About pH of an aqueous solution of SO2
Upon adding $\ce{SO2}$ gas to water, it shows its acidic oxide character by hydrolyzing to sulfurous acid:

$$\ce{SO2(g) + H2O(l) &lt;=&gt; H2SO3(aq)}$$

$\ce{H2SO3}$ is a weak diprotic acid in water, with the following acid dissociation equilibria:

$$\ce{H2SO3(aq) &lt;=&gt; H+(aq) + HSO3^{-}(aq)}\ \ \ \ \ \mathrm{K_{a1}=1.4\times 10^{-2}}$$
$$\ce{HSO3{}^{-}(aq) &lt;=&gt; H+(aq) + SO3{}^{2-}(aq)}\ \ \ \ \ \mathrm{K_{a2}=6.3\times 10^{-8}}$$

Thus, addition of $\ce{SO2}$ gas to water will produce an acidic solution.

What would happen if air were bubbled into the solution? Air is a mixture of mostly nitrogen and oxygen. It so happens that $\ce{SO2}$ reacts with oxygen to form $\ce{SO3}$:

$$\ce{2SO2(g) + O2(g) -&gt; 2SO3(g)}$$

Here, the sulfur atom in $\ce{SO2}$ is oxidized (oxidation number goes from +4 to +6) while the oxygen atoms in $\ce{O2}$ are reduced (oxidation number goes from 0 to -2). Now, $\ce{SO3}$ also happens to be an acidic oxide, and it hydrolyzes in water to form sulphuric acid:

$$\ce{SO3(g) + H2O(l) &lt;=&gt; H2SO4(aq)}$$

This time, $\ce{H2SO4}$ is a strong diprotic acid:

$$\ce{H2SO4(aq) &lt;=&gt;&gt; H+(aq) + HSO4^{-}(aq)}\ \ \ \ \ \mathrm{K_{a1} \approx 10^6}$$
$$\ce{HSO4{}^{-}(aq) &lt;=&gt; H+(aq) + SO4{}^{2-}(aq)}\ \ \ \ \ \mathrm{K_{a2}=1\times 10^{-2}}$$

Assuming that not much $\ce{SO2}$ is lost while bubbling air through the sulphurous acid solution, then you're effectively replacing a weak acid with a strong acid, which would cause the pH to decrease.

As a final remark, if the sulphurous acid solution were bubbled with pure nitrogen, then there would be no oxidation of $\ce{SO2}$ into $\ce{SO3}$, and after a while enough $\ce{SO2}$ will have been dragged away for the solution to become a less concentrated sulphurous acid, which would therefore have a higher pH.