Consider a solution of p^H = 6
Case-I : It is diluted by 100 times
Case-II : Hydronium concentration is decreased by 100 times.
If all the measurements are made at 298K, then the correct statement is

I) Dilution can never change the medium
$\therefore \left[H_3{O}^{+}\right]$ of solution =10^-8 M
$\left[H_3{O}^{+}\right]$ of  $H_{2}O={10}^{-7}M$
$\therefore Net\left[H_3{O}^{+}\right]=({10}^{-8}+{10}^{-7})=1.1\times {10}^{-7}$

$\therefore p^H=7-\log 1.1=6.956$
II) Decrease in  $\left[H_3{O}^{+}\right]$ can be achieved by adding alkali also, medium can change 

$\left[H_3{O}^{+}\right]={10}^{-8}$ 

$p^H=8$