Question

When 10 mL of an aqueous solution of \(Fe^{2+}\) ions was titrated in the presence of dil \(H_2SO_4\) using diphenylamine indicator, 15 mL of 0.02 M solution of \(K_2Cr_2O_7\) was required to get the end point. The molarity of the solution containing \(Fe^{2+}\) ions is \(x \times 10^{-2}\) M. The value of \(x\) is ________. (Nearest integer)

Hint:

In acidic medium, the n-factor of \(K_2Cr_2O_7\) is always 6 and \(KMnO_4\) is 5. Knowing these standard values speeds up titration calculations significantly.

Answer 1

Correct Answer: 18

Step 1: Understanding the Concept:
Titration between \(Fe^{2+}\) and \(K_2Cr_2O_7\) is a redox titration where \(Fe^{2+}\) is oxidized to \(Fe^{3+}\) and \(Cr_2O_7^{2-}\) is reduced to \(Cr^{3+}\) in acidic medium.
At the equivalence point, the number of equivalents of the reducing agent (\(Fe^{2+}\)) is equal to the number of equivalents of the oxidizing agent (\(K_2Cr_2O_7\)).
Step 2: Key Formula or Approach:
Equivalents of Reducing Agent = Equivalents of Oxidizing Agent
\[ (M_1 \times n_1) \times V_1 = (M_2 \times n_2) \times V_2 \]
Where \(n\) is the n-factor (change in oxidation state per molecule).
Step 3: Detailed Explanation:
1. Determine n-factors:
For \(Fe^{2+} \rightarrow Fe^{3+} + e^{-}\), n-factor (\(n_1\)) = 1.
For \(Cr_2O_7^{2-} \rightarrow 2Cr^{3+}\), the oxidation state of Cr changes from \(+6\) to \(+3\).
Since there are 2 Cr atoms in \(K_2Cr_2O_7\), n-factor (\(n_2\)) = \(2 \times (6 - 3) = 6\).

2. Calculate Molarity of \(Fe^{2+}\):
Given: \(V_1 = 10 \text{ mL}\), \(V_2 = 15 \text{ mL}\), \(M_2 = 0.02 \text{ M}\).
\[ M_1 \times 1 \times 10 = 0.02 \times 6 \times 15 \]
\[ 10 \times M_1 = 1.8 \]
\[ M_1 = 0.18 \text{ M} \]

3. Convert to requested format:
The molarity is \(x \times 10^{-2}\) M.
\[ 0.18 = 18 \times 10^{-2} \text{ M} \]
So, \(x = 18\).
Step 4: Final Answer:
The value of \(x\) is 18.