At ${100}^{\circ }\mathrm{C}\text{ the }{\mathrm{K}}_{\mathrm{w}}$of water is 55 times its value at 25°C.What will be the pH of neutral solution? $(\log 55=1.74)$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}\text{ at }{25}^{\circ }\mathrm{C}=1\times {10}^{-14} \\ \text{ At }{25}^{\circ }\mathrm{C} \\ {\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]={10}^{-14} \\ \text{ At }{100}^{\circ }\mathrm{C}\text{ (given) } \\ {\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=55\times {10}^{-14} \\ \because \text{ for a neutral solution } \\ \left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{OH}}^{-}\right] \\ \therefore {\left[{\mathrm{H}}^{+}\right]}^2=55\times {10}^{-14} \\ \text{ or }\left[{\mathrm{H}}^{+}\right]={\left(55\times {10}^{-14}\right)}^{1/2} \\ \because \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ \mathrm{On} \mathrm{taking} \log \mathrm{on} \mathrm{both} \mathrm{side} \\ -\log \left[{\mathrm{H}}^{+}\right]=-\log {\left(55\times {10}^{-14}\right)}^{1/2} \\ \mathrm{pH}=\frac{1}{2}-\log 55+14\log 10 \\ \mathrm{pH}=6.13 \end{gathered}$$