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Question
Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?
Solution
Ionic product,
`"K"_"w" = ["H"^+] ["OH"^-]`
Let `["H"^+] = x`
Since `["H"^+] = ["OH"^(-)], "K"_"w" = x^2`
`=> "K"_"w" "at 310K" " is" 2.7 xx 10^(-14)`
`therefore 2.7 xx 10^(-14) = x^2`
`=> x = 1.64 xx 10^(-7)`
`=> ["H"^+] = 1.64 xx 10^(-7)`
`=> "pH" = -log["H"^+]`
`= - log [1.64 xx 10^(-7)]`
= 6.78
Hence, the pH of neutral water is 6.78.
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