$A$ $0.1 \ N$ solution of an acid at room temperature has a degree of ionisation $0.1$. The concentration of $OH^{-}$ would be

When $300 \, mL$ of $M/3 \, HCl$,$200 \, mL$ of $M/2 \, HNO_3$,and $400 \, mL$ of $M/4 \, NaOH$ are mixed,and the total volume is made up to $1 \, dm^3$,what will be the $pH$ of the resulting solution?

Consider the following statements:
$(a)$ The $pH$ of a mixture containing $400 \, mL$ of $0.1 \, M \, H_2SO_4$ and $400 \, mL$ of $0.1 \, M \, NaOH$ will be approximately $1.3$.
$(b)$ Ionic product of water is temperature dependent.
$(c)$ $A$ monobasic acid with $K_a = 10^{-5}$ has a $pH = 5$. The degree of dissociation of this acid is $50 \%$.
$(d)$ The Le Chatelier's principle is not applicable to common-ion effect.
The correct statements are:

At $25\, ^oC$,the dissociation constant of $CH_3COOH$ and $NH_4OH$ in aqueous solution are almost the same. The $pH$ of a $0.01\, N$ $CH_3COOH$ solution is $4.0$ at $25\, ^oC$. The $pH$ of a $0.01\, N$ $NH_4OH$ solution at the same temperature would be:

The decreasing order of electrical conductivity of the following aqueous solutions is:
$0.1 \ M$ Formic acid $(a)$
$0.1 \ M$ Acetic acid $(b)$
$0.1 \ M$ Benzoic acid $(c)$

Hydrogen ion concentration of an aqueous solution is $1 \times 10^{-4} \ M$. The solution is diluted with an equal volume of water. The hydroxyl ion concentration of the resultant solution in terms of $mol \ dm^{-3}$ is