Question

A sample of 1.029g of KC8H5O4 was titrated with an NAOH solution of unknown concentration. If 22.83 mL of base was used to reach the endpoint, what was the concentration of NaOH? What mass of KC8H5O4 should be used such that a volume of NaOH between 35-40 mL is needed to reach the endpoint?

Please help, I procrastinated and its due tomorrow!!!

Answer 1

You probably intended this to be potassium hydrogen phthalate which I will rewrite, to make things simpler, as KHP. The H is the hydrogen neutralized by the NaOH.
NaOH + KHP ==> NaKP + H2O.
mols KHP = 1.029 grams/molar mass KHP.
mols NaOH = mols KHP.
mols NaOH = L x M. You know mols and you know L. M is the only unknown.

The second part of the problem is done by working backwards. Using the molarity of the NaOH calculated above, use mols = L x M and plug in , say 38 mL (0.038 L or about half way between 35 and 40). That allows you to calculate mols. Then use mols KHP and mols KHP = grams/molar mass. You know mols KHP and you know molar mass so calculate grams KHP.