Q.$H_2S$, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of $H_2S$ in water at STP is 0.195 m, calculate Henry's law constant.

The key idea is Henry’s Law: at constant temperature, the partial pressure of a gas above a liquid is proportional to its mole fraction dissolved in the liquid, $p = K_H \cdot x$.

Step 1: Convert the given solubility (0.195 molal) to a mole fraction.

Moles of water in 1 kg: $n_{water}=1000/18=55.56\,\text{mol}$.

Step 2: At STP, the partial pressure of $H_2S$ is $p=1\,\text{bar}$.

Step 3: Solve for $K_H$.

Henry's law states $p = K_H\, x$, where $p$ is the partial pressure of the gas above the solution, $x$ is the mole fraction of the dissolved gas, and $K_H$ is Henry's law constant.

1. Moles of solute and solvent.

A solubility of $0.195\ m$ means $0.195$ mol of $H_2S$ dissolved per $1$ kg of water.

2. Mole fraction of $H_2S$.

3. Apply Henry's law.

At STP the partial pressure of $H_2S$ above the (dilute, near-pure) solvent is taken as $p = 1\,\text{bar}$ (the standard reference pressure), so

Method: Henry's Law (mole-fraction form)

Henry's Law states that at constant temperature, the partial pressure of a gas above a liquid is directly proportional to its mole fraction dissolved in the liquid:

where:

• $p$ = partial pressure of the gas above the solution
• $K_H$ = Henry's law constant (same units as pressure)
• $x$ = mole fraction of the gas in solution

Steps

1. Convert molality to mole fraction. Solubility $m = 0.195\,\text{mol/kg}$ means 0.195 mol $H_2S$ per 1 kg (55.56 mol) of water.

2. Identify the pressure at STP.

   $P = 1\,\text{bar}$ (standard STP pressure).

3. Apply Henry's Law and solve for $K_H$.

Key Exam Point

For this chapter, Henry's Law is always applied in its mole-fraction form, $p=K_H x$ - never in a 'per kg of solvent' (molal) form. Always convert a given molality to a mole fraction before applying Henry's Law.

This question is a direct application of Henry's Law in terms of molality, and it's a classic trap for students.

✓ The Core Concept First

Henry's Law states:

At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

Mathematically:

where:

• $p$ = partial pressure of the gas (in bar or atm)
• $K_H$ = Henry's law constant (same units as pressure)
• $x$ = mole fraction of the gas in the solution

✗ Common Mistake #1: Using molality ($m$) directly as mole fraction

The error:

Students see "0.195 m" and plug it directly into $p = K_H \cdot m$.

Why it's wrong:

Henry's law uses mole fraction, not molality. Molality is moles of solute per kg of solvent — a different quantity.

How to avoid:

Always convert molality to mole fraction first.

Correct method:

• At STP, take $p = 1 \text{ bar}$ (see Mistake #2).
• 0.195 m means 0.195 moles of $H_2S$ in 1 kg of water.
• Moles of water in 1 kg = $\frac{1000}{18} \approx 55.56$ mol.
• Mole fraction of $H_2S$:

• Then:

Key takeaway: Always compute $x$ first.

✗ Common Mistake #2: Fumbling the pressure at STP

The error:

Students either ignore pressure entirely or assume some unrelated value.

Why it's wrong:

STP (Standard Temperature and Pressure) on the modern IUPAC definition — the one NCERT uses — is 273.15 K and 1 bar, so here $p = 1 \text{ bar}$. (Older texts defined STP with 1 atm; since the mole fraction is unchanged, the numerical answer is ≈ 286 either way — only the unit label follows your pressure convention.)

How to avoid:

Memorise:

• STP → $T = 273.15 \text{ K}$, $p = 1 \text{ bar}$ (older convention: 1 atm)
• NTP → $T = 293 \text{ K}$, $p = 1 \text{ atm}$
• SATP → $T = 298 \text{ K}$, $p = 1 \text{ bar}$

✗ Common Mistake #3: Using molarity instead of molality

The error:

Students treat "0.195 m" as 0.195 M (moles per litre).

Why it's wrong:

The unit "m" means molality (mol/kg solvent), not molarity (mol/L solution). They are not interchangeable, especially when density is not given.

How to avoid:

• "m" = molality
• "M" = molarity
• If the problem says "0.195 m", it's molality — use mass of solvent, not volume.

✗ Common Mistake #4: Ignoring the "negligible" approximation

The error:

Students write $x = \frac{0.195}{55.56}$ and forget to add 0.195 in the denominator.

Why it's wrong:

Mole fraction = $\frac{n_{solute}}{n_{solute} + n_{solvent}}$. If you omit $n_{solute}$ in the denominator, you get a slightly larger $x$ and a smaller $K_H$.

How to avoid:

Always include both terms in the denominator. For dilute solutions, the error is small — but in exams, full accuracy is expected.

✓ Final Correct Answer (for reference)

(≈ 286 atm on the older 1-atm STP convention — same number, different unit label.)

📌 Quick Checklist to Avoid Mistakes