If the endpoint in the titration of KHC8H4O4 solution with the NaOH solution is mistakenly surpassed, will the molar concentration of NaOH solution be reported too high or too low?

Answer 1

(1) mols KHC8H4O4 = grams/molar mass
(2)mols NaOH = mols KHC8H4O4 (since the mols ratio is 1:1 in the titration. Write and balance the equation to confirm that.)
(3) M NaOH = mols NaOH/L NaOH.

So you over titrate with the base. No mL are involved in equation 1 above. No change. No mL involved in equation 2. No change. But mL NaOH are involved in equation 3, that makes a larger denominator which make the answer smaller than it should be.