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#
