Calculate the $pH$ of a $0.002 \ M$ $KOH$ solution.

The ratio of $pH$ of $0.1 \, M \, H_2SO_4$ and that of $0.1 \, N \, H_2SO_4$ is

An aqueous solution of $HCl$ with $pH \ 1.0$ is diluted by adding an equal volume of water (ignoring the dissociation of water). The $pH$ of the $HCl$ solution would be (Given $\log 2 = 0.30$):

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

$50 \ mL$ water is mixed with $50 \ mL$ $0.01 \ M$ $Ba(OH)_2$. Calculate its $pH$.

Calculate the $pH$ of a mixed solution containing $300 \ mL$ of $0.05 \ M \ H_2SO_4$ and $150 \ mL$ of $0.05 \ M \ KOH$. (in $.301$)