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:

The $pH$ at which the maximum hydrate is present in a solution of oxaloacetic acid $(HOOC-CO-CH_2-COOH)$ with $pK_{a1} = 2.2$ and $pK_{a2} = 3.98$ is:

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 $pK_{b}(NH_{3}) = 4.74$)

$20 \ mL$ of $0.5 \ N \ HCl$ and $35 \ mL$ of $0.1 \ N \ NaOH$ are mixed. The resulting solution will

$30 \ cc$ 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

The $p^H$ of a mixture containing $10 \ mL$ of $0.1 \ N$ $NaOH$ and $10 \ mL$ of $0.05 \ N$ $H_2SO_4$ is ............