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Ionic product of water at 310 K is 2.7xx...

Ionic product of water at `310 K` is `2.7xx10^(-14)`. What is the `pH` of netural water at this temperature?

Text Solution

Verified by Experts

`K_w=[H_3O^+][OH^-]=2.7xx10^(-14)`
For neutral water, `[H_3O^+]=[OH^-]`
`[H_3O^+]^2 =2.7xx10^(-14)`
`[H_3O^+]=1.643xx10^(-7)`
`pH=-log (1.643xx10^(-7))`
=-(0.2156-7)=6.75
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