The ion-product constant for water at 45 ˚C is 4.0 × 10^–14.
What is the pH of pure water at this temperature?
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The ion product for water is given as follows.
`K_w = [H^+][OH^-]`
Usually in pure water `[H^+]= [OH^-]`
`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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