200 cc of x × 10⁻³ M potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium. Here x = ____

200 cc of x × 10⁻³ M potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium

Q. 200 cc of x × 10⁻³ M potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium.

Here x = _____.

Correct Answer: 375

Explanation

This is a redox titration based on the concept of equivalents.

First, write the relevant half-reactions in acidic medium.

For potassium dichromate:

$$ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O $$

So, 1 mole of K₂Cr₂O₇ accepts 6 moles of electrons.

For Mohr’s salt (Fe²⁺):

$$ Fe^{2+} \rightarrow Fe^{3+} + e^- $$

So, 1 mole of Fe²⁺ loses 1 mole of electrons.

Hence,

$$ 1 \text{ mole of } Cr_2O_7^{2-} \equiv 6 \text{ moles of } Fe^{2+} $$

Now calculate moles of Mohr’s salt used.

Given:

Volume = 750 cc = 0.75 L
Molarity = 0.6 M

$$ \text{Moles of Fe}^{2+} = 0.75 \times 0.6 = 0.45 $$

Required moles of K₂Cr₂O₇:

$$ = \frac{0.45}{6} = 0.075 $$

Now calculate moles of potassium dichromate from given data.

Volume = 200 cc = 0.2 L
Molarity = x × 10⁻³ M

$$ \text{Moles of } K_2Cr_2O_7 = 0.2 \times x \times 10^{-3} $$

At equivalence point:

$$ 0.2 \times x \times 10^{-3} = 0.075 $$

Solving:

$$ x = \frac{0.075}{0.2} \times 10^3 $$

$$ x = 375 $$

Therefore, the value of x = 375.