Question:
In an analytical laboratory, a $20.0\text{ mL}$ sample of an aqueous solution containing oxalic acid ($\text{H}_2\text{C}_2\text{O}_4$) requires exactly $16.0\text{ mL}$ of a $0.05\text{ M}$ potassium permanganate ($\text{KMnO}_4$) solution for complete oxidation in a hot, acidic medium ($\text{H}_2\text{SO}_4$). Calculate the molarity of the oxalic acid solution.
In an analytical laboratory, a $20.0\text{ mL}$ sample of an aqueous solution containing oxalic acid ($\text{H}_2\text{C}_2\text{O}_4$) requires exactly $16.0\text{ mL}$ of a $0.05\text{ M}$ potassium permanganate ($\text{KMnO}_4$) solution for complete oxidation in a hot, acidic medium ($\text{H}_2\text{SO}_4$). Calculate the molarity of the oxalic acid solution.
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Using normality factor ratios saves you from ever having to write out or balance long redox reactions during multiple-choice tests! Just keep these two standard acidic medium n-factors memorized as an invariant pair: Permanganate is always 5, Oxalate is always 2.
Updated On: May 29, 2026
- \( 0.010\text{ M} \)
- \( 0.040\text{ M} \)
- \( 0.100\text{ M} \)
- \( 0.250\text{ M} \)
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The Correct Option is C
Solution and Explanation
Concept:
This question can be solved using the Law of Chemical Equivalents, which dictates that at the stoichiometric endpoint of any reaction, the total gram equivalents of the oxidizing agent must equal the total gram equivalents of the reducing agent:
\[
\text{Equivalents of KMnO}_4 = \text{Equivalents of H}_2\text{C}_2\text{O}_4
\]
Using the relationship between Normality (\(N\)), Molarity (\(M\)), and the valence factor (\(n\)-factor), where \(N = M \times n\):
\[
M_1 \cdot n_1 \cdot V_1 = M_2 \cdot n_2 \cdot V_2
\]
Step 1: Determining the n-factors for both redox species in an acidic medium.
• For \(\text{KMnO}_4\) (Oxidizing Agent): In an acidic solution, the permanganate ion reduces from a \(+7\) state down to a \(+2\) state (\(\text{MnO}_4^- \to \text{Mn}^{2+}\)), involving a transfer of 5 electrons: \[ n_{\text{KMnO}_4} = 5 \]
• For \(\text{H}_2\text{C}_2\text{O}_4\) (Reducing Agent): The carbon atoms in the oxalate ion oxidize from a \(+3\) state up to a \(+4\) state inside carbon dioxide (\(\text{C}_2\text{O}_4^{2-} \to 2\text{CO}_2 + 2e^-\)). Since there are 2 carbon atoms performing this shift per molecule, the total electron release is \(2 \times 1 = 2\): \[ n_{\text{H}_2\text{C}_2\text{O}_4} = 2 \]
Step 2: Plugging the parameters into the equivalents equation.
Let \(M_x\) represent the unknown molarity of the oxalic acid solution. Setting up our equality: \[ (M \times n \times V)_{\text{KMnO}_4} = (M \times n \times V)_{\text{H}_2\text{C}_2\text{O}_4} \] \[ 0.05\text{ M} \times 5 \times 16.0\text{ mL} = M_x \times 2 \times 20.0\text{ mL} \]
Step 3: Isolating and computing $M_x$.
\[ 0.25 \times 16.0 = M_x \times 40.0 \] Since \(0.25 = \frac{1}{4}\), the left side simplifies cleanly: \[ \frac{1}{4} \times 16.0 = 4.0 \quad \implies \quad 4.0 = 40.0 \cdot M_x \] \[ M_x = \frac{4.0}{40.0} = 0.100\text{ M} \] Thus, the molar concentration of the oxalic acid is \(0.100\text{ M}\).
Step 1: Determining the n-factors for both redox species in an acidic medium.
• For \(\text{KMnO}_4\) (Oxidizing Agent): In an acidic solution, the permanganate ion reduces from a \(+7\) state down to a \(+2\) state (\(\text{MnO}_4^- \to \text{Mn}^{2+}\)), involving a transfer of 5 electrons: \[ n_{\text{KMnO}_4} = 5 \]
• For \(\text{H}_2\text{C}_2\text{O}_4\) (Reducing Agent): The carbon atoms in the oxalate ion oxidize from a \(+3\) state up to a \(+4\) state inside carbon dioxide (\(\text{C}_2\text{O}_4^{2-} \to 2\text{CO}_2 + 2e^-\)). Since there are 2 carbon atoms performing this shift per molecule, the total electron release is \(2 \times 1 = 2\): \[ n_{\text{H}_2\text{C}_2\text{O}_4} = 2 \]
Step 2: Plugging the parameters into the equivalents equation.
Let \(M_x\) represent the unknown molarity of the oxalic acid solution. Setting up our equality: \[ (M \times n \times V)_{\text{KMnO}_4} = (M \times n \times V)_{\text{H}_2\text{C}_2\text{O}_4} \] \[ 0.05\text{ M} \times 5 \times 16.0\text{ mL} = M_x \times 2 \times 20.0\text{ mL} \]
Step 3: Isolating and computing $M_x$.
\[ 0.25 \times 16.0 = M_x \times 40.0 \] Since \(0.25 = \frac{1}{4}\), the left side simplifies cleanly: \[ \frac{1}{4} \times 16.0 = 4.0 \quad \implies \quad 4.0 = 40.0 \cdot M_x \] \[ M_x = \frac{4.0}{40.0} = 0.100\text{ M} \] Thus, the molar concentration of the oxalic acid is \(0.100\text{ M}\).
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