A biochemist is trying to isolate a natural product that is a weak monoprotic ac

Question

A biochemist is trying to isolate a natural product that is a weak monoprotic acid. The pKa value of this compound is estimated to be 6.21 at 25degreeC. Preliminary experiments indicate that this compound has a distribution ratio of 75 between water and ethyl acetate at pH = 3.00. The conjugate base does not show any appreciable extraction at any pH. What fraction of this compound will be extracted from a 30.0 mL water sample into 20.0 mL of ethyl acetate at pH = 4.00? What fraction of the compound that enters the ethyl acetate at pH = 4.00 will go back into a fresh 30.0 mL portion of water at pH = 8.00? What overall fraction of the compound in the original sample will be present in the fresh portion of water after the biochemist has performed the back extraction in part b)?

Explanation / Answer

5.

pKa = 6.21

Ka = 6.166 x 10^-7

pH = 3

[H+] = 1 x 10^-3

Kd = Kp(1/(1 + H+/Ka)

75 = Kp(1/(1 + 0.003/6.166 x 10^-7))

Kp = 3.65 x 10^5

a. Kd at pH 4

= 3.65 x 10^5(1/(1 + 0.0004/6.166 x 10^-7)))

= 2237

So with 3 ml water and 20 ml ethylacetate

2237 = (x/20)/((1 - x)/30)

74.567 - 74.567x = 0.05x

x = 0.999 extracted

b. fraction that goes back at pH 8

Kd = 3.65 x 10^5(1/(1 + 1 x 10^-8/6.166 x 10^-7)))

= 359175

So with 3 ml water and 20 ml ethylacetate

359175 = (0.999 - x)/20((x)/30)

11972.5x = 0.999 - x

x = 8.34 x 10^-5 goes back in water layer

c. compound left in water = 0.0001

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.