posted by richard .
A 25.00-mL sample of 0.723 M HClO4 is titrated with a 0.273 M KOH solution. The H3O+ concentration after
the addition of 10.0 mL of KOH is ???? M.
Start with an equation
HClO4 -> H+ + ClO4-
so one mole of HClO4 gives one mole of H+ or H3O+
and one mole of HClO4 reacts with one mole of KOH
25.00 x 0.723 /1000 moles
moles of KOH added
10.0 x 0.273 /1000 moles
so moles of HClO4 left =
(25.00 x 0.723 /1000 moles)-(10.0 x 0.273 /1000 moles)
which is also the number of moles of H+
new volume = (25.0+10.0) ml = 35.0 ml
hence calculate the new concentration