What is the $pH$ of $0.005 \ M \ H_2SO_4$ solution?

The $pH$ value of a $\frac{N}{10} \ NaOH$ solution is:

Calculate the $pH$ of $0.01 \ M$ strong dibasic acid.

$0.023 \ g$ of sodium metal is reacted with $100 \ cm^{3}$ of water. The $pH$ of the resulting solution is

Assuming complete dissociation,the $pH$ of a $0.01 \, M$ $NaOH$ solution is equal to

Equal volumes of two solutions $A$ and $B$ of a strong acid having $pH=6.0$ and $pH=4.0$ respectively are mixed together to form a new solution. The $pH$ of the new solution will be in the range