Let c M be the concentration of H^+ in the acid solution of
pH =6
So
-log_10c=6
=>log_10c=-6
=>c=10^-6 M
If the solution is 100 times diluted then new concentration of H^+ in respect of acid will be [H^+]=10^-8 M
If we take negative 10 base logarithm of this [H^+]=10^-8 M to calculate the pH then pH=-log_10(10^-8)=8 This means the solution becomes basic (as pH>7) but this is not feasible.
An approximate calculation
if we add concentration of H^+ due to ionization of water 10^-7 M with the concentration of H^+ from acid then total concentration of H^+ =10^-8+10^-7=10^-7xx1.1 M
So pH=-log_10(10^-7 xx1.1)=7-log_10 1.1~~6.95
More correct approach
Let water on ionisation contributes H^+ and OH^- each in x M
So
[H_"water"^+]=[OH_"water"^-]=x M
As the ionic product of waterk_w=10^-14 is a constant
then
([H_"acid"^+]+[H_"water"^+])xx[OH_"water"^-]=10^-14
=>(10^-8+x)xx x=10^-14
=>x^2+10^-8x-10^-14=0
on solving we get
x~~9.5xx10^-8
So in acid solution
[H_"total"^+]=10^-8+9.5xx10^-8=10.5xx10^-8=1.05xx10^-7#
This gives
pH=-log_10[H_"total"^+]~~6.98
