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Q.
An acid solution has a $pH = 6$. It is diluted 100 times, the $pH$ of the resultant solution would be
Equilibrium
Solution:
For original undiluted solution,
$
pH =6
$
$
\left[ H _{3} O ^{+}\right]=10^{- pH }=10^{-6} M
$
The solution is diluted 100 times.
For the final diluted solution
$
\left[ H _{3} O ^{+}\right]=\frac{10^{-6} M }{100}=10^{-8} M
$
But there will be $\left[ H _{3} O ^{+}\right]$due to autoionisation of water.
In any way, since solution is acidic, $\left[ H _{3} O ^{+}\right]>10^{-7} M$ and $pH <7$
Hence, the $pH$ of solution becomes $6.95$.