Animated Solution for Chemistry - Basic Concepts in Chemistry: Comprehension Passage

From Flask to Aliquot

Imagine you are in a chemistry lab. You have just dissolved a $5.6\text{ g}$ sample containing iron into cold dilute $\text{HCl}$. This process converts the iron into ${\text{Fe}}^{2+}$ ions, creating a $250\text{ mL}$ stock solution. Let's assume the total number of moles of ${\text{Fe}}^{2+}$ in this large flask is $n$.

Now, we don't titrate the entire $250\text{ mL}$ at once. Instead, we carefully pipette out a $25.0\text{ mL}$ aliquot into a conical flask. Because the solution is perfectly homogeneous, the moles of ${\text{Fe}}^{2+}$ in this smaller portion will be exactly proportional to its volume.

Law of Equivalence

We now titrate this $25.0\text{ mL}$ aliquot with a $0.03\text{ M}$ solution of Potassium Permanganate (${\text{KMnO}}_4$). The titration reaches its endpoint when exactly $12.5\text{ mL}$ of ${\text{KMnO}}_4$ has been added.

In redox titrations, the most powerful tool at our disposal is the Law of Chemical Equivalence. It states that at the equivalence point, the equivalents of the reducing agent must perfectly match the equivalents of the oxidizing agent.

To use the Law of Equivalence, we must convert moles into equivalents by multiplying them by their respective n-factors (the number of electrons transferred per mole).

For iron, it oxidizes from ${\text{Fe}}^{2+}$ to ${\text{Fe}}^{3+}$, releasing exactly $1$ electron. Thus, its n-factor is $1$.

For permanganate in an acidic medium (provided by the $\text{HCl}$), the $\text{Mn}$ atom reduces from a $+7$ oxidation state in ${\text{MnO}}_4^{-}$ down to ${\text{Mn}}^{2+}$. This is a gain of $5$ electrons. Thus, the n-factor for ${\text{KMnO}}_4$ is $5$.

Let's plug our values into the equivalence equation. Remember that the moles of ${\text{KMnO}}_4$ can be found by multiplying its molarity by its volume in liters.

Solving the right side of the equation gives us the moles of ${\text{Fe}}^{2+}$ present in the $25\text{ mL}$ aliquot:

To find the total moles $n$ in the original $250\text{ mL}$ flask, we simply multiply by $10$:

The problem states that the total moles are $x\times 10^{-2}$. By comparing the two expressions, we find that $x=1.875$, which rounds to $1.88$.

Now that we know the total moles of iron in the sample, we can easily find its mass. We multiply the moles by the molar mass of iron ($56\text{ g/mol}$).

This $1.05\text{ g}$ is the actual amount of pure iron present in the original $5.6\text{ g}$ sample. To find the percentage by weight ($y$), we divide the mass of iron by the total mass of the sample and multiply by $100$.

Thus, the value of $y$ is $18.75$.

This problem assumes all the iron dissolved as ${\text{Fe}}^{2+}$. But what if the original sample contained a mixture of ${\text{Fe}}^{2+}$ and ${\text{Fe}}^{3+}$? ${\text{KMnO}}_4$ only oxidizes ${\text{Fe}}^{2+}$. To find the total iron content in such a scenario, you would first need to treat the solution with a reducing agent (like ${\text{SnCl}}_2$ or $\text{Zn/HCl}$) to convert all ${\text{Fe}}^{3+}$ into ${\text{Fe}}^{2+}$ before performing the titration. Always keep an eye out for these subtle variations in advanced problems!