Question

What is the concentration of nitrogen (in mg/L) in a Lake Tahoe at 25⁰C if the altitude of the lake is 6254 ft and atmospheric pressure at see level is 1.02 atm.

Consider K_H = 6.4 x 10^–9 mol L^-1 Pa^–1. Assume no salinity or other impacts of dissolution of nitrogen in the water.

Answer 1

According to Henry's law:

x/m = K_H * P .......... Equation 1

Where x/m is the solubility or concentration of N₂ gas in lake water

K_H is Henry's constant = 6.4*10^-9 mol L^-1 Pa^-1

P is pressure of N₂ gas = 1.02*(78/100) = 0.7956 atm (since percent of nitrogen in air = 78%)

Therefore, from equation 1:

x/m = 6.4*10^-9 mol L^-1 Pa^-1 * 0.7956*101325 Pa (since 1 atm = 101325 Pa)

i.e. x/m = 5.16*10^-4 mol L^-1

i.e. x/m = 5.16*10^-4 mol L^-1 * 28 g/mol (since the molar mass of N₂ = 28 g/mol)

i.e. x/m = 144.5*10^-4 g/L

i.e. x/m = 14.45*10^-3 g/L

i.e. x/m = 14.45 mg/L (since 1 mg = 10^-3 g)