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I am using alcohol to make an ester product. It reached an equilibrium, of which I calculated an equilibrium constant. I need help with the below question: Suppose that the alcohol you selected has a...

I am using alcohol to make an ester product. It reached an equilibrium, of which I calculated an equilibrium constant. I need help with the below question:

Suppose that the alcohol you selected has a small but significant solubility in water, while the ester has absolutely no solubility in water. Would this cause the calculated equilibrium constant to be too large or too small? Refer explicitly to the equilibrium constant expression in your answer.

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Answer #1

Since alcohol is soluble in water, so it will be a aqueous solution and ester is insoluble in water, so it will be treated like precipitate (solid) with concentration of unity i.e. 1.

For solids, concentration is always 1. Now,

Alcohol(aq) ---------> ester(s)

Equilibrium constant (Kc) = [ester]/[alcohol]

Since [ester] = 1 and it is mentioned that alcohol is small soluble. So, [alcohol] is small. Let say x.

Then, Kc = 1/x

Since x is a small number so, 1/x will be large number. So,

Kc will be large.

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