What is the $pH$ of the resulting solution when $20 \text{ mL}$ of $\frac{M}{10} \text{ NaOH}$ and $10 \text{ mL}$ of $\frac{M}{10} \text{ H}_2\text{SO}_4$ are mixed together?

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:

One litre of water contains $10^{-7} \ mole$ of hydrogen ions. The degree of ionization in water will be:

During the titration of acetic acid with an aqueous $NaOH$ solution,the neutralization graph shows a vertical line. This line indicates:

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$BaCO_3 \downarrow + 2HCl \longrightarrow BaCl_2 + CO_2 \uparrow + H_2O$

For the equilibrium $2H_2O \rightleftharpoons H_3O^{+} + OH^{-}$,the value of $\Delta G^o$ at $298 \ K$ is approximately ....... $kJ \ mol^{-1}$.