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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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mickey mickey
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11 years ago
15.0 mL of a 0.160 M vanadium solution is titrated with 20.0 mL of 0.024 M potassium permanganate to oxidise the vanadium solution to VO2+. Calculate the original oxidation state of the vanadium solution.
How do I do it? Please explain not just give the answer. Really appreciate any help.
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#1
11 years ago
The reaction are as follows:

MnO4- + 5e + 8H+  Rightwards Arrow Mn2+ + 4H2O
mole of MnO4- = 20 x 0.024 = 0.48 mmole
mole of electron = 0.48 x 5 = 2.4 mmole

V(x+) + H2O Rightwards Arrow VO2+ + 2H+ + ne
mole of V(x+) = 15 x 0.16 = 2.4 mmole
mole of electron = n x 2.4 mmole = 2.4n mmole

the mole of electron in the second reaction should be the same with the mole electron in the first reaction = 2.4 mmole

then the mathematic function become
2.4n = 2.4
n = 1
the second reaction become
V(3+) + H2O Rightwards Arrow VO2+ + 2H+ + 1e

Then the original oxidation state of vanadium is 3+
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