The value of K (@ 25 degree C) for compound A when partitioned between water and

Question

The value of K (@ 25 degree C) for compound A when partitioned between water and methylcyclopentane is 4.79. Starting with a solution of 32.9 g of compound A in 865 mL of water what quantity of compound A (in grams) would remain in the water layer after two successive extractions with methylcyclopentane each using 375 mL of methylcyclopentane? Please carry maximum number of significant figures through to final answer then round to one decimal place. Do not include units. The value of K_p (@ 25 degree C) for compound B when partitioned between water and benzene is 11.3. Starting with a solution of 215 g of compound B in 745 mL water what volume of benzene (in mL) would be required (in the context of a single extraction protocol) to extract 75.5 g of compound B from the original aqueous solution? Please carry maximum number of significant figures through to final answer then round to one decimal place. Do not include units. Consider the following flowchart and list of compound

Explanation / Answer

Question 1:

In this K is known as partition or distribution coefficient. It is defined as the concentration ratio of a chemical between the two media at equilibrium. It is has no units and is temperature dependent.

The partition will involve the distribution of a solute between two immiscible liquid phases.

There is continual interchange of solute between the liquid layers via the interface i.e. a dynamic equilibrium is formed.

Here in this Problem

In the first extraction

Let x = grams of A in Methylcyclopentane
32.9-x = grams of A in water

Korg/aq=4.79

Make a note that the volumes of both the phases are not equal.

Korg/aq = (x/375)/[(32.9-x)/865]=4.79

            =2.31x/(32.9-x)=4.79

            =4.79(32.9-x)=2.31x

        x= 22.19gm

So in the first extraction the amount of A remained in water = 32.9-22.19 = 10.71gm

In the second extraction

Let x = grams of A in Methylcyclopentane
10.71-x = grams of A in water

Korg/aq=4.79

Make a note that the volumes of both the phases are not equal.

Korg/aq = (x/375)/[(10.71-x)/865]=4.79

            =2.31x/(10.71-x)=4.79

            =4.79(10.71-x)=2.31x

        x= 7.23g

The amount of A in Methylcyclopentane = 7.23 gm

The amount of A in Water = 10.71-7.23 = 3.48 gm

So the remaining amount of A in water after second extraction with Methylcyclopentane is 3.5 gm.

Question 2:

Let the volume of benzene required to pull 75.5gm of B is x.

Then Korg/aq = (75.5/x)/((215-75.5)/745) = 11.3

            11.3 = (75.5*745)/(x*139.5)

                 x = 35.682ml

The required volume of benzene to extract 75.5gm of B is 35.7 ml

Question 3:

Box 1 contains G,D as those two are polar and hydrophilic.

Question 4:

Box 2 contains only C as it is nonreactive with aq.HCl and is lipophilic so stays in organic phase.

Question 5:

Box 3 contains E as G becomes E upon treatment with Aq.NaOH.

Question 6:

Box 4 contains F as D reacts with Aq.NaOH

Question 7:

Box 5 contains none of the above.

Question 8:

Box 6 contains C as it is nonreactive with Aq.NaHCO3

Question 9:

Box 7 contains none of the above.

Question 10:

Box 8 contains none of the above.

Box 5,7 and 8 contains none as the C is Nonreactive with Aq.NaHCO3.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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