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Ionic product of water at 310 K is 2 point 7 multiplied to 10 raise to power minus 14 What is the pH of neutral water at this temperature

7.65     Ionic product of water at 310 K is 2.7 × 10-14. What is the pH of neutral water at this temperature?

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M manish

We have the ionic product of water at 310 K is 2.7 \times 10^{-14}
It is known that,
ionic product K_w = [H^+][OH^-]

SInce [H^+]=[OH^-], therefore K_w = [H^+]^2

\Rightarrow K_w at 310 K is 2.7 \times 10^{-14}
\therefore K_w = 2.7\times 10^{-14} = [H^+]^2
 here we can calculate the value of [H^+] concentration.

[H^+] = \sqrt{2.7 \times 10^{-14}} = 1.64 \times 10^{-7}

Thus, p^H = -\log[H^+]
                   = -\log(1.64\times 10^{-7})
                  = 6.78

Hence the p^H of neutral water is 6.78

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