A fundamental Acid Base question.

[Kane]

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Before dissociation, 1L of water has is 55.6 M (1L of water weighs 1000g; 1 mol water weighs 18g; 1000/18 = 55.6). After dissociation, it's...

55.6 - 0.0000001

therefore:

K = 10^-14/55.6


I don't understand the "55.6 - 0.0000001" part.

Shouldn't it be: 55.6 - 2(0.0000001) ? 55.6 moles of water molecules minus 10^-7 H+ ions and 10^-7 OH- ions?

TIA

Kane

 

[Kane]

A little clarification:

The point of the example being that the molarity of water, clearly, isn't changed much by the dissociation of such a small # of particles.

But shouldn't the number of dissociated particles in the example be doubled to: 2(0.0000001) = 0.0000002 to include both H+ and OH-

 

[Shunwei]

A little clarification:

The point of the example being that the molarity of water, clearly, isn't changed much by the dissociation of such a small # of particles.

But shouldn't the number of dissociated particles in the example be doubled to: 2(0.0000001) = 0.0000002 to include both H+ and OH-

well, one molecule of water dissociates to form the two ions, so accounting for both with just 1 x 10-7 is correct. Remember both ions come from the same parent species.

 

[Kane]

i think you broke through my confusion.

we are only "losing" .0000001 moles of water molecules.. that then go on to break into .0000001 moles of H+ and .0000001 moles of OH-.

i was going to say something nasty about chemistry, but i'll be good.

thanks muchly.

 

[IsRaEl21]

My advice to u is to get the destroyer and read the explanations there are many very good acid base problems. After the destroyer practice some challenging achiever question. Believe me u will Ace ur chemistry after that.