At 250 degree C, the equilibrium constant for the dissociation of water is 10^ -11
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At 250 degree C, the equilibrium constant for the dissociation of water is 10^ -11

[From: ] [author: ] [Date: 12-11-12] [Hit: ]
pH = -log[H^+] = -log(3.16x10^-6) = 5.......
At this temperature, what is the pH of a neutral solution?

I need help ASAP!! Please respond if you can help!

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H2O(l) ⇌ H^+(aq) + OH^-(aq)

[H^+] = [OH^-] = x

Kw = [H^+][OH^-] = 1.0x10^-11

x² = 1.0x10^-11

x = [H^+] = 3.16x10^-6 M

pH = -log[H^+] = -log(3.16x10^-6) = 5.5
1
keywords: 10,250,dissociation,equilibrium,water,is,of,degree,for,constant,At,11,the,At 250 degree C, the equilibrium constant for the dissociation of water is 10^ -11
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