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At $100^{\circ} \mathrm{C}$ the $\mathrm{K}_w$ of water is 55 times its value at $25^{\circ} \mathrm{C}$. What will be the $\mathrm{pH}$ of neutral solution? $(\log 55=1.74)$
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Verified Answer
The correct answer is:
6.13
$$
\begin{aligned}
\mathrm{H}^{+} & =\sqrt{55 \times 10^{-14}} \\
\mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right] \\
\mathrm{pH} & =-\log \left[\sqrt{55 \times 10^{-14}}\right] \\
& =\frac{1}{2}\left[-\log \left(55 \times 10^{-14}\right)\right] \\
& =6.13
\end{aligned}
$$
\begin{aligned}
\mathrm{H}^{+} & =\sqrt{55 \times 10^{-14}} \\
\mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right] \\
\mathrm{pH} & =-\log \left[\sqrt{55 \times 10^{-14}}\right] \\
& =\frac{1}{2}\left[-\log \left(55 \times 10^{-14}\right)\right] \\
& =6.13
\end{aligned}
$$
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