Question

If the mol ratio of nitrogen and oxygen is `4:1`, what is the ratio of their solubilities in terms of mol fractions? Take `k_H = 3.3 xx 10^7 "torr"` for `"N"_2(g)` and `6.6 xx 10^7 "torr"` for `"O"_2(g)`.

Answer 1

`2:1`, `"N"_2:"O"_2`

You should consider whether the question wants `"N"_2:"O"_2` or `"O"_2:"N"_2`.

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There are two things you should be thinking about here:

• What is Henry's law in terms of mol fractions?
• What does knowing the ratio of `"N"_2:"O"_2` in air tell you about their partial pressures?

Setting up the solution is usually the hard part. Imagine a beaker of water that has air dissolved in it, but there is, say, parafilm on it so that it is a closed container. Kind of like this, but with transparent water:

What you are looking for is the ratio of the solubilities on the mol fraction scale. It may or may not be `4:1` or `1:4`.

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Henry's law in terms of mol fractions for the solute in particularly-dilute solutions is:

`bb(P_j = chi_(j(v))k_(H,j))`

where:

• `P_j` is the partial pressure of solute `j` above the solution.
• `chi_(j(v))` is the mol fraction of solute `j` in the vapor phase (`v`), that is, above the solution.
• `k_(H,j)` is the Henry's law constant of solute `j`, and `k_(H,j)` approaches the pure vapor pressure `P_j^"*"` as `chi_(j(l)) -> 1` (since the solute particles surround the solvent particles completely, and vaporize most easily).

Writing Henry's law for `"N"_2` and `"O"_2`:

`P_(N_2) = chi_(N_2(v))k_(H,N_2)`

`P_(O_2) = chi_(O_2(v))k_(H,O_2)`

But from Dalton's law of partial pressures for ideal gases, we have:

`P_(t ot) = P_(N_2) + P_(O_2)`

`= 0.78P_(t ot) + 0.21P_(t ot) + stackrel("Other")overbrace(0.01P_(t ot))`

Thus, we can rewrite Henry's law as:

`0.78P_(t ot) = chi_(N_2(v))k_(H,N_2)`

`0.21P_(t ot) = chi_(O_2(v))k_(H,O_2)`

Finding the ratio of their mol fraction solubilities, we then have:

`(0.78P_(t ot))/(0.21P_(t ot)) = (chi_(N_2(v))k_(H,N_2))/(chi_(O_2(v))k_(H,O_2))`

`=> (chi_(N_2(v)))/(chi_(O_2(v))) = 3.71(k_(H,O_2)/k_(H,N_2))`

`= 3.71((3.3 xx 10^7 "torr")/(6.6 xx 10^7 "torr"))`

`= 1.86`

That's using a mol ratio of `0.78:0.21`, as it is in reality, but using a mol ratio of `4:1` like given in the problem, we would have:

`color(blue)((chi_(N_2(v)))/(chi_(O_2(v)))) = 4((3.3 xx 10^7 "torr")/(6.6 xx 10^7 "torr"))`

`= color(blue)(2:1)`, `color(blue)("N"_2:"O"_2)`

If the question wants the ratio of `"O"_2` to `"N"_2`, then in that case it would be `1:2`, `"O"_2:"N"_2`.

In the end, it is just the mol ratio in air, weighted by the reciprocal ratio of the Henry's law constants.