TEXT:
11.3 Example 11.2
The example refers to figure 11.9 and gives a link to somewhere, but in fact it wants you to use figure 11.8 to get Henry's law constant.
The example is confusing because you (sort of) have to use Henry's law twice to solve the problem: first to get k from the solubility data in figure 11.8, and then again to calculate the actual solubility at a partial pressure of 0.17 atm. The example is also confusing because it uses data that explores the temperature dependence of solubility to illustrate Henry's law calculations, which deal with the pressure dependence (not temperature dependence) of solubility.
In this discussion, for ease of typing, I will use "s" to refer to solubility, and "k" to refer to the Henry's law constant. (There are many many other k's in this course. Try not to get too confused.)
Gas solubility changes with temperature. At any GIVEN temperature, the solubility of a gas depends on the partial pressure of that gas (Henry's law, s = kP where s=solubility and P=partial pressure of the gas). If the solubility is different at a different temperature, with the same P, then either k must depend on T, or there is some other factor missing from the equation that would tell us how s depends on T. It is common to assume that the temperature dependence of solubility is incorporated into k, in other words that you have to determine k for the temperature you are using, and that it will change at a different T. (There are no hidden factors, s=kP and k changes if you change T.)
This example indirectly gives you k as a function of T for different gases. Indirectly, because it actually gives you the solubility as a function of T, for a partial pressure of 1 atm. That boils down to the same thing, as we will see in part a) below.
So let's go through the reasoning that the author intended you to follow:
a) The figure tells us that at 30oC, the solubility of O2 is 1.15x10-3 M, when (according to the figure's caption) the partial pressure of O2 is 1 atm. (That's a lot of significant figures to get from this graph, but hey, confusion is the name of the game here.) Assuming that Henry's law applies, s=kP and we can use the solubility at P=1 atm from the graph to get k: k=s/P with s=1.15x10-3 M and P=1 atm.
b) Now that we have k for O2 at 30oC, we can use Henry's law again to calculate s if P=0.17 atm (calculation valid at 30oC), s=(1.15e-3)(1 atm).