Why are the moles of acid= to the moles of base during the titration of a weak acid and strong base?

Question

Weak acids only partially ionise, so only a small percentage of the acid actually reacts with the fully ionised base so wouldnt there actually be more acid than your calculations give?

PLZ HELP ANY DETAIL IS APPRECIATED

Answer 1

It doesn't matter if the acid and/or base is strong or weak, the moles will be equal upon neutralization provided the stoichiometry is 1:1, i.e. a monprotic acid and a mono basic base, or diprotic acid and dibasic base.  The fact that a weak acid only ionizes partially, does not negate the fact that the base (strong or weak) will still neutralize it on a mole for mole basis.  Take the example of acetic acid (a weak acid) reacting with sodium hydroxide (a strong base).  The equation is

CH3COOH + NaOH ==> CH3COONa + H2O producing a salt (sodium acetate) and water.  The number of mole of NaOH needed for complete neutralization will be the same as the number of moles of acetic acid present.  It is important to note that the pH at the point of neutralization will not be 7 in this case, because you are creating a buffer during the titration (a weak acid + the salt of that acid).  The sodium acetate formed will hydrolyze and will be alkaline, or the pH will be greater than 7.