Question

$pH$ of $10^{-10}MHCl$ solution at $25^{\circ }C$ is approximately

Answer 1

Answer: 7

$\left[H^{+}\right]$in $10^{-10}MHCl$ solution is $10^{-7}+10^{-10}\approx 10^{-7}$
Hence $pH=-\log 10^{-7}=+7$