How Does Distribution Coefficient Affect Solvent Extraction Efficiency?

In summary, the conversation discusses an extraction efficiency problem involving the distribution coefficient between dichloromethane and water for solute A. The first part of the conversation calculates the weight of A that would be removed from a solution of 10 g of A in 100 mL of water by a single extraction and by three successive extractions with 33 mL portions of dichloromethane. The second part of the conversation discusses how much dichloromethane would be required to remove 98.5% of A in a single extraction.
  • #1
dingol
3
0
Extraction efficiency problem PLease help...

Homework Statement



The distribution coefficient, k = (conc. in CH2Cl2/conc in H2O), between dichloromethane and water for solute A is 7.3.

What weight of A would be removed from a solution of 10 g of A in 100 mL of water by a single extraction with 100 mL of dichloromethane?

What weight of A would be removed by three successive extractions with 33 mL portions of dichloromethane?

How much dichloromethane would be required to remove 98.5% of A in a single extraction?
 
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  • #2
is this method right>>?>??

Basic numbers :

10 g of A total
100 ml of H2O, 100 ml of dichloromethan ( Iwilll abbreviate as dich.)
10 g of A in 100 ml of h2o by a single extraction w/ 100 mL of dich.?
if (x) g = the amount that stays in the water, 7.3(x)g of A must go into dich. phase.

Wt. of A in Dich. = 100 ml * 7.3(x)g/ml = 730 (x)g
Wt. of A in H2O = 100ml * (x) g/ml = 100(x)g

Total weight of A = 10 g (now partitioned between two layers)

add 100 (x) g + 730(x)g = 830 (x)g

830 (x)g = 10g
(x) = .0120481928

now, plugging x back into the equations:

wt. of A in Dich. = 730 (.0120481928)g = 8.7965 g

wt. of A in H20 = 100 (.0120481928)g = 1.205 g
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Pt. 2: what weight og A owuld be removed by 3 successive extractions with 33 ml portions of Dich.?

Everything is the same, except instead of one extraction there will be 3 extractions with 33 mL portion s of dich. instead of 100 mL.

Wt. of A in dich. = (33mL)(7.3)(x)g/mL = 240.9 (x)g
Wt. of A in H20 = 100mL (x)g/ml = 100 (x) g

Total weight = 10 g

after adding 100 x + 240.9 x = 340.9 x

340.9(x)g = 10 g

x= 0.0293341156

after substituting back into equations:

Wt. of A in dich. = (33mL)(7.3)(.0293341156)g/mL = 6.824697g
Wt. of A in H20 = 100mL (.0293341156)g/ml = 2.83341156 g

DIch. later is removed. and second round begins

Wt. of A in dich. = 240.9g
wt. of A in H20 = 100g

total wieght = 2.833 g

340.9 (x)g = 2.833g

(x) = .0083103549

Weight of A in Dich. = 240.9 (.0083103549) = 1.99947g
weight of A in H2O = 100 (.0083103549) = .86103549g

Now layer is removed and third round begins

weight of A in Dich. = 240.9
Wt. of A in water = 100

total weight = 340.9

there fore,

340.9 x = .83g

x= .00243g

wt. of A in dich = 240.9 * .00243g = .5855387g
wt. of A in H2o = 100 * .00243g = .243g

total % weight = 6.825 + 1.999+ .5855 = 9.04095 from 10 g

(9.04095 / 10) * (100%) = 94%
 
  • #3
continuation


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Pt 3:

How much dich. would be required to remove from 98.5% of A in a single extraction?

98.5% = 9.85g of total weight of A in dichloro

wt. of A in Dich = (x)(y) = .985g

x= x
y = orig wt. of dich.
z = x*y

z= 10g

wt. of A in Dich. = y * 7.3x = y7.3 x
wt. of A in water = 100mL * 7.3 x = 730x


to be continued...

 

Related to How Does Distribution Coefficient Affect Solvent Extraction Efficiency?

1. What is an extraction efficiency problem?

An extraction efficiency problem refers to the difficulty in extracting a desired substance from a mixture or solution with minimal loss or contamination.

2. What factors can affect extraction efficiency?

The efficiency of extraction can be affected by various factors such as the solubility of the substance, the type and amount of solvent used, temperature, and the nature of the substance being extracted.

3. How can I improve extraction efficiency?

To improve extraction efficiency, one can adjust the experimental conditions such as using a more suitable solvent, optimizing the extraction time, and controlling the temperature and pH of the solution. Additionally, using specialized equipment and techniques such as sonication, centrifugation, and solid-phase extraction can also improve efficiency.

4. What are the consequences of low extraction efficiency?

Low extraction efficiency can result in inaccurate or inconsistent results, as well as a waste of time, resources, and materials. It can also lead to the loss of valuable substances and affect the overall success of the experiment.

5. How can I troubleshoot extraction efficiency problems?

To troubleshoot extraction efficiency problems, one can review the experimental procedure, check for any errors or inconsistencies, and consider adjusting the experimental conditions. It may also be helpful to consult with other scientists or seek guidance from literature or online resources.

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