Question

Determine the solubility of N2 in water exposed to air at 25°C if the atmospheric pressure is 1.2 atm. Assume that the mole fraction of nitrogen is 0.78 in air and the Henry's law constant for nitrogen in water at this temperature is 6.1 × 10-4 M/atm.

Answer 1

Henry's law states that the concentration of a given gas dissolved in a solution is directly proportional to the partial pressure of the gas over the solution Henry's law can be calculated as follows C-solubility of gas M) kH-Henry's law constant (M/ atm) P-partial pressure of gas(atm) Here the partial pressure of the gas is to be calculated. The partial pressure is given by PxPartial pressure of gas Xx,-Molefraction of gas Pica-Total pressure Given that, The mole fraction of nitrogen gas,x,-0.78 Total pressure, P-1.2atm Therefore, Parital pressure of N2.PN.^Xx,XP -0.78x1.2 atm -0.94 atm Now using Henry's law the solubility of nitrogen in water can be calculated Given kH-6.1x101 M/atm On substitution, -6.1x104 M/atmx0.94atm 5.7x101 M 4 Thus the solubility of nitrogen in water is5.7x10 M