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Only recently I learned that the pH of water varies with temperature, from about 7.47 at 0 °C to 6.14 at 100 °C.

My question is, first, is there a geometric explanation for this, in the sense that water can perhaps arrange itself in a way that partial bonding of oxygen to a third hydrogen can happen without simultaneously producing hydroxide ions to balance things out?

And--related--is the spread of pH above enough to create a detailed profile for the variation of pH with temperature? If so is it linear?

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    $\begingroup$ Note that the change in pH does not necessarily mean that the solution is not neutral, only that the dissociation increases at higher temperature. $\endgroup$ – Zhe Nov 28 '17 at 18:04
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    $\begingroup$ basic means $\ce{[OH^-] > [H^+]}$, acidic means that $\ce{[H^+] > [OH^-]}$, and neutral means $\ce{[OH^-] = [H^+]}$ $\endgroup$ – MaxW Nov 28 '17 at 18:31
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    $\begingroup$ Water isn't really unique in that sense. The pKa of acids and the pKb of bases will vary with temperature. We just normally assume everything is at "room" temperature. For water curve see en.wikipedia.org/wiki/Water_(data_page)#Self-ionization $\endgroup$ – MaxW Nov 28 '17 at 18:33
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    $\begingroup$ Re: basicity means [OH−]>[H]+], i.e., a preponderance of [OH−] in a "neutral" solution. No if $\ce{[OH^-] > [H^+]}$ then the solution is basic. A neutral solution has $\ce{[OH^-] = [H^+]}$. But because of the temperature variation the autoionization constant of water isn't a "constant" but changes with temperature. But again for all temperatures pure water will be neutral and it will have $\ce{[OH^-] = [H^+]}$. $\endgroup$ – MaxW Nov 28 '17 at 19:48
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    $\begingroup$ Is the downvote to punish me for deleting comments?! Tough crowd! $\endgroup$ – daniel Nov 28 '17 at 20:06
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To summarize... in aqueous solutions:

  • Basic means $\ce{[OH−] > [H+]}$
  • Acidic means that $\ce{[H+] > [OH-]}$
  • Neutral means $\ce{[OH−] = [H+]}$

There is a for the Wikipedia article for the autodissociation of water. The chemical equation for the autodissociation is:
$$\ce{H2O <=> H^+ + OH^-}$$
The mathematical relationship between $K_d^t$, $\ce{[H^+]}$, $\ce{[OH^-]}$ and $t$ is: $$K_d^t =\ce{[H+][OH-]}\quad\text{for }t > 32\text{ C}$$ Note that $K_d^t$ changes with temperature $t$, but for pure water at any temperature, $t$, the water will be neutral, and $\ce{[OH−] = [H+]}$. The temperature dependency of $K_d^t$ also means that pH + pOH is not always 14.

The Wikipedia article shows that the relationship between $K_d^t$ and temperature is a curve. (The curve shown is a bit odd in that water boils at 100 C, so higher temperatures must be taken under pressure.)

Also regardless of temperature the $\text{pH}$ of an aqueous solution is defined as $\text{pH} = -\log{\ce{[H^+]}}$ so the pH does not depend on $\ce{[OH^-]}$.

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  • $\begingroup$ This is a good summary of the comment discussion. Can you also address the main question of the post? Do we know whether the change in pH is linear over that interval? +1. $\endgroup$ – daniel Nov 29 '17 at 4:11
  • $\begingroup$ Oh--I do have a question--the pH does not depend on $[OH^-]$? But doesn't pH+pOH=14? That is an explicit dependence. $\endgroup$ – daniel Nov 29 '17 at 4:29
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    $\begingroup$ @Danial - No, that is the point. pH + pOH is not always 14 for a neutral solution since $K_d^t$ changes with temperature. If $\ce{[H^+] = [OH-]}$ then $K_d^t = [H^+]^2$ and $pK_d^t = 2 \times \text{pH}$ $\endgroup$ – MaxW Nov 29 '17 at 4:58
  • $\begingroup$ Thanks, somehow I missed the picture. This is I think a good (and complete) answer. Appreciate your patience with my questions! Will accept this very shortly. $\endgroup$ – daniel Nov 29 '17 at 5:06

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