Question

The vanadium in a sample of ore is converted to \(\mathrm{VO}^{2+}\). The 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 \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$ \mathrm{V}(\mathrm{OH})_{4}^{+}(a q)+\mathrm{Mn}^{2+}(a q)+\mathrm{H}^{+}(a q) $$ 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.

Short answer

The mass of the ore sample is approximately 0.2608 grams.

Step-by-step solution

Balance the titration reaction

The unbalanced reaction is: \[ \mathrm{MnO}_{4}^{-}(a q)+\mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{V}(\mathrm{OH})_{4}^{+}(a q)+\mathrm{Mn}^{2+}(a q)+\mathrm{H}^{+}(a q) \] Using the oxidation states method, we find the oxidation numbers for each element: Mn: \(+7\) in MnO4-, \(+2\) in Mn2+ V: \(+4\) in VO2+, \(+5\) in V(OH)4+ The change in the oxidation numbers are: Mn: \(+7 \rightarrow +2\), reduction, change = 5 V: \(+4 \rightarrow +5\), oxidation, change = 1 Multiplying the oxidation half-reaction by 5 and adding both half-reactions, we get the balanced reaction: \[ 5\mathrm{VO}^{2+}(a q) + \mathrm{MnO}_{4}^{-}(a q) + 8\mathrm{H}^{+}(a q) \longrightarrow 5\mathrm{V}(\mathrm{OH})_{4}^{+}(a q) + \mathrm{Mn}^{2+}(a q) + 4\mathrm{H}_{2} \mathrm{O}(l) \]

Find the mole ratio between MnO4- and VO2+

From the balanced reaction, the mole ratio between MnO4- and VO2+ is 1:5. This means that 1 mole of MnO4- reacts with 5 moles of VO2+.

Calculate the moles of MnO4- used in the titration

The volume of MnO4- solution required is 26.45 mL and the concentration is 0.02250 M. To find the moles of MnO4-, we use the formula: Moles of MnO4- = volume × concentration Moles of MnO4- = (26.45 × 10^{-3} L) × 0.02250 M = 5.950 × 10^{-4} mol

Calculate the moles of vanadium in the sample

Using the mole ratio between MnO4- and VO2+ found in Step 2 (1:5), we can find the moles of vanadium in the sample: Moles of vanadium = (5.950 × 10^{-4} mol) × (5 mol vanadium/mol MnO4-) Moles of vanadium = 2.975 × 10^{-3} mol

Calculate the mass of vanadium in the sample

To find the mass of vanadium, use the formula: Mass = moles × molecular weight The molecular weight of vanadium is 50.94 g/mol. Thus, the mass of vanadium in the sample is: Mass of vanadium = (2.975 × 10^{-3} mol) × (50.94 g/mol) = 0.1516 g

Calculate the mass of the ore sample

Given that the mass percent of vanadium in the ore sample is 58.1%, we can find the mass of the ore sample using the formula: Mass of ore sample = (mass of vanadium) / (mass percent of vanadium) Mass of ore sample = (0.1516 g) / (0.581) = 0.2608 g So, the mass of the ore sample is approximately 0.2608 grams.

Key concepts

Vanadium Ore Analysis

When working with minerals and mining resources, it's vital to accurately determine the content and purity of the ore, like those rich in vanadium. In particular, vanadium ore analysis is essential for industries that manufacture steel, as vanadium is often used to improve the strength and toughness of the metal. A common analysis method is redox titration, which is a type of titration used to determine the concentration of an unknown sample through a redox reaction. In this type of analysis, a solution containing a known concentration and volume of titrant reacts with a substance in the sample leading to a change in oxidation state, which signifies the point at which equivalent quantities have reacted.

The exercise provided involved converting the vanadium in the ore sample to VO2+ ions. These ions were then titrated with MnO4- in an acidic solution, changing the oxidation state of the vanadium. The end result was the determination of the vanadium content in the sample based on the volume of titrant used. Accurate vanadium ore analysis via titration reveals vital economic information and ensures that subsequent usage of the ore meets the required specifications and standards for industrial processes.

Oxidation States Method

In redox chemistry, understanding the oxidation states method is crucial for balancing redox reactions. This technique revolves around the concept of oxidation states which signify the hypothetical charges that an atom would have if all its bonds were ionic. In a redox reaction, the total increase and decrease in oxidation state must balance out because the number of electrons lost must equal the number of electrons gained.

By comparing oxidation states before and after the reaction, you can determine which elements have been oxidized (increase in oxidation state) and which have been reduced (decrease in oxidation state). In the given exercise, determining the change in oxidation states for manganese and vanadium helped in identifying how many electrons were transferred in the reaction. This information is critical for balancing the equation, as it ensures mass and charge are conserved. The application of the oxidation states method in balancing equations is a fundamental skill for students as it aids in understanding the underlying chemical processes and provides the basis for calculations in stoichiometry.

Stoichiometry

The core of chemical calculations is stoichiometry, a segment of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In the context of the exercise, stoichiometry is used to calculate the amount of ore based on the redox titration results. Students should be familiar with concepts such as the mole ratio, which comes from the balanced chemical reaction and guides the calculation of how much vanadium is present.

Knowing the mole ratio allows you to understand the proportions in which reactants combine and products form. In this case, the mole ratio between MnO4- and VO2+ told us that one mole of MnO4- titrant reacts with five moles of VO2+ vanadium compound. With the stoichiometry of the balanced reaction, the volume and concentration of the MnO4- solution are used to calculate the moles of vanadium present, and subsequently, the mass of the vanadium and the ore. The step-by-step approach provided illustrates the methodical nature of stoichiometric calculations, ensuring a solid grasp of the procedure for determining quantities involved in chemical reactions.