Given:$\mathrm{K}_{\mathrm{w}}=9.6 \times 10^{-14} \mathrm{M}^{2}$
As we know that,
for a neutral solution,
$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]$
Now,
$\mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \times\left[\mathrm{OH}^{-}\right]$
$\Rightarrow \mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{2}\left(\because\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]\right)$
$\Rightarrow\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{2}=9.6 \times 10^{-14}$
$\Rightarrow\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\sqrt{9.6 \times 10^{-14}} \approx 3.1 \times 10^{-7}$
Hence at $60^{\circ} \mathrm{C}$, for a neutral aqueous solution, the $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$ is $3.1 \times 10^{-7}$.
Now, pH of the neutral solution-
$\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$
$\mathrm{pH}=-\log \left(3.1 \times 10^{7}\right)$
$\Rightarrow \mathrm{pH}=(7-\log (3.1))=6.51$
since pH of a neutral solution is less than 7 , the aqueous solution with a $\mathrm{pH}=7$ is basic in nature.
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$(i)\,\,RC \equiv \mathop C\limits^ \ominus $
$(ii)\,\,[IMAGE]$
$(iii)\,\,{R_2}C = \mathop C\limits^ \ominus H$
$(iv)\,\,{R_3}C - \mathop C\limits^ \ominus {H_2}$
Assertion $A:$ $5 f$ electrons can participate in bonding to a far greater extent than $4 f$ electrons
Reason $R:$ $5 f$ orbitals are not as buried as $4 f$ orbitals
In the light of the above statements, choose the correct answer from the options given below
$MnO_4^ - $ ${C_2}O_4^{2 - }$ ${H^ + }$