Question

In a similar experiment to the one you have performed a student is measuring the equilibrium for an organic acid that is soluble in water and in cyclohexane, a nonpolar organic solvent. When shaken with the solution of water and cyclohexane, it distributes between the organic and aqueous phase. The two phases are separated and titrated with 0.12 M NaOH. Calculate the equilibrium constant for an acid where the titration of the aqueous phase requires 6.1 mL of the base and the organic phase requires 5.84 mL of the base.

Answer 1

This is my attempt of solution, cause I have a little doubt about it, but I think this is the right answer.

According to M1V1 = M2V2

M₁ = 0.12 M; V₁ = 6.1 mL

M₂ = X; V₂ = 5.84 mL

(0.12 * 6.1) = (5.82)M₂

M₂ = 0.1258 M

Equilibrium constant (K_c) = Concentration of Organic phase / Concentration of aqueous phase

= 0.1258 / 0.12

Kc = 1.0483

If there's something to be fixed, or something else you need to understand tell me in a comment and I'll help you out.

Hope this helps