Question

\(\mathrm{KMnO}_{4}\) reacts with oxalic acid according to the equation \(2 \mathrm{KMnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{++}+10 \mathrm{CO}_{2}+\) \(8 \mathrm{H}_{2} \mathrm{O}\) Here \(20 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{KMnO}_{4}\) is equivalent to: (1) \(20 \mathrm{~mL}\) of \(0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (2) \(50 \mathrm{~mL}\) of \(0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (3) \(50 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (4) \(20 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)

Short answer

(3) 50 mL of 0.1 M H2C2O4

Step-by-step solution

- Assess the stoichiometry

Look at the balanced chemical equation: \[2 \mathrm{KMnO}_{4}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{++}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}\] This tells us that 2 moles of KMnO4 react with 5 moles of oxalic acid.

- Calculate the moles of KMnO4

Use the molarity equation, where molarity (M) = moles/volume in liters: Moles of KMnO4 = 0.1 M * 0.020 L = 0.002 moles.

- Determine the moles of oxalic acid required

From the stoichiometric ratio in the balanced equation, 2 moles of KMnO4 react with 5 moles of oxalic acid. Therefore: Moles of oxalic acid = \(\frac{5 \times \text{moles of KMnO}_{4}}{2}\). Using our earlier calculation: Moles of oxalic acid \(\approx\) \(\frac{5 \times 0.002}{2} = 0.005\) moles.

- Calculate the required volume of oxalic acid

Now use the molarity equation again for oxalic acid: Volume = \(\frac{\text{moles}}{\text{molarity}}\). For various given molarity options, volume can be calculated: For (1) - \(0.5 \text{M}\): \(Volume = \frac{0.005}{0.5} = 0.010\, L = 10\, mL\).For (2) - \(0.5 \text{M}\): Volume = \(\frac{0.005}{0.5} = 0.010 \text{ L} = 10 \text{ mL}\).For (3) - \(0.1 \text{M}\): Volume = \(\frac{0.005}{0.1} = 0.050\, L = 50\, mL\).For (4) - \(0.1 \text{M}\)Volume = \(\frac{0.005}{0.1} = 0.050 \text{ L} = 50 \text{ mL}\).

- Compare calculated volumes with options

From the calculations above, the closest matching option is (3) - 50 mL of 0.1 M oxalic acid.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the quantitative relationship between reactants and products in a chemical reaction. It is crucial for determining exact amounts needed for reactions. In our given exercise, for instance, we start by examining the balanced chemical equation:
\[2 \mathrm{KMnO}_{4}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{++}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}\]
This equation tells us that 2 moles of potassium permanganate (KMnO4) react with 5 moles of oxalic acid. This ratio—known as the stoichiometric ratio—allows us to scale up or down to any quantity of reactants and predict the respective amounts of products.

Redox Reactions

Redox reactions, or oxidation-reduction reactions, involve the transfer of electrons between two substances. One substance gets oxidized, losing electrons, and the other gets reduced, gaining electrons. In the given problem, KMnO4 is an oxidizing agent, and oxalic acid is a reducing agent.
KMnO4 gets reduced from a +7 oxidation state to a +2 oxidation state, forming Mn^2+. Meanwhile, oxalic acid (H2C2O4) gets oxidized, producing CO2. Understanding the flow of electrons helps in determining the proportions in the chemical equation and can aid in balancing it.

• Oxidation: \( \text{H}_{2} \text{C}_{2} \text{O}_{4} \rightarrow \text{CO}_{2} \)
• Reduction: \( \text{KMnO}_{4} \rightarrow \text{Mn}^{++} \)

Molarity Calculations

Molarity (M) is a measure of the concentration of a solute in a solution. It is defined as moles of solute per liter of solution. In the given exercise, we are dealing with the molarity of KMnO4 and oxalic acid. To solve the problem:

• First, calculate the moles of KMnO4 using its molarity and volume: \( \text{Moles of KMnO4} = \text{Molarity} \times \text{Volume in liters} \). For a 0.1 M solution of KMnO4 in 20 mL, the moles are: \( 0.1 \text{ M} \times 0.020 \text{ L} = 0.002 \text{ moles} \).
• Then, using the stoichiometry from the balanced equation, convert moles of KMnO4 to moles of oxalic acid.
• Finally, determine the required volume of oxalic acid solution with a given molarity. For example, if we need 0.005 moles of oxalic acid and have a 0.1 M solution: \( \text{Volume} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.005}{0.1} = 0.050 \text{ L} = 50 \text{ mL} \).

Chemical Equations

Chemical equations are symbolic representations of chemical reactions, where the reactants are written on the left and the products on the right, separated by an arrow. A balanced chemical equation has the same number of each type of atom on both sides.
For example, the given reaction: \(2 \mathrm{KMnO}_{4}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{++}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}\) tells us that 2 molecules of potassium permanganate react with 5 molecules of oxalic acid and 16 hydrogen ions to produce 2 manganese ions, 10 carbon dioxide molecules, and 8 water molecules.
Balancing equations ensures the conservation of mass and charge, which is fundamental in all chemical reactions.