Question

H2S, a toxic gas with rotten fish smell,is used for the qualitative analysis.If the solubility of H2S in water at STP is 0.195m, calculate the henry's law constant

Answer 1

Answer:

solubility of H2S in water at STP is 0.195 m, i.e., 0.195 mol of H2S is dissolved in 1000 g of water.

Moles of water = 1000g / 18g mol-1

= 55.56 mol

∴Mole fraction of H2S, x = Moles of H2S / Moles of H2S+Moles of water

0.195 / (0.195+55.56)

= 0.0035

At STP, pressure (p) = 0.987 bar

According to Henry's law:

p= KHx

⇒ KH = p / x

= 0.0987 / 0.0035 bar

= 282 bar