Calculate mass of potassium hydroxide (KOH) available in 25.0 ml solution given that the titration endpoint is 5.20 ml of 0.2 N hydrochloric acid (HCl) solution.

General information:

During titration, the strong acid (HCl) reacts with the strong base (KOH) as the following:

HCl + KOH → KCl + H₂O

Accordingly, each mole (equivalence) of  KOH is neutralized with 1 mole (equivalence) of the acid. Therefore,  the number of moles of HCl used to reach the endpoint (equivalence point) is equal to the number of moles of KOH available in the solution. Molarity of HCl solution is the same as its normality because it has one equivalence for each mole.

Molecular weight (molar mass) of KOH = 56.11 g/ mole

Calculations:
Endpoint volume in liters = 5.2 ml × 0.001 (liter/ ml) = 0.0052 liter
Number of moles of HCl used to reach end point = 0.2 (mole/ liter) × 0.0052 liter = 0.00104 mole
Number of moles of KOH available in solution = 0.00104 mole
Mass of KOH available in solution =  0.00104 mole × 56.11 (g/ mole) = 0.0584 g

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