Question 29.2: The Henry’s law constant for oxygen dissolved in water is 4....
The Henry’s law constant for oxygen dissolved in water is 4.06 \times 10^{9} \mathrm{~Pa} /\left(\mathrm{mol}\right. of \mathrm{O}_{2} per total mol of solution) at 293 \mathrm{~K}. Determine the solution concentration of oxygen in water that is exposed to dry air at 1.013 \times 10^{5} \mathrm{~Pa} and 293 \mathrm{~K}.
Henry’s law can be expressed in terms of the mole fraction units by
p_{A}=H^{\prime} x_{A}
where H^{\prime} is 4.06 \times 10^{9} \mathrm{~Pa} /\left(\mathrm{mol}\right. of \mathrm{O}_{2} / total mol of solution ).
From example 24.1, we recognize that dry air contains 21 mol% oxygen. By Dalton’s law
p_{A}=y_{A} P=(0.21)\left(1.013 \times 10^{5} \mathrm{~Pa}\right)=2.13 \times 10^{4} \mathrm{~Pa} .
The equilibrium mole fraction of the liquid at the interface is computed by Henry’s law
\begin{aligned} x_{A}=\frac{P_{A}}{H} & =\frac{2.13 \times 10^{4} \mathrm{~Pa}}{4.06 \times 10^{9} \mathrm{~Pa} /\left(\mathrm{mol} \mathrm{O}_{2} / \mathrm{mol} \mathrm{soln}\right)} \\ & =5.25 \times 10^{-6}\left(\mathrm{~mol} \mathrm{O}_{2} / \mathrm{mol} \text { soln }\right) \end{aligned}
For one cubic meter of very dilute solution, the moles of water in the solution will be approximately
\begin{aligned} n_{\text {water }} & =\left(1 \mathrm{~m}^{3}\right)\left(1 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\right)\left(\frac{1}{0.018 \mathrm{~kg} / \mathrm{mol}}\right) \\ & =5.56 \times 10^{4} \mathrm{~mol} \end{aligned}
The total moles in the solution is essentially the moles of water because the concentration of oxygen is quite low. Accordingly, the moles of oxygen in one cubic meter of solution is
\begin{aligned} n_{\text {oxygen }} & =\left(5.25 \times 10^{-6} \mathrm{~mol} \mathrm{O}_{2} / \mathrm{mol} \text { soln }\right)\left(5.56 \times 10^{4} \mathrm{~mol} \text { soln }\right) \\ & =0.292 \mathrm{~mol} \text { of } \mathrm{O}_{2} \end{aligned}
The saturation concentration is
\left(0.292 \mathrm{~mol} / \mathrm{m}^{3}\right)(0.032 \mathrm{~kg} / \mathrm{mol})=9.34 \times 10^{-3} \mathrm{~kg} \mathrm{O}_2 / \mathrm{m}^{2}(9.34 \mathrm{mg} / \mathrm{L})