The vanadium in an ore sample is converted to VO2 The VO2io

: The vanadium in an ore sample is converted to VO²⁺. The VO²⁺ions produced from the sample are then titrated with MnO₄^-in acidic solution to form V(OH)₄⁺ and manganese(II) ion.

Write the balanced redox half-cells and final equation for the reaction occurring in the titration.

To titrate the solution, 26.45 mL of 0.02250 M MnO₄^-was required. How many moles of vanadium were present in the solution?

If the original sample was 58.1% V by mass, what was the mass of the ore sample?

Solution

VO2+ + 3H2O => V(OH)4+ + 2H+ + e- -1.00 V

MnO4− + 8H+ + 5e− => Mn2+ + 4H2O +1.51 V

5VO2+ + 15H2O => 5V(OH)4+ + 10H+ + 5e- X5

adding and canceling electrons

5VO2+ + 15H2O + MnO4− + 8H+ => 5V(OH)4+ + 10H+ Mn2+ + 4H2O

eliminating the common ions and water

5VO2+ + 11H2O + MnO4- => 5 V(OH)4+ + 2H+ + Mn2+

1 mole of MnO4- reacts with 5 moles of VO2+

moles of MnO4- at the eq point = 0.0225 * 26.45 / 10000 = 0.000595125

moles of VO2+ at the eq point = 5 X 0.000595125
= 0.00297 moles of VO2+ in solution
moles of V in solution = = 0.00297
0.00297 x 50.94 g of V = 0.151g

if this is 58.1 % of the sample the ore sample weighed 0.151 X 100 / 58.1 =0.260g