Question

26 Calculate the equilibrium concentration of dissolved oxygen in 15°C water at 1 atm, and again at 2,000 m elevation. Suppose the gas above the soda in a bottle of soft drink is pure CO2 at a pressure of 2 atm. Calculate (CO2) at 25°C. 27

Answer 1

Ans 26) We know,

Amount of oxygen in air = 21 %

=> Partial pressure of oxygen in air (P) = 1 atm x 0.21 = 0.21 atm

Henry law constant for oxygen, K_H = 0.0015236 mol / L-atm

=>Equilibrium concentration = K_H P = 0.0015236 mol/L-atm x 0.21 atm = 0.00032 mol/L

1 mol O₂ = 32 gm or 32000 mg

=> Equilibrium concentration in mg/L = 0.00032 mol/L x 32000 mg/mol = 10.24 mg/L

Now

We know, at elevation H, atmospheric pressure , P_a = 1 - 0.000115 H

So, at H = 2000 m , P_a = 1 - 0.000115(2000) = 0.77 atm

   => Partial pressure of oxygen in air (P) = 0.77 atm x 0.21 = 0.1617 atm

=>Equilibrium concentration = K_H P = 0.0015236 mol/L-atm x 0.1617 atm = 0.000246 mol/L

=> Equilibrium concentration in mg/L = 0.000246 mol/L x 32000 mg/mol = 7.87 mg/L

Ans 27) We know, according to Henry law,

[CO₂] = K_H P_CO2

Henry law constant for CO₂ at 25^o C = 0.033363 mol/L-atm

P_CO2 = 2 atm (given)

=> [CO₂] = 0.033363 mol / L-atm x 2 atm = 0.0667 mol/L

Molar mass of CO₂ = 44 gm/mol

=> [CO₂] = 0.0667 mol/L x 44 gm/mol = 2.93 gm/L