\endgroup$ Commented Nov 2, 2014 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.