Question

If $K_2Cr_2O_7$ ($200 \text{ cm}^3$, $x \times 10^{-3} \text{ M}$) reacts with $0.6 \text{ M}$, $750 \text{ cm}^3$ Mohr's salt then find value of x?

• A. 375
• B. 750
• C. 150
• D. 500

Hint:

In redox titrations, the fundamental equation is $N_1 V_1 = N_2 V_2$, where Normality $N = M \times n$-factor. For $K_2Cr_2O_7$ in acid, the $n$-factor is 6.

Answer 1

The Correct Option is A

The reaction is between $K_2Cr_2O_7$ (oxidant) and Mohr's salt ($Fe^{2+}$, reductant) in acidic medium.
For $K_2Cr_2O_7$: $Cr(+6) \rightarrow Cr(+3)$, $n$-factor ($n_1$) is $2 \times 3 = 6$.
For Mohr's salt ($Fe^{2+}$): $Fe^{2+} \rightarrow Fe^{3+}$, $n$-factor ($n_2$) is 1.
At equivalence point: Milli-equivalents of $K_2Cr_2O_7$ = Milli-equivalents of $Fe^{2+}$.
$M_1 V_1 n_1 = M_2 V_2 n_2$.
$(x \times 10^{-3}) \times (200) \times 6 = (0.6) \times (750) \times 1$.
$x \times 10^{-3} \times 1200 = 450$.
$x \times 1.2 = 450$.
$x = \frac{450}{1.2} = 375$.