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Suppose solute A has a distribution coefficient of 1.0 between water and diethyl ether. Demonstrate that if 100 mL of a solution of 5.0g of A in water were extracted with two 25mL portions of ether, a smaller amount of A would remain in the water than if the solution were extracted with one 50-mL portion of ether.

Respuesta :

Manetho

Answer:

Explained

Explanation:

a) performing extraction of 25 ml of ether each time

K= 1= [tex]\frac{\frac{5-x}{25} }{x/100}[/tex]

solving we get x= 4 g dissolved in water

5-x=1 g dissolved in ether

second extraction

K= 1 = [tex]\frac{\frac{4-x}{25} }{x/100}[/tex]

solving this we get x= 3.2g dissolved in water

4-x= 3.2 g dissolved in ether

combining the two extractions 0.97 g of A extracted into ether that is 97% is extracted

b) performing one extraction 50% ether

distribution coefficient , K is the concentration of solute dilute to that to the solvent water

K= 1.0= [tex]\frac{\frac{5-x}{25} }{x/100}[/tex]

solving we get

x= 3.33 g dissolved in water

and 5-x= 1.67 dissolved in ether

ajeigbeibraheem

The explanation below shows that we have successfully demonstrated that the smaller amount of A would remain in the water if 100 mL of a solution of 5.0g of A in water were extracted with two 25mL portions of ether than if the solution were to be extracted with one 50-mL portion of ether.

From the given information:

Let consider the solubility of ether and solubility in water = S

This is because the distribution coefficient of 1.0 shows us that the solute is equally soluble in both solvents.

Similarly;

  • Suppose the volume of ether is represented by = [tex]\mathbf{V_e}[/tex]
  • the volume of water = [tex]\mathbf{V_w}[/tex]
  • the total mass amount for the substance = 5.0 g

Then, the mass amount in the water can be computed as:

[tex]=\mathbf{5\Big( \dfrac{V_w}{V_e+V_w} \Big)}[/tex]

Also, the mass amount in the ether can be computed as:

[tex]=\mathbf{5\Big( \dfrac{V_e}{V_e+V_w} \Big)}[/tex]

Now, if we start with the second extraction of one 50 mL portion of ether, we have:

[tex]=\mathbf{5\Big( \dfrac{50}{50+100} \Big)}[/tex]

[tex]=\mathbf{5\Big( \dfrac{50}{150} \Big)}[/tex]

= 1.67 grams in ether.

The amount of water remaining from this extraction is:

= (5.00 - 1.67) grams

= 3.33 grams in water

In the extraction of two 25-mL portions of ether;

we will first do the 1st extraction followed by the 2nd;

i.e.

[tex]=\mathbf{5\Big( \dfrac{25}{25+100} \Big)}[/tex]

[tex]=\mathbf{5\Big( \dfrac{25}{125} \Big)}[/tex]

= 1 grams in ether

The amount of water remaining in this extraction is (5 - 1)= 4 grams.

Now for the 2nd extraction for 25 mL portion, the mass of the water has reduced to 4 grams

∴

[tex]=\mathbf{4\Big( \dfrac{25}{25+100} \Big)}[/tex]

[tex]=\mathbf{4\Big( \dfrac{25}{125} \Big)}[/tex]

= 0.8 grams in ether.

The total grams in ether for the extraction of two 25-mL portions of ether;

= (1.0 + 0.8) grams

= 1.8 grams in ether

The remaining amount of grams in water will be:

= 5.0 g - 3.8 grams

= 3.2 grams in water

Therefore, we can conclude that we have successfully demonstrated that the smaller amount of A would remain in the water if 100 mL of a solution of 5.0g of A in water were extracted with two 25mL portions of ether than if the solution were to be extracted with one 50-mL portion of ether.

Learn more about Laboratory techniques here:

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