Calculate the solubility of o2 in water at a partial pressure of o2 of 120 torr at 25 ̊c. the henry's law constant for o2 at 25 ̊c is 1.3 x 10-3 mol/l atm. how do you expect the solubility to change if the temperature were decreased?

1) 2.054 x 10⁻⁴ mol/L.

2) Decreasing the temperature will increase the solubilty of O₂ gas in water.

Explanation:

1) The solubility of O₂ gas in water:

2) The effect of decreasing temperature on the solubility O₂ gas in water:

A. The solubility of O₂ in water : S = 2.052 . 10⁻⁴ mol/L

B. If the temperature were decreased then the solubility of O₂ will increase

In the transfer from the gas phase to the liquid phase diffusion process occurs.  

The solubility of the gas itself decreases with increasing temperature  

The partial pressure of the gas itself is influenced by the concentration of the gas and temperature  

Henry's Law states that the solubility of a gas is proportional to its partial pressure  

Can be formulated  

S = kH. P.  

S = gas solubility, mol / L  

kH = Henry constant, mol / L.atm  

P = partial gas pressure  

The partial gas pressure is proportional to the percentage of the gas volume in the mixture  

Henry's law constant = 1.3 x 10⁻³mol / l atm.  

P = 120 torr = 0.15789 atm

So  

S = 1.3 x 10⁻³ . 0.15789

S = 2.052 . 10⁻⁴ mol/L

If the temperature were decreased then the solubility of O₂ will increased

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Keywords: Henry's Law, partial pressure, gas solubility  

the correct answer would be d. hope this !