Chapter 6, Problem 144IP

The vanadium in a sample of ore is converted to VO²⁺. The VO²⁺ ion is subsequently titrated with MnO₄⁻ in acidic solution to form V(OH)₄⁺ and manganese(II) ion. The unbalanced titration reaction is

$Mn{O}_4-\left(aq\right)+V{O}^{2+}+\left(aq\right)+H_{2}O\left(l\right)\to V{\left(OH\right)}_4{}^{+}\left(aq\right)+M{n}^{2+}\left(aq\right)+H^{+}\left(aq\right)$

To titrate the solution, 26.45 mL of 0.02250 M MnO₄⁻ was required. If the mass percent of vanadium in the ore was 58.1 %, what was the mass of the ore sample? Hint: Balance the titration reaction by the oxidation states method.