Only \( 2 \, \text{mL} \) of \( \text{KMnO}_4 \) solution of unknown molarity is required to reach the end point of a titration of \( 20 \, \text{mL} \) of oxalic acid (\( 2 \, \text{M} \)) in acidic medium. The molarity of \( \text{KMnO}_4 \) solution should be ______ \( \text{M} \).
Correct Answer: 8
Approach Solution - 1
In an acidic medium, the reaction between potassium permanganate (\(\text{KMnO}_4\)) and oxalic acid (\(\text{H}_2\text{C}_2\text{O}_4\)) can be represented as:
\[ 2 \text{KMnO}_4 + 5 \text{H}_2\text{C}_2\text{O}_4 + 6 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2\text{O} \]
Using the Concept of Equivalents:
According to the principle of equivalents:
\[ \text{equivalents of KMnO}_4 = \text{equivalents of H}_2\text{C}_2\text{O}_4 \]
Calculate Equivalents for Each Solution:
For oxalic acid (\(\text{H}_2\text{C}_2\text{O}_4\)):
\[ \text{Molarity} \times \text{Volume} \times \text{n-factor} = 2 \times 20 \times 2 = 80 \, \text{meq} \]
where \(\text{n-factor} = 2\) for oxalic acid.
For \(\text{KMnO}_4\):
\[ M \times 2 \times 5 = 10M \, \text{meq} \]
where \(\text{n-factor} = 5\) for \(\text{KMnO}_4\).
Equating Equivalents:
\[ 10M = 80 \]
Solving for \(M\):
\[ M = \frac{80}{10} = 8 \, \text{M} \]
Conclusion:
The molarity of the \(\text{KMnO}_4\) solution is \(8 \, \text{M}\).
Approach Solution -2
- Volume of oxalic acid solution \( = 20 \, \text{mL} = 0.020 \, \text{L} \)
- Molarity of oxalic acid \( = 2 \, \text{M} \)
- Volume of \( \text{KMnO}_4 \) solution used \( = 2 \, \text{mL} = 0.002 \, \text{L} \)
Step 2: Reaction and Stoichiometry.
The balanced redox reaction in acidic medium is:
\[ 5 \, \text{H}_2\text{C}_2\text{O}_4 + 2 \, \text{KMnO}_4 + 6 \, \text{H}_2\text{SO}_4 \longrightarrow 10 \, \text{CO}_2 + 2 \, \text{MnSO}_4 + K_2\text{SO}_4 + 8 \, \text{H}_2\text{O} \] Mole ratio of oxalic acid to potassium permanganate is:
\[ 5 : 2 \]
Step 3: Calculate moles of oxalic acid.
\[ \text{Moles of oxalic acid} = 2 \times 0.020 = 0.04 \, \text{mol} \]
Step 4: Calculate moles of \( \text{KMnO}_4 \) required.
\[ \text{Moles of KMnO}_4 = \frac{2}{5} \times 0.04 = 0.016 \, \text{mol} \]
Step 5: Calculate molarity of \( \text{KMnO}_4 \) solution.
\[ M = \frac{0.016}{0.002} = 8 \, \text{M} \]
Final Answer:
\[ \boxed{8} \]
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