At $25^{\circ} C$,the ionic product of water is $10^{-14}$. The free energy change for the self-ionization of water in $kCal \ mol^{-1}$ is close to

The dissociation constant of pure water at $18\,^oC$ will be .......... (Degree of dissociation of pure water at $18\,^oC = 1.8 \times 10^{-9}$)

If $0.1 \, M, 100 \, mL$ $NaOH$ solution is mixed with $0.05 \, M, 100 \, mL$ $H_2SO_4$ solution,the $pH$ of the resulting solution is:

If $20 \, mL$ of $0.25 \, N$ strong acid and $30 \, mL$ of $0.2 \, N$ strong base are mixed,then the resulting solution is

The hydrogen ion concentration in a mixture of $10 \ mL$ of $0.1 \ M$ $H_2SO_4$ and $10 \ mL$ of $0.1 \ M$ $KOH$ solution in water,is . . . . . . $M$.

The first ionization constant of $H_2S$ is $9.1 \times 10^{-8}$. Calculate the concentration of $HS^{-}$ ion in its $0.1 \ M$ solution. How will this concentration be affected if the solution is $0.1 \ M$ in $HCl$ also? If the second dissociation constant of $H_2S$ is $1.2 \times 10^{-13}$,calculate the concentration of $S^{2-}$ under both conditions.