I came across the following (solved) problem in my textbook:

The Henry's law constant for $\ce{O2}$ dissolved in water is $\pu{4.34 \times 10^4 atm}$ at $\pu{25 ^\circ C}$. If the partial pressure of oxygen in air is $\pu{0.4 atm}$, calculate the concentration (in terms of molarity, $\pu M$) of dissolved oxygen in water, in equilibrium with air at $\pu{25 ^\circ C}$.

The following is my attempt at a solution to the above problem:

From Henry's law, we have that

$$p{\ce{O2}} = K_\mathrm H \cdot x_\ce{O2}$$

where $x_\ce{O2}$ is the mole fraction of oxygen dissolved in the solution, $K_\mathrm H$ is the Henry's law constant, and $p_{\ce{O2}}$ is the partial pressure of oxygen over the liquid.

Given that the Henry's law constant for a mixture of water and some amount of dissolved oxygen is $\pu{4.34 \times 10^4 atm}$, this gives us

$$x_\ce{O2} = p_{\ce{O2}}/K_\mathrm H = 0.4/(4.34 \times 10^4) = 9.2 \times 10^{-6}$$

To calculate the concentration of oxygen in this mixture in terms of its molarity, I need to know the number of moles of oxygen present in the solution and also the number liters of water in which the oxygen is dissolved.

To calculate the number of moles of dissolved oxygen, I can use the following equation:

$$x_\ce{O2} = \frac{n_\ce{O2}}{n_\ce{O2} + n_\ce{H2O}}$$

However, no information regarding the number of moles of the solvent, water, has been provided.

Hence, I checked the solutions provided by my textbook and found that they have arbitrarily assumed that we are dealing with $\pu{1 L}$ of water.

My question is this: are we allowed to arbitrarily take the volume of water as $\pu{1 L}$ even when absolutely nothing has been mentioned about this in the problem?


1 Answer 1


The gist is that as long as you have a specified mole fraction, the concentration of oxygen does not depend on the amount of water present. This is what is called an "intensive property". Intensive properties don't depend on how much stuff there is, and therefore, you can arbitrarily specify any volume you like. In order to make the calculations simpler, this specified volume is usually some unit quantity, like $\pu{1 L}$.

It is somewhat analogous to determination of an empirical formula, in that it doesn't matter how many grams of the compound is present, you will always end up with the same empirical formula. A common tactic is to assume $\pu{100 g}$ of compound is present, such that if there is $x\%$ of carbon by mass, there is $x\ \pu{g}$ and therefore $x/\pu{12 mol}$ of carbon.

Here, if you have $\pu{2 L}$ of water, then $n_{\ce{H2O}}$ is double of that if you only have $\pu{1 L}$. But since you have twice as much water, to maintain the mole fraction (that you determined via Henry's law), you have to dissolve twice as much oxygen - otherwise the mole fraction won't be the same. So $n_{\ce{O2}}$ is double of that if you only have $\pu{1 L}$. The factor of 2 will cancel out in the equation you gave:

$$\begin{align} x_{\ce{O2}} &= \frac{2n_{\ce{O2}}}{2n_{\ce{O2}} + 2n_{\ce{H2O}}} \\ &= \frac{n_{\ce{O2}}}{n_{\ce{O2}} + n_{\ce{H2O}}} \end{align}$$

If you are really uncomfortable with assuming the volume, then simply assign an unknown value: $k$ litres, for example. You should find that $k$ cancels out and you are left with the exact same value for the molar concentration of $\ce{O2}$.

There is, perhaps, a more "elegant" way to prove this. Note that

$$\begin{align} \frac{1}{x_{\ce{O2}}} &= \frac{n_{\ce{O2}} + n_{\ce{H2O}}}{n_{\ce{O2}}} \\ &= 1 + \frac{n_{\ce{H2O}}}{n_{\ce{O2}}} \\ &= 1 + \frac{V_{\ce{H2O}} \times [(\pu{1000 g L-1})/(\pu{18 g mol-1})]}{n_{\ce{O2}}} \\ &= 1 + \frac{\pu{55.56 mol L-1}}{c_{\ce{O2}}} \end{align}$$

where $V$ is the volume and $c = n/V$ the molar concentration. You can see that the molar concentration is uniquely determined by the mass fraction $x$ and the exact value of $V$ doesn't matter at all.


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