A 25.00-mL sample of 0.723 M HClO4 is titrated with a 0.273 M KOH solution. The H3O+ concentration after

Question

the addition of 10.0 mL of KOH is ???? M.

Answer 1

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

Answer 2

.43914