Question

The equilibrium constant for the esterification of ethanol and aceticacid in an aqueous solution CH3COOH(aq) + C2H5OH(aq) -----> CH3COOC2H5(aq) + H2O(l) amounts to 0.566 at 298 K. The reaction mixture initially contains 0.040 mol dm^−3 of acetic acid and 0.025 mol dm^−3 of ethanol. Find themolar concentration of ethyl acetate at the equilibrium. Assume that the pH is low enough that CH3COOH is not dissociated at all and thatthe activity of water equals 1 just like for pure liquid water (after all,the solution is quite dilute). Hint:This problem requires you to keep an extra large number ofsignificant digits in all intermediate calculations: otherwise, your finalanswer will be very inaccurate. Do not forget to check your answer in the end.

Answer 1

CH3COOH(aq) + C2H5OH(aq) ---------------------> CH3COOC2H5(aq) + H2O(l)

The activity of water is equal to 1 and CH3COOH doesn't disintegrate at lower pH value

Initial concentration of CH3COOH = 0.040 M

Initial concentration of C2H5OH = 0.025 M

Let the final concentration of CH3COOC2H5 be x

So, the final concentration of CH3COOH will be (0.040-x) and C2H5OH will be (0.025-x)

K = [Products]/[Reactants] = [CH3COOC2H5]/([CH3COOH][C2H5OH]) = x/((0.040-x)(0.025-x)) = 0.566 [water will be ignored since activity is 1]

Solving the quadratic equation we get

x/(x^2-0.065x+0.001) = 0.566, so the final value of x will be 0.00054608

The molar concentration of ethyl acetate at equilibrium will be 0.00054608M