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#

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**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 water#k_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#