The pH of normal rain water is 5.6 whereas that of a rain called "acid" can be down to 3. Calculate the ratio of the molarities of oxonium ions in the normal rain relative to the acid rain.
$pH$ $=$ $5.6$ $[H_3O^+]$ $=$ $10^{-5.6}$ $pH$ $=$ $3$ $[H_3O^+]$ $=$ $10^{-3}$ $\frac{[H_3O^+]_{normal}}{[H_3O^+]_{acide}}$ $=$ $\frac{10^{-5.6}}{10^{-3}}$ $=$ $10^{-2.6}$ $=$ $3.2\cdot 10^{-3}$