The molarity of an aqueous solution of potassium hydroxide, KOH, is determined by titration against a 0.197 M HI solution. If 28.4 mL of the base are required to neutralize 18.6 mL of HI, what is the molarity of the KOH solution?

Answer 1

Assuming HI + KOH = KI + H2O,
one mole of KOH is needed to neutralize ach mole of HI

18.6ml of .197M HI contains .00366 moles of HI
So, we need .00366 moles of KOH in 28.4mL of solution.

.0284x = .00366
x = .129, so the KOH is .129M