Why is water taken in calculating K, equilibrium constant

For the given question,
[![enter image description here](https://i.stack.imgur.com/c6Mil.jpg)](https://i.stack.imgur.com/c6Mil.jpg)


I was trying to solve it but I was not considering the concentration of water in the calculation (because water is a pure liquid and its concentration is supposed to be taken as 1), but then I needed the volume to be able to calculate the answer. I saw the solution,


[![enter image description here](https://i.stack.imgur.com/NSCiK.jpg)](https://i.stack.imgur.com/NSCiK.jpg)


Is it taken here because the reactants are organic and thus water would probably not be the solvent?




The question they gave you is somewhat flawed, in that there are four compounds and you don't know the range of their mutual solubility. 


To really answer the question, you would need information such as from [Solubility, liquid–liquid equilibrium and critical states for the quaternary system acetic acid–ethanol–ethyl acetate–water at 293.15 K](http://www.sciencedirect.com/science/article/pii/S0378381211004675) *Fluid Phase Equilibria* vol. 313, pages 46–51. 


The reactants, ethanol and acetic acid, are completely miscible; however, ethyl acetate only has limited solubility in water. 


When you evaluate the equilibrium constant, all the reacting compounds, including water, which are in a single phase, must be included. 


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All reactants and products are in the liquid phase and mixed together in one vessel, so the total volume at equilibrium, $V$, is the same for each concentration we need to find $K$. Also, all reactants and products are in one to one molar ratio in the balanced chemical equation.


Therefore, if we start with $\pu{3 mol}$ of each reactant, and reach equilibrium hen there are $\pu{2 mol}$ of the ester formed, we can be sure that there were also $\pu{2 mol}$ of water produced, and $\pu{2 mol}$ of each reactant onsumed, leaving only $\pu{1 mole}$ of each reactant in the vessel.


So, at equilibrium, the concentration of each reactant is $\frac{\pu{1 mol}}{V}$, and the concentration of each product is $\frac{\pu{1 mol}}{V}$, where $V$ = the magnitude of the total volume in completely arbitrary units, as it cancels out of the equation along with the units:
$$K=\frac{[(\frac{2}{V})\cdot{(\frac{2}{V})]}}{[(\frac{1}{V})\cdot{(\frac{1}{V})]}}= 4 $$


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