pH curves concentration help

I don't quite get this question because if the conc is halved then it is two times less effective, so surely you will need twice more of the base to reach the equivalence point. I don't get why two times less base is needed if you have a lower conc of base.

Thanks

Can't fulyl answer the question but doesn't this have to do with a balanced equation?

The concentration of the acid is being halved, not the alkali.

It is: CH3COOH + NaOH -> CH3COO-Na+ + H2O

But is doesn't matter what the stoichiometry is, the important thing is that if it requires X mol of acid to neutralise Y mol of alkali, it would require Y/2 mol of alkali to neutralise X/2 mol of acid.

ASSUMING that when the concentration of acid were halved, the volume wasn't also changed.

The concentration of the acid is being halved, not the alkali.

Initial pH will be (a bit) higher
AND
Vertical bit on the graph will happen when half the volume of NaOH is added (compared to original graph).

your homework = done

(or at least this bit)