# Does the amount of acid change when I add water to vinegar?

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?

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