The flocculation value of HCl for arsenic sulphide sol. is 30 m mol L^-1. If H2SO4 is used for the f

Q: Why do we only care about the $H^+$ ions and ignore the $Cl^-$ or $SO_4^{2-}$ ions?

Arsenic sulphide ($As_2S_3$) forms a negatively charged sol because it preferentially adsorbs sulphide ($S^{2-}$) ions. To neutralize and coagulate a negative sol, we need positive ions (cations). The anions ($Cl^-$ or $SO_4^{2-}$) are repelled and play no role in the coagulation process.

Q: What exactly does a 'flocculation value' of 30 mean?

The flocculation value is the minimum concentration of an electrolyte, expressed in millimoles per litre (mmol/L), required to completely coagulate a sol in a specific time. A value of 30 means you need 30 millimoles of the active ion per litre of the sol.

Q: Why did we divide the moles of $H^+$ by 2 when switching to $H_2SO_4$?

One molecule of $HCl$ provides one $H^+$ ion, but one molecule of $H_2SO_4$ provides two $H^+$ ions. Since $H_2SO_4$ is twice as effective at delivering the required $H^+$ ions, you only need half as many molecules of $H_2SO_4$ to get the same total number of $H^+$ ions.

Q: How do I know if a sol is positively or negatively charged if it's not mentioned?

It's a standard convention you must memorize for JEE. Metal sulphides (like $As_2S_3$, $CdS$, $Sb_2S_3$) are generally negatively charged. Metal hydroxides (like $Fe(OH)_3$, $Al(OH)_3$) are generally positively charged.

Q: What if the question asked for the volume of a 0.1 M $H_2SO_4$ solution instead of the mass?

Once you find that you need 3.75 mmol of $H_2SO_4$, you would use the molarity formula: $\text{Volume} = \frac{\text{Moles}}{\text{Molarity}}$. So, $\text{Volume} = \frac{3.75 \text{ mmol}}{0.1 \text{ M}} = 37.5 \text{ mL}$.

Question

The flocculation value of $	ext{HCl}$ for arsenic sulphide sol. is $30 	ext{ m mol L}^{-1}$. If $	ext{H}_2	ext{SO}_4$ is used for the flocculation of arsenic sulphide, the amount in grams of $	ext{H}_2	ext{SO}_4$ is $250 	ext{ mL}$ required for the above purpose is ...... (molecular mass of $	ext{H}_2	ext{SO}_4 = 98 	ext{ g /mol}$)

Sigma Percentile · Accepted Answer

### The Nature of the Beast: Arsenic Sulphide Sol

Imagine you are looking at a beaker filled with a cloudy, yellowish liquid. This is our arsenic sulphide ($As_2S_3$) sol.

Before we do any math, we must understand the physical reality of this mixture. Arsenic sulphide is a classic example of a **negatively charged sol**.

Why is it negative? Because during its formation, the colloidal particles preferentially adsorb sulphide ($S^{2-}$) ions from the surrounding medium.

According to the **Hardy-Schulze rule**, to force these negatively charged particles to clump together and settle down—a process called coagulation or flocculation—we need to introduce an oppositely charged ion.

In our case, we need a positive ion (a cation). The problem mentions hydrochloric acid ($HCl$) and sulphuric acid ($H_2SO_4$), both of which provide the mighty hydrogen ion ($H^+$) as the active coagulating agent.

### The Hardy-Schulze Rule in Action

The problem states that the **flocculation value** of $HCl$ is $30 	ext{ mmol L}^{-1}$.

What does this number actually mean? The flocculation value is simply the minimum concentration of an electrolyte required to completely coagulate a sol.

Since one molecule of $HCl$ dissociates to give exactly one $H^+$ ion, a flocculation value of 30 means we need exactly **$30 	ext{ millimoles}$ of $H^+$ ions** to coagulate one full litre of the arsenic sulphide sol.

### Scaling Down the Volume

But wait, we don't have a massive one-litre beaker. We only have a smaller sample of $250 	ext{ mL}$.

We need to scale down our requirement proportionally. First, let's convert the volume into litres:

$$V = 250 	ext{ mL} = 0.25 	ext{ L}$$

Now, we calculate the exact number of $H^+$ moles needed for this specific volume by multiplying the concentration by the volume:

$$n_{H^+} = 30 	ext{ mmol L}^{-1} 	imes 0.25 	ext{ L} = 7.5 	ext{ mmol}$$

So, to coagulate our $250 	ext{ mL}$ sample, we need exactly **$7.5 	ext{ millimoles}$ of $H^+$ ions**.

### The Sulphuric Acid Twist

Here is where many students make a silly mistake. The question doesn't ask for the mass of $HCl$; it asks for the mass of sulphuric acid ($H_2SO_4$).

We must look at the stoichiometry of sulphuric acid. When one molecule of $H_2SO_4$ dissociates, it releases **two** $H^+$ ions:

$$H_2SO_4 ightarrow 2H^+ + SO_4^{2-}$$

Because sulphuric acid is twice as effective at delivering the required hydrogen ions, we will only need half as many molecules of it compared to $HCl$.

Let's calculate the millimoles of $H_2SO_4$ required:

$$n_{H_2SO_4} = \frac{n_{H^+}}{2} = \frac{7.5 	ext{ mmol}}{2} = 3.75 	ext{ mmol}$$

### The Final Calculation

We are in the home stretch. We know exactly how many moles of sulphuric acid we need, but the question asks for the answer in grams.

We need to convert millimoles to grams using the given molar mass of $H_2SO_4$, which is $98 	ext{ g mol}^{-1}$.

First, remember that $3.75 	ext{ mmol}$ is equal to $3.75 	imes 10^{-3} 	ext{ moles}$. Now, we multiply by the molar mass:

$$	ext{Mass} = 3.75 	imes 10^{-3} 	ext{ mol} 	imes 98 	ext{ g mol}^{-1}$$

$$	ext{Mass} = 0.3675 	ext{ g}$$

Finally, we round off our result to a reasonable number of decimal places.

Rounding $0.3675$ gives us our final, elegant answer:

**$	ext{Mass} \approx 0.37 	ext{ g}$**

And there we have it! By carefully tracking the charges, the volumes, and the stoichiometry, we've successfully navigated through the chemistry of colloids.

Enter Numerical Value:

Visualized Solution

$Flocculation Value$

$n_{H^{+}} for 250 mL$

$n_{H_{2} SO_{4}} Required$

$Mass of H_{2} SO_{4}$

$Final Answer$

The Sigma Insight: Colloids, Micelles and and Emulsions

The Nature of the Beast

The Hardy-Schulze Rule in Action

Scaling Down the Volume

The Sulphuric Acid Twist

The Final Calculation

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