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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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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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