The vanadium in a sample of ore is converted to \(\mathrm{VO}^{2+}\). The \(\mathrm{VO}^{2+}\) ion is subsequently titrated with \(\mathrm{MnO}_{4}^{-}\) in acidic solution to form \(\mathrm{V}(\mathrm{OH})_{4}{ }^{+}\) and manganese(II) ion. The unbalanced titration reaction is \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(t) \longrightarrow\) \(\mathrm{V}(\mathrm{OH})_{4}^{+}(a q)+\mathrm{Mn}^{2+}(a q)+\mathrm{H}^{+}(a q)\) To titrate the solution, \(26.45 \mathrm{~mL}\) of \(0.02250 \mathrm{M} \mathrm{MnO}_{4}^{-}\) 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.

First, we need to assign oxidation numbers to each element in the unbalanced equation: \(\mathrm{MnO}_{4}^{-}(a q)(+7)+\mathrm{VO}^{2+}(a q)(+5)+\mathrm{H}_{2}\mathrm{O}(t)(0) \longrightarrow\) \(\mathrm{V}(\mathrm{OH})_{4}^{+}(a q)(+4)+\mathrm{Mn}^{2+}(a q)(+2)+\mathrm{H}^{+}(a q)(+1)\) Next, we will balance the equation: 1. Balance the vanadium atoms: There is one V atom on each side, so it's already balanced. 2. Balance the manganese atoms: There is one Mn atom on each side, so it's also already balanced. 3. Balance the oxygen atoms: Add 8 \(\mathrm{H}^{+}\) on the right side of the reaction. 4. Balance the hydrogen atoms: Add 4 \(\mathrm{H}_{2}\mathrm{O}\) on the left side of the reaction. 5. Balance the charges: Add 5 \(\mathrm{e}^{-}\) on the right side of the reaction. Balanced equation: \(\mathrm{MnO}_{4}^{-}+8\mathrm{H}^{+}+5\mathrm{e}^{-}+\mathrm{VO}^{2+}+4\mathrm{H_{2}O} \longrightarrow\) \(\mathrm{V}(\mathrm{OH})_{4}^{+}+\mathrm{Mn}^{2+}+8\mathrm{H}^{+}\)

We are given the volume and concentration of \(\mathrm{MnO}_{4}^{-}\) solution. We can use this information to calculate the moles of \(\mathrm{MnO}_{4}^{-}\). As the stoichiometry between \(\mathrm{MnO}_{4}^{-}\) and \(\mathrm{VO}^{2+}\) is 1:1 in the balanced equation, the moles of V in \(\mathrm{VO}^{2+}\) are equal to the moles of \(\mathrm{MnO}_{4}^{-}\). Moles of \(\mathrm{MnO}_{4}^{-}\) = Volume x Concentration Moles of \(\mathrm{MnO}_{4}^{-}\) = \(26.45\,\mathrm{mL} * 0.02250\,\mathrm{M}\) Moles of \(\mathrm{MnO}_{4}^{-}\) = 0.000595125 mol As the stoichiometry is 1:1, moles of \(\mathrm{VO}^{2+}\) = 0.000595125 mol

Now that we have the moles of V in \(\mathrm{VO}^{2+}\), we can calculate the mass of vanadium in the ore sample. To do this, we will first find the molar mass of vanadium in g/mol (refer to the periodic table for atomic weights). Molar mass of V = 50.94 g/mol Mass of vanadium = Moles of V x Molar mass of V Mass of vanadium = 0.000595125 mol x 50.94 g/mol Mass of vanadium = 0.030309225 g

We are given the mass percent of vanadium in the ore as \(58.1\%\). Using this information, we can find the mass of the ore sample. Mass of ore sample = (Mass of vanadium) ÷ (Mass percent of vanadium) Mass of ore sample = 0.030309225 g ÷ 0.581 Mass of ore sample = 0.052161 g The mass of the ore sample is approximately 0.0522 g.