Bridging the gap between waters selfionizatio and Le Chatelier


by christian0710
Tags: bridging, chatelier, selfionizatio, waters
christian0710
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#1
Feb17-14, 07:00 PM
P: 156
Hi i have a question regarding the connection between Le chatelier and Waters self Ionization. I'd really appreciate if someone could give me heads up if my understanding is correct, or point me in the right direction if I'm off. Here we go:

We know that water self-ionizes from the following equilibrium reaction
2H2O = OH(-) + H3O(+)

And the equilibrium constant is
[OH][H+]=kw=10^-14

So letís assume we add 1mol HCl to 1Liter water:

According to the equilibrium constant the concentration of OH(-) is
[OH]=Kw/[1mol] =10^-14

So from this, can we conclude the following (Is this correctly understood?) :
1) In very acidic solutions (1mole of HCl) waters self-ionization still occurs and the same amount of OH(-) is still produced as in a neutral solution BUT a big amount of the added H3O(+) reacts with OH to form water.
2) According to Le chatelier, by adding 1 mole of HCl, Waters self-ionization equilibrium should shift to the left (due to the high concentration of H+) such that only 10^-14 [OH] Is produced from the reaction between water molecule for the Ksp to be 10^-14. This means that waters self-ionization happens on a SMALLER scale so only 2x10^-14 water molecules split per unit time, and not 2*10^-7 such as at pH=7,.

So the equilibrium constant always remains constant, but unlike most equilibrium expressions involving both reactants and products, the concentrations of reactions at equilibrium do not need to be equivalent - [OH]=10^-14 and [H+]=1Molar which are not at all equivalent but they still make the equilibrium constant true.
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Borek
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#2
Feb18-14, 03:09 AM
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Quote Quote by christian0710 View Post
So the equilibrium constant always remains constant
Correct.

but unlike most equilibrium expressions involving both reactants and products, the concentrations of reactions at equilibrium do not need to be equivalent
You can safely assume concentrations of products are (almost) never identical. This is only approximation that works for a simple systems.

Other than that nothing cries out loud "you are wrong!".
christian0710
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#3
Feb18-14, 03:26 AM
P: 156
Thank you so much!! I'm glad i finally understand this :D


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