How do you solve this chemistry problem?

Question

A quantity of 18.68 mL of KOH solution is needed to neutralize 0.4218g KHP. What is the concentration (in molarity) of the KOH solution?

Answer 1

to neutralize in this context means that at endpoint ( color change of indicator ) which is an estimate of the true equivalence point, that the mmles of base = mmoles of the KHPH

mmoles =( mL ) ( M) and mmoles = wt/mMW

so VolKOH ( MKOH) = 0.4218grams/0.2042 grams/mmole

M KOH = 0.4218grams/( 0.2042grams/mmol)(18.68mL )

note the gramscancel and the amswer has units of mmole/mL = M

sooo M KOH = 0.1106

Proof 18.68 mL x 0.1106 M = 2.066 mmoles

         0.4218 grams of KHPH / 0.2042grams/mmole= 2.0656 mmoles

Answer 2

I guess KHP should be potassium hydrogen phthalate (M= 204.22 g/mol)

Reaction is:

KOH + KHP -> K2P + H2O

=> c*V=m/M=> c= m/(M*V)

Numerics:

c= 0.4217 g/(204.22g/mol*18.68 e-3 L)= 0.11053 mol/L