Animated Solution for Chemistry - Basic Concepts in Chemistry: The mass of potassium dichromate crystals required to oxidise $750{\text{ cm}}^3$ of $0.6\text{ M}$ Mohr's salt solution is (molar mass $=392$)

Imagine you are standing in a chemistry lab, holding a flask filled with a light green solution of Mohr's salt. Your mission? To completely oxidize it using bright orange potassium dichromate crystals. This isn't just a simple mixing task; it's a precise dance of electrons, and we need to figure out exactly how much dichromate to invite to the party. Let's break down the chemistry and the math behind this classic redox titration.

Decoding the Reactants

First, let's understand our players. Mohr's salt is a double salt with the formula ${\text{FeSO}}_4\cdot ({\text{NH}}_4{)}_2{\text{SO}}_4\cdot 6{\text{H}}_2\text{O}$. It looks intimidating, but in the context of a redox reaction, most of it is just a spectator. The true hero here is the ferrous ion, ${\text{Fe}}^{2+}$. This ion is eager to lose an electron and become the more stable ferric ion, ${\text{Fe}}^{3+}$.

On the other side, we have potassium dichromate (${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$). In an acidic medium, the dichromate ion (${\text{Cr}}_2{\text{O}}_7^{2-}$) is a powerhouse oxidizing agent. It is hungry for electrons and will snatch them from any willing donor, like our ${\text{Fe}}^{2+}$.

Redox at Play

To find out how these two interact, we need to look at their half-reactions.

For the oxidation of iron:

Each ferrous ion gives up exactly one electron.

For the reduction of dichromate:

Notice the massive appetite of the dichromate ion—it requires six electrons to fully reduce its two chromium atoms from a +6 to a +3 oxidation state.

Stoichiometry

Nature demands balance. If one dichromate ion needs six electrons, and one iron ion can only provide one, it's clear that we need a team of six iron ions to satisfy a single dichromate ion.

By multiplying the iron half-reaction by 6 and adding it to the dichromate half-reaction, we get our master stoichiometric relationship:

This tells us the golden rule for this titration: 6 moles of ${\text{Fe}}^{2+}$ are chemically equivalent to 1 mole of ${\text{Cr}}_2{\text{O}}_7^{2-}$.

Now, let's look at the numbers given in the problem. We have $750{\text{ cm}}^3$ (which is $0.75\text{ L}$) of a $0.6\text{ M}$ Mohr's salt solution.

Using the molarity formula, we can find the exact number of moles of ${\text{Fe}}^{2+}$ we are dealing with:

Since 6 moles of ${\text{Fe}}^{2+}$ require 1 mole of dichromate, we can easily find the moles of dichromate needed by dividing by 6:

Navigating the Distractor

Here is where the question tries to trick you. It casually mentions "(molar mass = 392)" right at the end. If you are rushing, you might blindly multiply your 0.075 moles by 392. Don't fall for it!

That 392 g/mol is the molar mass of Mohr's salt, not potassium dichromate. We don't even need the molar mass of Mohr's salt because we were already given its molarity and volume. To find the mass of potassium dichromate crystals required, we must use its own molar mass.

Let's calculate the molar mass of ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$:

Finally, we convert the moles of dichromate into mass:

And there we have it! We need exactly 22.05 grams of potassium dichromate crystals to completely oxidize the Mohr's salt solution. This perfectly matches option (c). By staying alert to the stoichiometry and avoiding the distractor trap, a seemingly complex redox problem becomes a straightforward logical sequence.