The concentration of $[H^{+}]$ and concentration of $[OH^{-}]$ of a $0.1 \ M$ aqueous solution of $2\%$ ionised weak acid is [Ionic product of water $= 1 \times 10^{-14}$]

The first and second ionization constants of a weak dibasic acid $H_{2}A$ are $8.1 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. $0.1 \text{ mol}$ of $H_{2}A$ was dissolved in $1 \text{ L}$ of $0.1 \text{ M}$ $HCl$ solution. The concentration of $HA^{-}$ in the resultant 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

If $pK_{a} = -\log K_{a} = 4$ for a weak acid $HX$ and $K_{a} = C\alpha^{2}$,then the Van't Hoff factor when $C = 0.01 \ M$ is

Dilution processes of different aqueous solutions with water are given in $LIST-I$. The effects of dilution of the solutions on $[H^{+}]$ are given in $LIST-II$. (Note: Degree of dissociation $(\alpha)$ of weak acid and weak base is $ << 1$; degree of hydrolysis of salt $ << 1$; $[H^{+}]$ represents the concentration of $H^{+}$ ions)

$LIST-I$	$LIST-II$
$P$. ($10 \ mL$ of $0.1 \ M$ $NaOH$ + $20 \ mL$ of $0.1 \ M$ acetic acid) diluted to $60 \ mL$	$1$. The value of $[H^{+}]$ does not change on dilution
$Q$. ($20 \ mL$ of $0.1 \ M$ $NaOH$ + $20 \ mL$ of $0.1 \ M$ acetic acid) diluted to $80 \ mL$	$2$. The value of $[H^{+}]$ changes to half of its initial value on dilution
$R$. ($20 \ mL$ of $0.1 \ M$ $HCl$ + $20 \ mL$ of $0.1 \ M$ ammonia solution) diluted to $80 \ mL$	$3$. The value of $[H^{+}]$ changes to $1/\sqrt{2}$ times of its initial value on dilution
$S$. $10 \ mL$ saturated solution of $Ni(OH)_2$ in equilibrium with excess solid $Ni(OH)_2$ is diluted to $20 \ mL$ (solid $Ni(OH)_2$ is still present after dilution)	$4$. The value of $[H^{+}]$ changes to $\sqrt{2}$ times of its initial value on dilution

Match each process given in $LIST-I$ with one or more effect$(s)$ in $LIST-II$. The correct option is:

At $25^{\circ}C$,the dissociation constants of acetic acid and $HCN$ are $1.5 \times 10^{-5}$ and $4.5 \times 10^{-10}$ respectively. What is the equilibrium constant for the following reaction? $(CN^- + CH_3COOH \rightleftharpoons HCN + CH_3COO^-)$