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:

We cam calculate the solubility of O₂ in water using Henry's law: Cgas = K P,where, Cgas is the solubility if gas,K is henry's law constant (K for O₂ at 25 ̊C is 1.3 x 10⁻³ mol/l atm),P is the partial pressure of O₂ (P = 120 torr / 760 = 0.158 atm).Cgas = K P = (1.3 x 10⁻³ mol/l atm) (0.158 atm) = 2.054 x 10⁻⁴ mol/L.

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

Decreasing the temperature will increase the solubilty of O₂ gas in water.When the temperature increases, the solubility of O₂ gas in water will decrease because the increase in T will increase the kinetic energy of gas particles and increase its motion that will break intermolecular bonds and escape from solution.Decreasing the temperature will increase the solubility of O₂ gas in water will because the kinetic energy of gas particles will decrease and limit its motion that can not break the intermolecular bonds and increase the solubility of O₂ gas.

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

Further eplanation

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

Learn more  

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the partial pressure of hydrogen gas  

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the solubility of a gas  

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

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