The $pH$ of $0.01 \ N$ lime water is:

$0.1 \ (M) \ HCl$ and $0.1 \ (M) \ H_2SO_4$ each of volume $2 \ mL$ are mixed and the volume is made up to $6 \ mL$ by adding $2 \ mL$ of $0.01 \ (N) \ NaCl$ solution. The $pH$ of the resulting mixture is

$100 \, mL$ of $0.04 \, N \, HCl$ aqueous solution is mixed with $100 \, mL$ of $0.02 \, N \, NaOH$ solution. The $pH$ of the resulting solution is

What will be the resultant $pH$ when $100 \ mL$ of an aqueous solution of $HCl$ $(pH = 2)$ is mixed with $200 \ mL$ of an aqueous solution of $NaOH$ $(pH = 12)$?

If $0.4 \ g$ of $NaOH$ is present in $1 \ L$ of solution,then the $pH$ of the solution is .....

Assuming that $Ba(OH)_{2}$ is completely ionised in aqueous solution under the given conditions,the concentration of $H_{3}O^{+}$ ions in $0.005 \, M$ aqueous solution of $Ba(OH)_{2}$ at $298 \, K$ is $..... \times 10^{-12} \, mol \, L^{-1}$. (Nearest integer)