15 mL of aqueous solution of Fe2+ in acidic medium completely reacted with 20 mL of 0.03 M aqueous C

Q: Why do we use normality instead of molarity in titration equations?

Normality accounts for the actual number of electrons transferred (the n-factor). Since one molecule of dichromate accepts 6 electrons while one iron ion only gives 1, they don't react in a 1:1 molar ratio. Normality levels the playing field so $N_1V_1 = N_2V_2$ always holds true.

Q: How did we get an n-factor of 6 for dichromate?

In $\text{Cr}_2\text{O}_7^{2-}$, each Chromium atom is in a +6 oxidation state. It reduces to $\text{Cr}^{3+}$, meaning each atom gains 3 electrons. Since there are 2 Chromium atoms in the ion, the total electron transfer is $2 \times 3 = 6$.

Q: Does the acidic medium matter?

Absolutely! The medium dictates the final oxidation state of the products. For example, permanganate ($\text{KMnO}_4$) has an n-factor of 5 in acidic medium, but 3 in neutral/weakly basic medium, and 1 in strongly basic medium. Dichromate is typically used in acidic mediums where its n-factor is 6.

Q: Why didn't we convert milliliters to liters?

Because we have volume on both sides of the equation ($N_1V_1 = N_2V_2$). As long as the units are consistent on both sides, the conversion factor ($10^{-3}$) cancels out. We are effectively equating milliequivalents instead of equivalents.

Q: What if the question asked for the mass of iron instead of molarity?

Once you find the molarity ($0.24$ M), you can find the moles by multiplying by the volume in liters ($0.24 \times 0.015$). Then, multiply the moles by the molar mass of iron ($56 \text{ g/mol}$) to get the mass.

Question

15 mL of aqueous solution of $	ext{Fe}^{2+}$ in acidic medium completely reacted with 20 mL of 0.03 M aqueous $	ext{Cr}_2	ext{O}_7^{2-}$. The molarity of the $	ext{Fe}^{2+}$ solution is ...... $	imes 10^{-2}$ M (Round off to the nearest integer).

Sigma Percentile · Accepted Answer

### The Titration Battlefield

Imagine a classic titration setup in a chemistry lab. In our Erlenmeyer flask, we have $15 	ext{ mL}$ of an unknown Iron(II) solution, $	ext{Fe}^{2+}$. Suspended above it in a burette is our oxidizing agent: $20 	ext{ mL}$ of a $0.03 	ext{ M}$ dichromate solution, $	ext{Cr}_2	ext{O}_7^{2-}$.

When these two mix in an acidic medium, a fierce exchange of electrons begins. The dichromate acts as an electron vacuum, pulling electrons away from the iron. To find the unknown concentration of our iron solution, we must balance the chemical books.

### The Master Key: Law of Equivalence

In stoichiometry, you cannot simply equate the moles of two reactants unless they react in a perfect 1:1 ratio. This is where the **Law of Equivalence** becomes our master key. It states that for a complete reaction, the equivalents (or milliequivalents) of the reducing agent must exactly equal the equivalents of the oxidizing agent.

Mathematically, this is expressed as:
$$N_1 V_1 = N_2 V_2$$

Since Normality ($N$) is simply Molarity ($M$) multiplied by the $n$-factor, we can expand our master equation to:
$$M_1 	imes n_1 	imes V_1 = M_2 	imes n_2 	imes V_2$$

### Decoding the n-factors

The $n$-factor is the heart of redox reactions. It represents the total number of electrons transferred per molecule.

Let's look at the dichromate ion, $	ext{Cr}_2	ext{O}_7^{2-}$. In an acidic medium, it reduces to $	ext{Cr}^{3+}$. The oxidation state of Chromium drops from $+6$ to $+3$, which is a gain of $3$ electrons per atom. Because there are two Chromium atoms in a single dichromate ion, the total electron transfer is:
$$n_2 = 3 	imes 2 = 6$$

Now, let's examine Iron. $	ext{Fe}^{2+}$ oxidizes to $	ext{Fe}^{3+}$. The oxidation state increases by exactly $1$, meaning it loses one electron. Therefore, the $n$-factor for our Iron solution is simply:
$$n_1 = 1$$

### The Final Crunch

We have all our pieces. Let's substitute them into our equivalence equation. Let $M_1$ be the unknown molarity of the $	ext{Fe}^{2+}$ solution.

$$M_1 	imes 1 	imes 15 = 0.03 	imes 6 	imes 20$$

Now, let's crunch the numbers. On the right side, $6 	imes 20 = 120$. Multiplying that by $0.03$ gives us $3.6$.

$$15 M_1 = 3.6$$

Dividing $3.6$ by $15$ gives us the molarity:
$$M_1 = \frac{3.6}{15} = 0.24 	ext{ M}$$

The question specifically asks for the answer in the format of $x 	imes 10^{-2} 	ext{ M}$. We can easily rewrite $0.24$ as $24 	imes 10^{-2}$.

So, our final integer answer is **24**.

Animated Solution for Chemistry - Basic Concepts in Chemistry: 15 mL of aqueous solution of $Fe^{2 +}$ in acidic medium completely reacted with 20 mL of 0.03 M aqueous $Cr_{2} O_{7}^{2 -}$ . The molarity of the $Fe^{2 +}$ solution is ...... $\times 1 0^{- 2}$ M (Round off to the nearest integer).

Enter Numerical Value:

Visualized Solution

$Redox Titration Setup$

$Law of Equivalence$

$n -factor of Dichromate$

$n -factor of Iron$

$Substitution$

$Calculation$

$Final Formatting$

$The Way Forward$

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Titration Battlefield

The Master Key

Decoding the n-factors

The Final Crunch

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