Among the following

Diluting an acid by adding more amount of water increases the pH of the solution towards 7 making it less acidic.

pH = 5 using pH = $-log\left[H^{+}\right]$, then [$H^{+}$] =${10}^{-5}$  

pH = 3 using pH = $-log\left[H^{+}\right]$, then [$H^{+}$] =${10}^{-3}$ & therefore pH = 5 is 100 times more basic than a solution with pH = 3.  

pH = 8 using pH = $-log\left[H^{+}\right]$, then [$H^{+}$] =10^{-8}  

pH = 11 using pH = $-log\left[H^{+}\right]$, then [$H^{+}$] =${10}^{-11}$ & therefore pH = 8 is 1000 times more acidic than a solution with pH = 11.

[$O{H}^{-}$] = KOH = ${10}^{-9}$

pOH = $-log\left[O{H}^{-}\right]$ = -log[${10}^{(-9)}$] = 9

pH + pOH = 14

pH = 14-9 = 5