H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant.
Henry's law constant (K_H) = 285 bar
Explanation
This numerical problem applies Henry's law (p = K_H × x) to calculate the Henry's law constant for H2S. The solubility is given in molality (0.195 m), which means 0.195 moles of H2S dissolved in 1 kg of water. First, we calculate the mole fraction of H2S in the solution. At STP, the pressure is 1 bar. Using the relation K_H = p/x, we can find the Henry's law constant.
Solution Steps
Step 1: Calculate moles of water in 1 kg (1000 g) of water. n(H₂O) = 1000 g / 18 g mol⁻¹ = 55.5 mol
Step 2: Calculate mole fraction of H₂S. Given solubility = 0.195 m means n(H₂S) = 0.195 mol Mole fraction, x(H₂S) = n(H₂S) / [n(H₂S) + n(H₂O)] x(H₂S) = 0.195 / (0.195 + 55.5) = 0.195 / 55.695 = 0.0035
Step 3: Apply Henry's law to find K_H. At STP, partial pressure p = 1 bar From Henry's law: p = K_H × x Therefore, K_H = p / x = 1 bar / 0.0035 K_H = 285 bar