How will overshooting the end point of an acid-base titration by adding too much base to KHP affect the calculated molarity of the base?

Question

I would think you would need to add more KHP, then more base to bring it to the endpoint. So the molarity calculated should be the same.
Is that correct?

Answer 1

Or if you didn't correct this error, the molarity of the base would be higher than it should be.
Is that correct?

Answer 2

Oops, I mean the molarity would decrease because moles base / larger quantity in L = a small # therefore decreased molarity.
Correct?

Answer 3

I find it helpful to write each step of the procedure and see how the question affects each.
1. mols KHP = grams/molar mass
2. mols NaOH = mols KHP
3. M NaOH = mols NaOH/L NaOH

So you did steps 1 and 2 right but on 3 added too much L NaOH. That's in the denominator, too big number means too small M.