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 $(c)$

For a $0.1 \, M$ aqueous solution of a weak acid undergoing $2\%$ ionization,the concentration of $[H^{+}]$ and $[OH]^{-}$ will be respectively: (Ionic product of water $= 1 \times 10^{-14}$)

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:

Estimate the approximate $pK_a$ of $0.5 \ M$ $CH_3COOH$. Degree of dissociation (ionisation) is $0.15$ $(\log 1.32 = 0.12)$.

The $pK_{b}$ of dimethylamine and $pK_{a}$ of acetic acid are $3.27$ and $4.77$ respectively at $T(K)$. The correct option for the $pH$ of dimethylammonium acetate solution is :

The moles of $H^{+}$ from $H_2O$ alone in a $1 \ L$,$\sqrt{5} \times 10^{-7} \ M$ $HCl$ solution at $25 \ ^\circ C$ is ( $\sqrt{5} = 2.23$ )