Question

The 30 ml of water with zero dissolved oxygen is placed in a 40 ml vial and sealed. The rest of the vial initially contains air with 21.0% oxygen by volume. What will be the equi-librium concentration of oxygen in the water, and the partial pressure of the oxygen in the air in the vial?

Answer 1

Given in question,

Volume of water = 30 mL

Volume of vial = 40 mL

Percentage of oxygen = 21%

To Find,

Equilibrium conc. of oxygen in water =?

Partial pressure of oxygen in air in the vial =?

Solution:-

volume of air = vol. of vial - vol. of water = 40 mL -30 mL =10 mL

Volume of O₂ present = 21% of 10 mL = (21/100) * 10 = 0.21 * 10 = 2.1 mL

No. of moles of O₂ = 2.1 mL/ 22400 mL mol^-1 = 9.37 * 10 ^-5 moles [ 1 mole of gas = 22400 mL]

Formula for partial pressure is,

P = nRT / total volume of gas

Po₂ = {(9.37 * 10 ^-5 moles) * (0.0821 atm L mol^-1 K^-1) * (293.15 K) } / 10 mL

= 2.26 * 10^-2 atm

From Henry law,

Concentration of molecule in solution is,

[gas(aq)] = P /K_H

[ P=partial pressure , K_H= equilibrium between the aqueous gas and the free gas, gas(aq) = Concentration of dissolved gas in water ]

[ K_H is experimental and for O₂ its value is 769.23 atm mol^-1]

Thus , putting the value in Henry equation,

[gas(aq)] = (2.26 * 10^-2 atm) / (769.23 atm mol^-1)

= 2.9 * 10^-5 M

So,

Equilibrium conc. of oxygen in water = 2.9 * 10^-5 M

Partial pressure of oxygen in air in the vial = 2.26 * 10^-2 atm