water at 70°C? (Kw at 70°C = 1.53 Ⓡ 10-13) What is the pOH of pure...

Question

Question

Answer 1

Answer #1

H2O (l) <................> H+ (aq) + OH- (aq)

Kw = [H+][OH-]

or

1.53 * 10^-13 = [H+][OH-]

at equilibrium, [H+] = [OH-]

thus

1.53 * 10^-13 = [OH-]^2

or

[OH-] = 3.91 * 10^-7 M

pOH = - log[OH-]

or

pOH = - log (3.91 * 10^-7 )

or

pOH = 6.408

thus

option B) 6.408 is the answer.

Answer 2

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