Animated Solution for Chemistry - Surface Chemistry: 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is

Have you ever wondered how water filters remove impurities, or how gas masks protect against toxic fumes? The secret lies in a fascinating phenomenon called adsorption. Unlike absorption, where a substance is drawn deep inside another (like a sponge soaking up water), adsorption is strictly a surface game. Molecules stick to the exterior of a material.

In this classic JEE problem, we are going to witness adsorption in action using activated charcoal and acetic acid. Let's dive into the math and the molecules!

Imagine you are standing in a lab. You have a flask containing $50\text{ mL}$ of an acetic acid ($C{H}_{3}COOH$) solution. The initial strength of this solution is given as $0.06\text{ N}$ (Normal).

To this flask, you add $3\text{ g}$ of activated charcoal. Activated charcoal is essentially carbon that has been treated to have a massive network of microscopic pores, giving it an incredibly large surface area. This makes it a perfect "sticky trap" for molecules.

Counting the Molecules

Before the charcoal starts doing its job, we need to know exactly how much acetic acid is swimming around in the bulk solution. We can calculate the initial number of millimoles ($n_i$) using the relationship between normality and volume.

For acetic acid, the n-factor is $1$ (since it donates one $H^{+}$ ion), which means its Normality is exactly equal to its Molarity.

Let's substitute our known values:

So, we start with exactly $3\text{ mmol}$ of acetic acid.

We let the flask sit for an hour. During this time, the acetic acid molecules are constantly colliding with the charcoal. Many of them get trapped on the charcoal's vast surface.

When we filter the solution and test it again, the new strength is $0.042\text{ N}$. The concentration has dropped! Let's find out how many millimoles are left in the solution ($n_f$).

We started with $3\text{ mmol}$, and now we only have $2.1\text{ mmol}$ floating freely. Where did the rest go? They were adsorbed!

Converting to Mass

We know that $0.9\text{ mmol}$ of acetic acid is stuck to the charcoal. But the options are in milligrams. We need to convert moles to mass.

The molar mass ($M_w$) of acetic acid ($C{H}_{3}COOH$) is $60\text{ g/mol}$.

So, a total of $54\text{ mg}$ of acetic acid was pulled out of the solution.

I know it is tempting to look for $54\text{ mg}$ in the options and tick it. But wait! Read the question carefully. It asks for the amount of acetic acid adsorbed per gram of charcoal.

We used $3\text{ g}$ of charcoal to adsorb those $54\text{ mg}$. To find the efficiency of just $1\text{ g}$ of charcoal, we must divide the total mass adsorbed by the mass of the charcoal used.

And there we have it! Every single gram of that activated charcoal managed to trap $18\text{ mg}$ of acetic acid. This beautiful interplay of concentration, volume, and surface area is what makes physical chemistry so incredibly powerful.