When $100 \ mL$ of $1 \ M$ $NaOH$ solution is mixed with $10 \ mL$ of $10 \ M$ $H_2SO_4$,the resulting mixture will be

The dissociation of water at $25 \, ^\circ C$ is $1.9 \times 10^{-7} \%$ and the density of water is $1.0 \, g/cm^3$. The ionisation constant of water is

$30 \ cc$ of $\frac{M}{3} \ HCl$,$20 \ cc$ of $\frac{M}{2} \ HNO_{3}$ and $40 \ cc$ of $\frac{M}{4} \ NaOH$ solutions are mixed and the volume was made up to $1 \ dm^{3}$. The $pH$ of the resulting solution is

Given,$pH$ of a solution $A$ is $3$ and it is mixed with another solution $B$ having $pH$ $2$. If both are mixed in equal volume,then the resultant $pH$ of the solution will be

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.

One mole of ammonia was completely absorbed in one litre solution each of $(a)$ $1 \ M \ HCl$,$(b)$ $1 \ M \ CH_{3}COOH$,and $(c)$ $1 \ M \ H_{2}SO_{4}$ at $298 \ K$. The decreasing order for the $pH$ of the resulting solutions is (Given $pK_{b}(NH_{3}) = 4.74$)