Understanding pH - FAQs for the PF Forum

[Stephanus]

Dear PF Forum,
Actually I want to study how antioxidant can benefit for our health. But, my understanding in chemistry is very limited. Perhaps I have to go back and study pH, first.
I have read many sources in the internet.
And this is one of them:
http://antoine.frostburg.edu/chem/senese/101/acidbase/faq/what-is-pH.shtml

...where log is a base-10 logarithm and [H+] is the concentration of hydrogen ions in moles per liter of solution

So, let me understand this first.
1. In 1 liter of water, there are approximately 1000/18.016 (weight of H₂O, including isotope) = 55.51 moles of H₂O. Is it true?
2. For 1 liter water which have pH = 0 (if this is chemically possible) there are 1 moles H⁺ about 1 .008 gram?
2. In 1 liter water pH = 1, there are 1/10 moles or 0.1 gram H⁺?
3. Neutral water, pH = 7, there are 1 x 10^-7 moles H⁺ or about 0.1 microgram H⁺?
4. If number 3 is true. Why is it neutral? Because there are 1.7 microgram Hydroxide as well?

Thanks for helping me. I might have some questions more to ask.

 

[BvU]

You've got it !
1.yes
2.yes. we wouldn't call it water any more because of it extreme acidity ... and because usually we also require a charge balance there have to be a lot of negative ions present too.
2b.yes, same.
3.yes
4.yes

It all boils down to an equilibrium between $H_2O \leftrightarrows H^+ + OH^-$ in the liquid. At room temperature the equilibrium constant $k = {[H^+] \times [OH^-] \over [H_2O]} \approx 10^{-7}$.
[edit] big mistake, see Borek #5 below (thanks !). Should be $\approx 10^{-14}$
There are some minor details, but you've got the idea.

 

[Stephanus]

Thanks BvU for the answer.

Okay, I'll requote the web again.
http://antoine.frostburg.edu/chem/senese/101/acidbase/faq/what-is-pH.shtml

- where log is a base-10 logarithm and [H+] is the concentration of hydrogen ions in moles per liter of solution
- A pH of 7 is considered "neutral",
- because the concentration of hydrogen ions is exactly equal to the concentration of hydroxide (OH-) ions produced by dissociation of the water

If pH is the concentration of hydrogen ions per liter, why pH 7 is neutral?
There are 0.1 microgram H⁺ in it, right?
Answer: Because there are 1.7 microgram OH as well.

Okay..., perhaps I should ask again.
What if in 1 liter water, there are free H⁺ about 0.1 microgram but the amount of OH^- is not 1.7 microgram, but 0.85 microgram.
What is its pH?
Thanks for any answer.

 

[Martin0001]

Thanks BvU for the answer.

Okay, I'll requote the web again.
http://antoine.frostburg.edu/chem/senese/101/acidbase/faq/what-is-pH.shtml
If pH is the concentration of hydrogen ions per liter, why pH 7 is neutral?
There are 0.1 microgram H⁺ in it, right?
Answer: Because there are 1.7 microgram OH as well.

Okay..., perhaps I should ask again.
What if in 1 liter water, there are free H⁺ about 0.1 microgram but the amount of OH^- is not 1.7 microgram, but 0.85 microgram.
What is its pH?
Thanks for any answer.

You need to remember that water need to be electrically neutral, so in *pure* water such situation in normal circumstances is *not* possible.
Should other anions like Cl(-) be also present, to keep stuff neutral, such water would become slightly acidic.
Equlibrium would be achieved and most of OH(-) would recombine with part of H(+) to reconstitute water.
Some excess of H(+) would be left.

 

[Borek]

What if in 1 liter water, there are free H+ about 0.1 microgram but the amount of OH- is not 1.7 microgram, but 0.85 microgram.

Ain't going to happen. BvU already mentioned why - there is an equilibrium between these things, given by

H⁺ + OH^- ↔ H₂O

And (approximately)

[H^+][OH^-] = 10^{-14}

(we ignore concentration of water as it is constant in most cases, and the value of 10^-7 posted by BvU seems to be a typo). If you try to introduce amounts of H⁺ and OH^- that don't fit this equation they will react till the equilibrium is reached.

Neutral solution doesn't mean pH = 7, actually it means [H⁺] = [OH^-]. This is typically close to 7, but the exact pH of the neutral solution depends on the temperature - see the table here: http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product

 

[Stephanus]

Ahh, so there you are. Borek. I would like to thank Ruud Gullit, in "Give Credits to Mentor", but I forgot your user name.
Okay, let me summarize your reply.
A. There is an equilibrium between these things, given by H⁺ + OH^- ↔ H₂O
B. And (approximately) [H⁺][OH^-] = 10^-14
So, let me ask again.
This Equilibrium is not [H⁺] = [OH^-], but [H⁺][OH^-] = 10^-14
Supposed in 1 litre water there's
1 * 10^-4 moles H⁺
1 * 10^-6 moles OH^-
So the equilibrium is not H⁺ - OH^-
That H⁺ decreases by 0.000001 becomes 0.000099 and OH^- decreases by 0.000001 all depleted, but.
([H⁺] - n) ([OH^-] - n) = 10^-14
Using ABC equation formula, I find that n = 0.00000099989899.
So H⁺ becomes: 0.00009900010101
OH^- becomes: 0.00000000010100999795
Multiplying those numbers: 0.00009900010101 * 0.00000000010100999795 = 10^-14
Is equilibrium like this?

C. Concentration of water it is constant in most cases
D If you try to introduce amounts of H⁺ and OH^- that don't fit this equation they will react till the equilibrium is reached.
E. Neutral solution doesn't mean pH = 7, actually it means [H⁺] = [OH^-].

Please see B section: Is that how the equilibrium reached?
Thanks for any help.

 

[Borek]

([H⁺] - n) ([OH^-] - n) = 10^-14

Yes, you got it right.