(D) At $25\, ^oC$,$K_w = 10^{-14}$.At $100\, ^oC$,$K_w = 55 \times 10^{-14}$.For a neutral solution,$[H^+] = [OH^-] = \sqrt{K_w}$.$[H^+] = \sqrt{55 \t

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(D) At $25\, ^oC$,$K_w = 10^{-14}$.At $100\, ^oC$,$K_w = 55 \times 10^{-14}$.For a neutral solution,$[H^+] = [OH^-] = \sqrt{K_w}$.$[H^+] = \sqrt{55 \times 10^{-14}} = \sqrt{55} \times 10^{-7}$.$pH = -\log [H^+] = -\log (\sqrt{55} \times 10^{-7}) = -[\frac{1}{2} \log 55 - 7]$.$pH = 7 - \frac{1}{2} \log 55 = 7 - \frac{1.74}{2} = 7 - 0.87 = 6.13$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Water hydrolysis and pH scale

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At $100\, ^oC$,the $K_w$ of water is $55$ times its value at $25\, ^oC$. What will be the $pH$ of a neutral solution at this temperature? $(\log 55 = 1.74)$

NEET 2013 All NEET

A
$7$
B
$7.87$
C
$5.13$
D
$6.13$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Water hydrolysis and pH scale

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At $100\, ^oC$,the $K_w$ of water is $55$ times its value at $25\, ^oC$. What will be the $pH$ of a neutral solution at this temperature? $(\log 55 = 1.74)$

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