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Hydrogen sulfide (HS), a toxic gas with a rotten egg-like smell, is used for qualitative analysis

Hydrogen sulfide (HS), a toxic gas with a rotten egg-like smell, is used for qualitative analysis. If the solubility of hydrogen sulfide in water at standard temperature and pressure (STP) is 0.195 molal (m), calculate Henry's law constant.

Full Answer Section

2. Convert molality to mole fraction: Molality ( $m$ ) = 0.195 mol H₂S / kg water

First, convert kg of water to moles of water: Molar mass of water (H₂O) = 2 * 1.008 g/mol (H) + 16.00 g/mol (O) = 18.016 g/mol Moles of water in 1 kg (1000 g) = 1000 g / 18.016 g/mol $\approx$ 55.51 mol H₂O

Now, calculate the mole fraction of H₂S ( $X_{H 2 S}$ ): $X_{H 2 S} = moles of H ^{2} S + moles of H ^{2} O moles of H ^{2} S$ $X_{H 2 S} = 0.195 mol + 55.51 mol 0.195 mol$ $X_{H 2 S} = 55.705 0.195 \approx 0.00350$

3. Apply Henry's Law: The most common form of Henry's Law using mole fraction is: $P = K_{H} \cdot X$

Where:

$P$ is the partial pressure of H₂S, which at STP is 1 atm.
$X$ is the mole fraction of H₂S in the solution.
$K_{H}$ is Henry's Law constant.

Rearranging to solve for $K_{H}$ : $K_{H} = X P$ $K_{H} = 0.00350 1 atm$ $K_{H} \approx 285.7 atm$

The units for Henry's Law constant can vary. If we express solubility in mol/L and pressure in atm, the units for $K_{H}$ would be atm/(mol/L) or L·atm/mol. Since we used mole fraction, which is dimensionless, and pressure in atm, the constant $K_{H}$ will have units of atm.

Alternatively, some forms of Henry's Law use $C = K_{H'} \cdot P$ , where $K_{H'}$ has units of mol/(L·atm). If we were to convert molality to molarity, assuming the solution is dilute enough that 1 kg of solvent is approximately 1 L of solution (which is reasonable for dilute aqueous solutions): 0.195 mol H₂S / 1 kg H₂O $\approx$ 0.195 mol H₂S / L solution (approximately) Then, $K_H' = C/P = (0.195 \text{ mol/L}) / (1 \text{ atm}) = 0.195 \text{ mol/(L·atm)}$ . This is the inverse of the constant calculated above. Given the prompt and typical values for $K_{H}$ , the form $P = K_{H} \cdot X$ is often implied when working with mole fractions, resulting in $K_{H}$ having pressure units.

Using the common definition of $P = K_{H} \cdot X$ :

$K_{H} \approx 286 atm$ (rounded to 3 significant figures)

The final answer is $286 atm$ .

Sample Answer

Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The formula is often expressed as:

$C = K_{H} \cdot P$

Where:

$C$ is the concentration of the gas in the liquid (often in mol/L or mol fraction).
$K_{H}$ is Henry's Law constant.
$P$ is the partial pressure of the gas above the liquid.

We are given the solubility in molality ( $m$ ), which is moles of solute per kilogram of solvent. To use Henry's Law, we need to convert molality to mole fraction or molarity, and determine the partial pressure at STP.

1. Define STP (Standard Temperature and Pressure): According to IUPAC, STP is typically defined as:

Temperature (T) = 0 °C = 273.15 K
Pressure (P) = 1 bar = 100 kPa = 0.986923 atm

For gas solubility problems, it's common to use 1 atm (101.325 kPa) as the standard pressure. Let's use 1 atm for our calculation, as it's a common convention for Henry's Law problems unless specified otherwise.