Question

Fish rely on dissolved O2 in the water for life. Calculate the mole fraction of dissolved O2 in water at 25 degree Celsius and compare the molarities of O2. Assume O2 makes up 20% of the atmosphere and use Henry's constant. K: 3.27x10^-7 and K': 773 atm*kg/mol.

Answer 1

According to Henry's law, the mass of O2 gas dissolved in water, i.e. the concentration of O2 in water is directly proportional to its partical pressure in the gas phase.

i.e. p(O2) = K_H*c(aq), where c(aq) is the concentration of O2 in the aqueous phase

K_H(pc) = p(O2)/c(aq)

773 = 0.2/c(aq) [since O2 makes up 20% of the atmosphere]

Therefore, c(aq) = 0.2/773, i.e. 2.59*10^-4 mol/Kg

Hence n(O2) = 2.59*10^-4 mol

Given that

K_H(cc) = c(gas)/c(aq); where c(gas) is the concentration of O2 in the gas phase

3.27*10^-7 = c(gas)/2.59*10^-4

c(gas) = 8.47*10^-11 mol/Kg

The molefraction of dissolved O2 in water = n(O2)/{n(O2) + n(H2O)}

i.e. x(O2) = 2.59*10^-4/(1000/18) [since no. of moles of water in 1Kg of water = 1000/18)

         = 4.662*10^-6