# Several tricky examples of equilibrium constant expression

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1. ok so we ignore [h2o] in dissociation constant but we include it in the expression for esterification

but why is that?

and what about hydrolysis?

2) here’s a reaction:

A(l) + B(l) –> C(g) + D(l)

textbook shows that the equilibrium constant expression for this reaction is K=[C]

why isn’t it K=([C][D])/([A][B])? would it be like that if A, B and D were (aq)?

3) let’s take dissociation constant for a weak HX acid – the [HX] is the total concentration of the acid or it’s the concentration of undissociated HX molecules?

In: Chemistry

The activity of pure solids and liquids is equal to 1. In aqueous solutions, we treat the solvent (water) as pure water. Thus, it has no effect on the quotient. This includes the dissociation of acids in water, as well as the most common examples of hydrolysis.

The same goes for the liquid phase reactants and product in the second part of your question. A, B, and D are pure liquid phases, so their activity is equal to 1. Consider that the force driving the reaction is the partial pressure of the gas.

In other words, your equations are right, but [H2O], [A], [B], and [D] are all equal to 1, so they cancel out.

In esterification, water is not the solvent. Esterification typically occurs in an organic solvent. Thus, H2O is not treated as pure water; rather, it’s a solute.

>the [HX] is the total concentration of the acid or it’s the concentration of undissociated HX molecules?

[HX] is the activity of HX in the equilibrium state, so it’s the latter.

All that said, I’m not a tenured professor. You’d be better off attending office hours if possible.