P=hx where p=pressur =1atm..... x =solubility = 0.195..... h=henry's law so
h =p/x = 1/0.195=5.12

Solubility of H2S in water at STP

The solubility of a gas in a liquid is a measure of how much of the gas can dissolve in the liquid at a given temperature and pressure. In this case, we are interested in the solubility of hydrogen sulfide (H2S) in water at standard temperature and pressure (STP).

Henry's Law

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. Mathematically, it can be represented as:

C = k * P

where C is the concentration of the dissolved gas in the liquid, k is Henry's Law constant, and P is the partial pressure of the gas.

Calculating Henry's Law constant

To calculate Henry's Law constant, we need to know the solubility of H2S in water at STP and the partial pressure of H2S in the gas phase.

Given that the solubility of H2S in water at STP is 0.195 M, we can use this value as the concentration (C) in Henry's Law equation. The partial pressure of H2S in the gas phase is not given directly, but since we are at STP, we can assume that the partial pressure of H2S is equal to the pressure of the gas phase, which is 1 atm.

Plugging in the values into the equation, we have:

0.195 M = k * 1 atm

Solving for k, we find that the Henry's Law constant for H2S in water at STP is 0.195 M/atm.

Explanation and interpretation

Henry's Law constant represents the solubility of a gas in a liquid under specific conditions. In this case, a Henry's Law constant of 0.195 M/atm means that for every 1 atm of partial pressure of H2S in the gas phase, 0.195 moles of H2S will dissolve in 1 liter of water at STP.

This value can be used to predict the solubility of H2S in water at different pressures. For example, if the partial pressure of H2S is increased to 2 atm, we can use Henry's Law to determine the new concentration of H2S in the water.

C = k * P
C = 0.195 M/atm * 2 atm
C = 0.39 M

Therefore, at a partial pressure of 2 atm, the concentration of H2S in water would be 0.39 M.

Summary

In summary, the solubility of H2S in water at STP is 0.195 M. This can be used to calculate Henry's Law constant, which represents the solubility of the gas in the liquid under specific conditions. Henry's Law constant for H2S in water at STP is 0.195 M/atm, indicating that for every 1 atm of partial pressure of H2S in the gas phase, 0.195 moles of H2S will dissolve in 1 liter of water.