Change in H+ ion concentration of water on boiling [closed]

I know that on boiling water, pH scale decreases but water remains neutral. At 100°C the pH=6.14 is neutral. And at this temperature also, the H+ ion concentration and OH- ion concentration are equal. My question is what happens to the H+ ion concentration on boiling relative to the concentration at 25°C? Does it increase or decrease or remain same?

closed as unclear what you're asking by Mithoron, A.K., andselisk♦, Todd Minehardt, TyberiusApr 15 at 19:10

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• It seems like you answered your own question: as you said, at 100 °C the $\mathrm{pH} = 6.14$, and $[\ce{H+}] = 10^{-\mathrm{pH}}$. – andselisk Apr 14 at 8:29
• @andselisk You mean [H+] increases, but then [OH-] too increases equally. So in this case how is it that decrease in pH implies increase in [H+] as now the pH scale itself has changed? – suhridi sen Apr 14 at 9:43
• I'm afraid I don't understand what you are trying to imply. The pH scale itself is a superficial term (e.g. typical claims that pH can only take 0 - 14 values and pH 7 is always neutral) to the point it's almost a meme. – andselisk Apr 14 at 9:56
• So according to your previous comment, [H+] increases as pH=6.14 . But I don't want to get that one word answer, I would rather like to have an explanation for my question i.e., how does [H+] changes with change in 'pH scale'(not pH I repeat)? – suhridi sen Apr 14 at 12:15

It is an interesting question, however there is a misconception. You are giving too much attention to the word neutral. The word neutral is just a qualitative version. Numbers and equations are more accurate way to discuss science.

You have to see how water's $$\pu{pH}$$ at 100 °C is calculated. How do you get those numbers? Lets us do it algebraically, so that the word neutral does not matter.

Autoionization of water at temperature T:$$\ce{ H2O <=> H+ + OH-}$$, if the conc. of $$\ce{H+}$$ is x, the quantity of $$\ce{OH-}$$ is also x, if you have accepted the ionic product of water is

$$K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = x^2= 51.3\times{10^{-14}}$$

Solve for $$x = [\ce{H+}] => -log [\ce{H+}] = 6.14$$

So the numbers prove that $$\pu{pH}$$ has changed. Water has become slightly acidic. If you associate the word neutral = 7.0 at 25 °C. However, if you use the word neutral with respect to water temperature, then $$\pu{pH}$$ of 6.14 is neutral at 100 oC. Ultimately, the numbers Kw and pH matter, nobody will quibble about the word neutral in serious research.

Water is the most amazing solvent in Nature. People are still exploring its properties and it is yet to be fully understood. For example, I just heard in a conference that someone was trying to understand what happens on the surface of micron sized water droplets. "Pure" water showed very odd chemical effects which have never been shown before.

Read this well written Wikipedia article: https://en.wikipedia.org/wiki/PH#cite_note-2

• – Mithoron Apr 14 at 16:52