Problem 67 - General Chemistry: Principles and Modern Applications

Question

The titration of \(5.00 \mathrm{mL}\) of a saturated solution of sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) at \(25^{\circ} \mathrm{C}\) requires \(25.8 \mathrm{mL}\) of \(0.02140 \mathrm{M} \mathrm{KMnO}_{4}\) in acidic solution. What mass of \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in grams would be present in \(1.00 \mathrm{L}\) of this saturated solution? \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\mathrm{MnO}_{4}^{-} \longrightarrow_{\mathrm{Mn}^{2+}}+\mathrm{CO}_{2}(\mathrm{g}) \quad\) (not balanced)

Step-by-Step Solution

Answer

The mass of sodium oxalate in a liter of this saturated solution is 36.992 g.

Step 1: Balance the Equation

The balanced redox reaction equation is:\[5\, \mathrm{C}_{2} \mathrm{O}_{4}^{2-} + 2\, \mathrm{MnO}_{4}^{-} + 16\, \mathrm{H}^{+} \longrightarrow 10\, \mathrm{CO}_{2} + 2\, \mathrm{Mn}^{2+} + 8\, \mathrm{H}_{2} \mathrm{O}\]

Step 2: Calculate moles of \(\mathrm{KMnO}_{4}\)

From the balanced equation we observe that 5 moles of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) react with 2 moles of \(\mathrm{MnO}_{4}^{-}\). Now, using the Molarity (M) of \(\mathrm{KMnO}_{4}\), which is moles/L, we can find the number of moles of \(\mathrm{KMnO}_{4}\) used:\[ n(\mathrm{KMnO}_{4}) = M(\mathrm{KMnO}_{4}) \times V(\mathrm{KMnO}_{4}) = 0.02140\, \mathrm{M} \times 25.8 \, \mathrm{mL} \times (1\, \mathrm{L}/1000\, \mathrm{mL}) = 0.00055242 \, \mathrm{moles}\]

Step 3: Calculate moles of \(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4}\)

From stoichiometry, we know that the moles of sodium oxalate is given by \( n(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4}) = n(\mathrm{MnO}_{4}) \times (5/2) = 0.00055242\, \mathrm{moles} \times (5/2) = 0.00138105\, \mathrm{moles}\)

Step 4: Find Concentration of \(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4}\)

The molarity of sodium oxalate can be determined using the equation \(M = n/V\), thus:\[ M(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4}) = n(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4})/V(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4}) = 0.00138105 \, \mathrm{moles} / 5.00 \, \mathrm{mL} \times (1\, \mathrm{L}/1000\, \mathrm{mL}) = 0.27621\, \mathrm{M}\]

Step 5: Find Mass of \(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4}\)

From the concentration, the mass of sodium oxalate in 1 L of solution can be found. First, we need to convert the molarity to moles and then to grams using the molar mass (Mm) of sodium oxalate, which is 133.9983 g/mol:\[ Mass = M \times V \times Mm = 0.27621 \, \mathrm{M} \times 1.00 \, \mathrm{L} \times 133.9983 \, \mathrm{g/mol} = 36.992 g\] Note that the volume used is 1L, which is the volume of solution asked in the problem.

Key Concepts

Redox Reaction

In a redox reaction, electrons are transferred between two substances. The substance that donates electrons is oxidized, while the one that accepts electrons is reduced. In the titration provided, the oxalate ion (\(\mathrm{C_{2}O_{4}^{2-}}\)) donates electrons and gets oxidized, whereas the permanganate ion (\(\mathrm{MnO_{4}^{-}}\)) accepts electrons and is reduced. This transfer of electrons results in the conversion of \(\mathrm{C_{2}O_{4}^{2-}}\) to \(\mathrm{CO_{2}}\) gas, and \(\mathrm{MnO_{4}^{-}}\) to \(\mathrm{Mn^{2+}}\) ions. Redox reactions are vital in understanding numerous chemical processes such as energy production. These reactions are characterized by changes in oxidation states, highlighting the importance of keeping track of electron flow.

Stoichiometry

Stoichiometry involves calculating the relative quantities of reactants and products involved in a chemical reaction. It is based on the law of conservation of mass, where the amount of each element in the reactants equals that in the products. In this exercise, stoichiometry is used in the balanced chemical equation \(5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} + 2 \mathrm{MnO}_{4}^{-}\), which indicates that 5 moles of oxalate react with 2 moles of permanganate. From stoichiometry, after finding moles of \(\mathrm{KMnO}_{4}\), it relates directly to determine moles of \(\mathrm{Na}_{2}\mathrm{C}_{2} \mathrm{O}_{4}\) through a ratio of \(\frac{5}{2}\). The ability to convert between moles using molar ratios is essential to determine the amounts required for reactions.

Molarity Calculation

Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution (\(\mathrm{M} = \text{moles/L}\)). Calculating molarity gives an idea of how much solute is present in the solution. In this exercise, we calculate the molarity of sodium oxalate in the initial 5.00 mL sample, determined as \(0.27621\, \mathrm{M}\).To find it, use the formula \(M = \frac{n}{V}\), where \(n\) is moles of solute and \(V\) is volume in liters. With this data, you can deduce further calculations or conversions needed to solve related problems.

Chemical Equations Balancing

Balancing chemical equations is crucial for accurately representing chemical reactions. It involves making sure the number of atoms of each element is the same on both sides of the equation, which preserves mass adherent to the law of conservation of mass. For the initial unbalanced equation, \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-} + \mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+} + \mathrm{CO}_{2}+\ldots\), we end up with the balanced equation given in Step 1:\[5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} + 2 \mathrm{MnO}_{4}^{-} + 16 \mathrm{H}^{+} \rightarrow 10 \mathrm{CO}_{2} + 2 \mathrm{Mn}^{2+} + 8 \mathrm{H}_{2} \mathrm{O}\]Prediction and understanding of this balancing provide accuracy not only in academic scenarios but also in lab settings where exact ratios are required for experiments.