Question

55.2 mL of 0.976 M perchloric acid is added to 45.9 mL of potassium hydroxide, and the resulting solution is found to be acidic. 18.5 mL of 2.08 M sodium hydroxide is required to reach neutrality. What is the molarity of the original potassium hydroxide solution?

Answer 1

The reaction:

HClO4 + KOH --> H2O + KClO4

acidic...

NaOH is added ... HClO4 is remaining

HClO4 + NaOH = NaClO4 + H2O

find:

moles of NaOH used = MV = 2.08*18.5 = 38.48 mmol of NaOH used in excess HClO4 neutralization

mmol o fHClO4 used (total initially) = MV = 0.976 *55.2 = 53.8752 mmol of HClO4 initially

then

mmol of HClO4 used in KOH = 53.8752-38.48 = 15.3952 mmol of HClO4

now... this was used in KOH

mmol of KOH = 15.3952 mmol

[KOH] = mmol/mL = 15.3952 /45.9

[KOH] = 0.335407 M