Please help me visualize H2O autoionization concepts

  • #1
Firstly, I have just read that at 25degC, 1 out of every 10,000,000 H2O molecules breakes into [itex]H^{+}[/itex] and [itex]OH^{-}[/itex].

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:

[itex][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][/itex]

Thus 1 Liter of H2O= 3.3E25 H2O molecules

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

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

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

Are these calculations correct? Is it correct to divide 3.3E25/10E7 to find the number of H^{+} ions?
 
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Answers and Replies

  • #2
Borek
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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.
 
  • #3
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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  • #4
Borek
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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.
 
  • #5
-------------------------

Thanks just did that. Got 1 out of every 5.58x10^{9} H2O molecules will dissociate into H^{+} and OH^{-}

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

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.

--------------------------

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

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

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

For some reason, it feels as though if [itex]10^{-7}H[/itex] and [itex]10^{-7}OH[/itex] are dissociated, then logically there can only be [itex]10^{-7}[/itex] 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:

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

[itex]10^{-7}molH2O/L →10^{-7}molH/L + 10^{-7}molOH/L[/itex] ?
 
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  • #6
Borek
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Wrong again. It is not a sum:

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

This is so called water ion product.
 
  • #7
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Wrong again. It is not a sum:

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

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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