The reaction equation is first order with respect to [H+]
when PH1 = -㏒[H+]1 so, when PH = 6
So by substitution:
∴ 6 = -㏒[H+]1
∴[H+]1 = 1 x 10^-6
and when PH2 = -㏒[H+]2 so, when PH2 = 2
so by substitution:
∴ 2 = -㏒[H]2
∴[H]2 = 1 x 10^-2
So the rate of reaction changes by the factor of:
[H2]2/[H]1 = (1 x 10^-2) / (1 x 10^-6) = 10000
It is 10000 times faster when PH decreases from 6 to 2