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The ion-product constant for water at 45 ˚C is 4.0 × 10^–14.What is the pH of pure...

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lak-86 | Student, Undergraduate | Salutatorian

Posted August 18, 2013 at 9:55 AM via web

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The ion-product constant for water at 45 ˚C is 4.0 × 10^–14.
What is the pH of pure water at this temperature?
(A) 6.7

(B) 7.0

(C) 7.3

(D) 13.4

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1 Answer | Add Yours

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted August 18, 2013 at 10:03 AM (Answer #1)

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The ion product for water is given as follows.

`K_w = [H^+][OH^-]`

Usually in pure water `[H^+]= [OH^-]`

Therefore;

`K_w = [H^+]^2`

`4xx10^(-14) = [H^+]^2`

`2xx10^(-7) = [H^+]`

`PH = -log[H^+]`

`PH = -log(2xx10^(-7))`

`PH = 7-log2`

`PH = 7-0.3010`

`PH = 6.7`

So the PH of water sample at 40C is 6.7. The correct answer is at (A)

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