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Q.
A $1 \,L$ solution of $pH =1$, is diluted up to $10$ times. What volume of a solution with $pH =2$ is to be added in diluted solution so that $pH$ does not change?
Equilibrium
Solution:
After dilution $\left[ H ^{+}\right]=10^{-2}$
$pH =2$
Let $V$ L solution of $pH =2$ is added in original solution so that $pH$ remains fixed
$\therefore\left[ H ^{+}\right]=\frac{10^{-2} x+V \times 10^{-2}}{10+V}=10^{-2}$
This result is independent of volume taken