A 250 mL sample of 0723 M HClO4 is titrated with a 0273M KOH

A 25.0 mL sample of 0.723 M HClO₄ is titrated with a 0.273M KOH solution. What is the [H⁺] (molarity) before any base is added?

Solution

0.025Litres @ 0.723 mol/litre = 0.018075 moles H+

each mole of KOH releases 1 mole OH- ion
0.010 litres @ 0.273 mol/litre = 0.00273 moles OH-

each mole of OH- ion destroys 1 mole of H+ ion
0.018075 - 0.00273 = 0.015345 moles of H+ remain

the moles of H+ are now in 35 millilitres of solution
0.015345 moles / 0.035 litres =

your answer: [H+] = {H3O+] = 0.438 Molar