posted by Kerapetsi .
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?
(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.