If $4.0 \, g$ of $NaOH$ is present in $1 \, L$ of solution,then its $pH$ will be:

The concentration in $M$ of $OH^{-}$ in $0.001 \ M$ $H_2SO_4$ is

$25 \ mL$ of $0.1 \ M$ $HCl$ solution is diluted till $500 \ mL$. Calculate the $pH$ of the dilute solution. (in $.3$)

$1 \ L$ of $0.02 \ M$ aqueous $HCl$ is mixed with $1 \ L$ of $0.01 \ M$ aqueous $H_2SO_4$ solution. Assuming complete dissociation and no change in the volume upon mixing,the $pH$ of the resultant solution is $(\log_{10} 2 = 0.3)$.

Equal volumes of two solutions of $pH = 3$ and $pH = 5$ are mixed. The resultant $pH$ is

The $pH$ of a $H_2SO_4$ solution is $5.0$. If more $H_2SO_4$ is added to this solution,the $pH$ becomes $2.0$. The concentration of $H^{+}$ ions will become how many times the initial concentration?