Question

$\text{KMnO}_4$ oxidizes $\text{M}^{2+}$ to $\text{M}^{4+}$ in acid medium. $500\text{ mL}$ of $0.02\text{M}$ $\text{M}^{2+}$ solution requires $500\text{ mL}$ of $x\text{ M}$ $\text{MnO}_4^-$ solution for complete oxidation. The value of $x$ is:}

• A. $16 \times 10^{-3}$
• B. $8 \times 10^{-3}$
• C. $4 \times 10^{-3}$
• D. $2 \times 10^{-3}$

Hint:

For acidic $\text{KMnO}_4$ titrations, always remember that the $n$-factor of $\text{MnO}_4^-$ is $5$. Most numerical questions can then be solved directly using the equivalence relation $M_1V_1n_1=M_2V_2n_2$.

Answer 1

The Correct Option is B

Concept: In a redox titration, the total number of gram-equivalents of the reducing agent consumed is equal to the total number of gram-equivalents of the oxidizing agent consumed. Thus, \[ \text{Equivalents of Reducing Agent} = \text{Equivalents of Oxidizing Agent} \] Using the relation \[ \text{Equivalents}=M\times V\times n \] we obtain \[ M_1V_1n_1=M_2V_2n_2 \] where $n$ denotes the $n$-factor.

Step 1: Determine the $n$-factor of $\text{M^{2+}$} The metal ion is oxidized according to \[ \text{M}^{2+}\rightarrow \text{M}^{4+}+2e^- \] Since two electrons are lost, \[ n_1=2 \]

Step 2: Determine the $n$-factor of permanganate ion In acidic medium, \[ \text{MnO}_4^- + 8H^+ +5e^- \rightarrow \text{Mn}^{2+}+4H_2O \] One mole of permanganate gains five electrons. Therefore, \[ n_2=5 \]

Step 3: Apply the equivalence relation Given: \[ M_1=0.02\,\text{M} \] \[ V_1=500\,\text{mL} \] \[ M_2=x\,\text{M} \] \[ V_2=500\,\text{mL} \] Substituting into \[ M_1V_1n_1=M_2V_2n_2 \] gives \[ 0.02\times 500\times 2 = x\times 500\times 5 \] The common factor $500$ cancels: \[ 0.02\times 2=5x \] \[ 0.04=5x \] \[ x=\frac{0.04}{5} \] \[ x=0.008\,\text{M} \] \[ x=8\times 10^{-3}\,\text{M} \] Hence, the required molarity of $\text{MnO}_4^-$ solution is \[ \boxed{8\times 10^{-3}\,\text{M}} \]