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I am working on my scholarship exam practice. I believe this exam assumes high school + first year university knowledge. And I'm not quite sure what I did wrong. Could you please have a look?

The solubility of oxygen in 1.0 L water is 28 mL at 25 °C and 1.0 atm. How much oxygen can be dissolved in 1.0 L of water at 25 °C and 4.0 atm?

The answer remains 28 mL but IMO I thought when pressure goes up 4 times while other factors are constant, the solubility would do that as well. Please help on where I missed and what topics I should focus on specifically.

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As noted by Wikipedia,

"Henry's Law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid."

So you are absolutely right. 4 times the pressure of oxygen should result in 4 times as much oxygen being dissolved.

The answer key for the problem must be wrong.

User Delta_G made an interesting observation/correction. The problem is asking about the volume of gas that dissolves, not the concentration of oxygen in the water. So 28 ml at 4 atm of oxygen gas contains four times as much oxygen as 28 ml at 1 atm of the gas. So in either case 28 ml of the gas phase dissolves into the water.

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    $\begingroup$ Do we need to think about Boyle's law too? In the problem as written it gives the amount of oxygen in mL. At 4 times the pressure, the same volume of oxygen contains a lot more oxygen. $\endgroup$
    – Delta_G
    May 12, 2019 at 17:57
  • $\begingroup$ @Delta_G - ah ha! I half-read the problem and was just thinking about the fact that the water would contain four times as much oxygen. But you are absolutely right. If at 1 atm of oxygen 28 ml of oxygen gas dissolves, then when you increase the pressure to 4 atm, then 28 ml of the 4 atm oxygen gas will dissolve too. So 4 times as much oxygen in the water, but only 28 ml of the gas phase dissolves in either case. // Thank you for the correction! $\endgroup$
    – MaxW
    May 12, 2019 at 18:14
  • $\begingroup$ Let me reiterate your point again. If we increase the pressure, the solubility as in grams or moles would increase but not for volume. Please correct me if I am not on the right track. How does this link to Boyle's law though? Since Boyle's law state that pressure is inversely proportional to the volume, providing that other variables are constant. This means volume would decrease if we increase the pressure. $\endgroup$
    – Trey Anupong
    May 12, 2019 at 18:49
  • $\begingroup$ @TreyAnupong - That's the point. If 18 ml oxygen gas at 1 atm contains X grams of gas, then 18 ml of gas at 4 atm contains 4X grams of gas. So 4 times as much pressure, 4 times as much mass per volume in the gas. $\endgroup$
    – MaxW
    May 12, 2019 at 18:58
  • $\begingroup$ Sorry if my question is amateur, why would we only set the volume constant though? If we keep the mass of gas constant, the volume would now decrease. For example, if 28 mL oxygen gas at 1 atm contains X grams of gas, then for the same amount (X g) 7 mL of gas at 4 atm could be dissolved in water. $\endgroup$
    – Trey Anupong
    May 12, 2019 at 19:08
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This is a confusing question because, while solubilities can be reported in mL/L, there can be ambiguity when choosing a pressure during conversion to this unit, for instance using the following equation to convert from molarity $c$ to volume/volume units:

$$ \rho = \frac{cRT}{p}$$

In this online data page, for instance , in some columns the solubility is reported in mL/L, converted to this unit using throughout a pressure of $\pu{1 atm}$ (the source of the data cannot be verified), even as the partial pressure of oxygen $p_{O_2}$ is increased above $\pu{1 atm}$.

In the OP the volume refers presumably to an equivalent volume of oxygen gas at the (partial) pressure of the gas above the liquid.

Using the numbers from the OP, assuming the gas is ideal, then

$ c=\frac{101325\times28\times 10^{-6}}{8.3145\times298.15} \pu{M} =\pu{ 1.1 \times 10^{-3}M}$

when $p = \pu{1 atm}$.

On the other hand, if $p = \pu{4 atm}$

$ c=\frac{4\times101325\times28\times 10^{-6}}{8.3145\times298.15} \pu{M} =\pu{ 4.6\times 10^{-3} M}$

so the solubility is the same ($\pu{28 mL/L}$) if described in terms of volume at the given pressure, but $\times 4$ greater when regarded as a molar concentration.

Note by the way that according to a number of sources the solubility at $\pu{25 ^\circ C}$ is $\pu{258 \mu M}$ (~$\pu{8.2 mm Hg}$) at $p_{O_2}=\pu{1 atm}$, and $\pu{1.0 mM}$ at $p_{O_2}=\pu{4 atm}$.

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