By how many pH units does rain vs distilled water change a soil pH test?
Pure water (rain as well as distilled water) in equilibrium with the atmosphere ($p_{\ce{O2}}=10^{-3.5}\ \mathrm{atm}$) can be calculated to contain about
$$\begin{align}
\mathrm{pH}=-\log[\ce{H+}]&amp;=5.65\\
-\log[\ce{HCO3-}]&amp;=5.65\\
-\log[\ce{CO3^2-}]&amp;=10.3\\
-\log[\ce{H2CO3^*}]&amp;=5.0\\
-\log[\ce{CO2}]&amp;=5.0\\
-\log[\ce{H2CO3}]&amp;=7.8\\
\end{align}$$ 

(The calculation can be found in: Stumm, W.; Morgan, J. J. Aquatic Chemistry, Third Edition; John Wiley &amp; Sons: New York, NY, 1996; pp 159–161.)

Note that $[\ce{H2CO3^*}]$ is the total analytical activity of dissolved $\ce{CO2}$, i.e. $[\ce{H2CO3^*}]=[\ce{CO2(aq)}]+[\ce{H2CO3}]$. $[\ce{H2CO3}]$ is the concentration of true $\ce{H2CO3}$.

Thus, $[\ce{H2CO3^*}]=10^{-5}$ and $c(\ce{H2CO3^*})=10^{-5}\ \mathrm{mol/l}$; i.e. a few µmol/l.