Asked 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.
the addition of 10.0 mL of KOH is ???? M.
Answers
Answered by
Dr Russ
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
starting moles
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
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
starting moles
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
Answered by
jim
.43914
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