A solution of  $pH=5$ diluted to 100 times. The $pH$ of new solution is 

 $pH=5\Rightarrow \left[H^{+}\right]={10}^{-5}M$
100 times dilution decreases the concentration of $\left[H^{+}\right]$  becomes equal to  ${10}^{-7}M$. New pH of acidic 
Solution  $=-\log {10}^{-7}=+7$ (not possible)
Total $\left[H^{+}\right]={10}^{-7}+{10}^{-7}$  (from water) $=2\times {10}^{-7}$ 
Actual new  $pH=-\log (2\times {10}^{-7})=-\log 2-\log {10}^{-7}=-0.3+7=6.7$