Answer
For a variation of 2.00 in pH, there is a 100 times variation in $[H^+]$.
Work Step by Step
$[H^+] = 10^{-pH}$
$[H^+] = 10^{-(pH + 2) } = 10^{-pH} \times 10^{-2}$
$[H^+] = 10^{-(pH - 2) } = 10^{-pH} \times 10^{2}$
If the pH increases by 2, the [H+] gets 100 times smaller, and if the pH decreases by 2, the [H+] gets 100 times greater.
