Please help me visualize H2O autoionization concepts

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At 25°C, approximately 1 in every 5.58 billion water molecules dissociates into H+ and OH- ions, contrary to the initial claim of 1 in 10 million. The concentration of H+ and OH- in neutral water is 10^-7 M, leading to about 6.02 x 10^16 dissociated molecules per liter. The discussion clarifies that the ion product of water (K_w) is 10^-14, which results from multiplying the concentrations of H+ and OH- rather than summing them. This highlights the importance of understanding the relationship between dissociation and concentration in water chemistry. The calculations and concepts presented emphasize the need for precision in interpreting ionization in water.
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Firstly, I have just read that at 25degC, 1 out of every 10,000,000 H2O molecules breakes into H^{+} and OH^{-}.

If this true, then would this be correct:

Assuming no ionization occurs yet..

If 1 liter of H2O = 1000grams, and 1mol H2O = 18.015g, then by Dimensional Analysis:

[1L*\frac{1000g}{1L}=1000g H2O] => [1000g H2O \frac{1mol}{18.015g}=55.5mol H2O] =>[55.5mol H2O * \frac{6.022E23}{1mol} = 3.3E25 H2O molecules]

Thus 1 Liter of H2O= 3.3E25 H2O molecules

Then, since 1 out of every 10,000,000 H2O molecules ionizes into H^{+} and OH^{-}, then:

\frac{3.3x10^{25}H2Omolecules}{10^{7}}= 3.3x10^{18} H^{+}ions

That number, 3.3x10^{18} just feels like it's disproportionately too many H^{+}

Are these calculations correct? Is it correct to divide 3.3E25/10E7 to find the number of H^{+} ions?
 
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What you did looks OK to me, but the information about 1 out of 107 molecules dissociating is not. Somebody got it wrong - concentration of H+ (and OH-) in the neutral water at 25°C is 10-7 M, that's not equivalent to 1 molecule in 107 being dissociated.
 
This is where I saw that particular 1 out of 10E7 value:

I stopped watching about a minute into the video when she said 10^7 moles of H2O on one side of the equation and then on the other side of the equation she had 10^{-7) or something to that affect. I couldn't see how those numbers were calculated.
 
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Not that difficult to check by yourself. Concentration of H+ is 10-7 M which means in a liter of water 10-7 moles of water dissociated. That's 10-7*6.02*1023=6.02*1016 molecules. There are 1000/18*6.02*1023=3.34*1025 molecules in 1 L of water - just divide these numbers to see one in how many dissociated.
 
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Thanks just did that. Got 1 out of every 5.58x10^{9} H2O molecules will dissociate into H^{+} and OH^{-}

I also did \frac{55.5molH2O}{10^{-7}}= 5.5x10^{9}

Since both values are the same, I take it that for the first result, I can say 1 molecule out of every 5.5E9 molecules is dissociate. And guess that for the second result, I can say 1 mol out of 5.5E9 moles will dissociate.

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If the above is correct, there is another aspect that is not making much sense.

There is this equation 10^{-14}M =10^{-7}M +10^{-7}M. Does this mean that out of 1 Liter of H2O there will be 10^{-14} H2O molecules that dissociate into H and OH?

If so, is it correct to say that out of 55.5 moles of H2O, there will be 10^{14}moles which dissociate.

For some reason, it feels as though if 10^{-7}H and 10^{-7}OH are dissociated, then logically there can only be 10^{-7} original H2O molecules that could have dissociated.

What I think I mean is, if you have 10 H2O molecules, then you can only have 10 H and 10 OH molecules in total. If you split all 10 H2O molecules, you get 10 H and 10 OH. So,

10(H2O) →10(H) + 10(OH), Not 10(H2O) →5(H)+5(OH). So why then do we have:

10^{-14}molH2O/L → 10^{-7}molH/L + 10^{-7}molOH/L instead of

10^{-7}molH2O/L →10^{-7}molH/L + 10^{-7}molOH/L ?
 
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Wrong again. It is not a sum:

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

This is so called water ion product.
 
Borek said:
Wrong again. It is not a sum:

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

This is so called water ion product.

In fact, the product should even have units -- 10–14 M2 -- which makes it even clearer that the two concentrations must be multiplied together
 

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