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Ionic product of water(Kw) is given by products of molar concentrations of hydronium ions and hydroxide ions, that is, [H+][OH-].

Ionic product of water is the product of dissociation constant of water and molar concentration of water, which is 55.5 Molar. Now for a general acid dissolved in water, HA(aq)----> H+ (aq) + A- (aq) Dissociation equilibrium constant(Ka) is given by

([H+][A-])/[HA]

by law of equilibrium. But if this reaction is written in another way, HA(aq) + H2O(liq) ----> H3O+(aq) + A- (aq)

Then the equilibrium constant (Ka) expression changes and becomes Ka= ([H3O+][A-])/([HA][H2O]) That is, an extra term of concentration of H2O appears in the denominator. Now concentration of H+ and H3O+ is same for obvious reasons. But [H2O] is (55.5)molar. So the values of equilibrium constant from the two different expressions will be different. Why is this so? What is wrong? Just writing the reaction with an extra H2O term on reactants side, shouldn't be changing its dissociation constant, right?

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  • $\begingroup$ Have a look for 'activities' in your textbook, how they are related to concentrations and the explanation about their values for solids and pure liquids. $\endgroup$
    – porphyrin
    Commented Oct 15 at 9:05
  • $\begingroup$ I know that activity of pure liquids and solids is taken as 1. But in that case, ionic product of water should simply be equal to equilibrium constant of its dissociation(Ka). But its not the case. Ionic product(Kw)= (Ka of water) * [H2O] and [H2O]=55.5M $\endgroup$
    – NPC
    Commented Oct 15 at 9:26
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    $\begingroup$ This question is similar to: What is the real pKa of water?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. $\endgroup$
    – Andrew
    Commented Oct 15 at 11:36
  • $\begingroup$ Ka=a(H+).a(A-)/(a(HA).a(H2O)) = a(H+).a(A-)/(a(HA), Ka,c=[H+][OH-]/[HA] // Kw=a(H+).a(OH-)/a(H2O) = a(H+).a(OH-), Kw,c=[H+][OH-]. $\endgroup$
    – Poutnik
    Commented Oct 15 at 11:42

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Remember that chemistry is a human construct; nature doesn't tell us how to do chemistry, we just come up with conventions that we think make the most sense. Conventionally (by which I mean, essentially always), liquids and solids are excluded from equilibrium constant expressions because "adding more" liquid or solid usually doesn't affect equilibrium (from a kinetics perspective, it doesn't really make the reactions happen faster). (Look up some equilibrium constant expressions and you will see what I mean.) I guess if you found a way to increase the concentration of water, that would make reactions with water in the rate law faster, but that's not something that's typically done in a laboratory; usually, "adding more" of a liquid just involves pouring more of a liquid into a container, which doesn't actually affect the concentration of the liquid. Adding more of a gas or more of an aqueous solute to a container, however, actually increase the concentration of that gas or solute, so it does affect kinetics and equilibrium, so gases and solutes are part of the equilibrium constant expression.

The $K_w$ of water is conventionally given as $10^{-14}$, which is simply $\ce{[H3O+]*[OH-]}$. It doesn't involve the concentration of water.

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