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

Step 1: Use the equivalence principle.
At the endpoint of a redox titration the equivalents of oxidant equal the equivalents of reductant, written as \[ M_1 V_1 n_1 = M_2 V_2 n_2 \] where $n$ is the number of electrons exchanged.
Step 2: Find the n-factor of the metal ion.
Since $\text{M}^{2+} \rightarrow \text{M}^{4+} + 2e^-$, the metal loses two electrons, so $n_1 = 2$.
Step 3: Find the n-factor of permanganate.
In acid, $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$, so it gains five electrons and $n_2 = 5$.
Step 4: List the known values.
Here $M_1 = 0.02$, $V_1 = 500$ mL, $V_2 = 500$ mL, and $M_2 = x$.
Step 5: Substitute and cancel volumes.
\[ 0.02 \times 500 \times 2 = x \times 500 \times 5 \] and dividing out the common 500 gives $0.02 \times 2 = 5x$, that is $0.04 = 5x$.
Step 6: Solve for $x$.
So $x = \frac{0.04}{5} = 0.008 = 8 \times 10^{-3}$ M, which is option 2.
\[ \boxed{x = 8 \times 10^{-3}\ \text{M}} \]