In a similar experiment to the one you have performed a student is measuring the equilibrium...

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 9.9 mL of the base and the organic phase requires 4.76 mL of the base.

I can't figure out how to solve this problem. We did something similar in my lab but i can't remember. Show all work please.

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Expert Solution

solution:

(M_NaOH x total volume) / (volume required for aqueous layer ) = Molarity of aqueous layer [ i,e concentration in aqueous layer C1]

and ( M_NaOH x total volume ) / (volume required for organic layer ) = Molarity of organic layer [ i,e concentration in organic layer C2 ]

from above relation C _aqueous / C organic = K_d [ where Kd = partition coefficient ]

Kd = C1 /C2 = [ ( 0.12 x 14.66 ) / 9.9] / [ (0.12 x 14.66 ) / 4.76 ] = 0.17769 / 0.3695 = 0.4808

hence the equilibrium constant = distribution coefficient = 0.4808 this is the answer

queen_honey_blossom answered 1 year ago

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