Consider the following statements. The correct statement is(are)

The correct option is D Ionic product of water is temperature dependent.
The explanation (I),
millimoles of $H^{+}=400\times 0.1\times 2=80$
Millimoles of $O{H}^{-}=400\times 0.1\times 2=40$
(Limiting reagent)

$\therefore$ Millimoles of $H^{+}$ left $=80-40=40$
$\left[H^{+}\right]=\frac{40}{400+400}=\frac{40}{800}=\frac{1}{20}M$

$⟹pH=-log\left[H^{+}\right]=-log\left(\frac{1}{20}\right)$
$=-log1+log2+log10$
$=-0+0.301+1$
$⟹1.30$
Hence, the statement I is correct.

In statement II, the ionic product of $H_{2}O$ is temperature dependent.
$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}(mol/L{)}^{2}at{25}^{o}C$
With increase in temperature, dissociation of $H_{2}O$ units into $H^{+}andO{H}^{-}$ ions will also increase. As a result, the value of ionic product, $\left[H^{+}\right]\times \left[O{H}^{-}\right]$ will be increased.

Hence, the statement II is correct.

In statement III, for a weak monobasic acid HA,
$\underset{(1-\alpha )CM}{HA}\rightleftharpoons {\underset{\alpha CM}{H}}^{+}+{\underset{\alpha CM}{A}}^{-}$
$⟹$ pH of the solution is,
$\left[H^{+}\right]={10}^{-5}M=\alpha C$
$⟹K_a=\frac{\alpha C\times \alpha C}{(1-\alpha )C}=\frac{{10}^{-5}\times \alpha }{1-C}$
$⟹{10}^{-5}=\frac{{10}^{-5}\times \alpha }{1-\alpha }$
$⟹\alpha =0.5$
$⟹\alpha \%=50$

In statement IV, Le- Chatelier's principle is applicable to common ion effect. Because, in presence of common ion (given) by strong electrolyte (say, $N{a}^{+}{A}^{-}$), the product of the concentration terms in RHS increases. For the weaker electrolyte, HA (say) the equilibrium shifts to the LHS,
$HA\rightleftharpoons H^{+}+A^{-}$

As a result dissociation of HA gets supressed. Hence, the statement IV is incorrect.