Calculating the Distribution Coefficient of A in Ligroin and Water

In summary: If yes, how many mL of ligroin are needed to extract a certain weight of A in a given volume of water?
  • #1
Mivz18
44
0
Problem:

The distribution coefficient, k = (conc. in ligroin/ conc. in water), between ligroin and water for solute A is 7.5 . What weight of A would be removed from a solution of 10g of A in 100mL of water by a single extraction with 100 mL of ligroin? What weight of A would be removed by four successive extractions with 25mL portions of ligroin? How much ligroin would be required to remove 98.5% of A in a single extraction?

I'm having so much trouble with this problem. I plugged in the numbers giving to find the first part. After calculations, I found out that depending on the way you divide and multiply to find the weight removed, I'm assuming that 750 grams of A is removed. However, I'm not sure, not even sure if I'm doing this right. ANY SUGGESTIONS PLEASE??
 
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  • #2
I remember that a formula called Jensen's formula is used to calculate this kind of problems. It involves logarithmic expressions. Please try to find it by yourself, as I lost the source. However, it will surely help you.

Best wishes.
 
  • #3
Mivz18 said:
Problem:

The distribution coefficient, k = (conc. in ligroin/ conc. in water), between ligroin and water for solute A is 7.5 . What weight of A would be removed from a solution of 10g of A in 100mL of water by a single extraction with 100 mL of ligroin? What weight of A would be removed by four successive extractions with 25mL portions of ligroin? How much ligroin would be required to remove 98.5% of A in a single extraction?

I'm having so much trouble with this problem. I plugged in the numbers giving to find the first part. After calculations, I found out that depending on the way you divide and multiply to find the weight removed, I'm assuming that 750 grams of A is removed. However, I'm not sure, not even sure if I'm doing this right. ANY SUGGESTIONS PLEASE??

You have 10g of A in 100mL water + 100 mL ligroin.

Let the amount of A in water be x. Then the amount in ligroin would be 7.5x (since the volume of ligroin is the same as that of water). The total amount, 8.5x is equal to 10g. This tells you what x is. And x is the required number since it the the weight of A left in water. Another important number is the fraction of A extracted, which in this case is 1/8.5

Can you try the second part, where the volume of water is different from the volume of ligroin ?
 

FAQ: Calculating the Distribution Coefficient of A in Ligroin and Water

What is the distribution coefficient?

The distribution coefficient, also known as the partition coefficient, is a measure of the relative solubility of a substance in two immiscible phases, typically water and an organic solvent. It is calculated by dividing the concentration of the substance in one phase by the concentration in the other phase.

Why is it important to calculate the distribution coefficient?

The distribution coefficient is important in various fields of science, such as chemistry, biochemistry, and environmental science. It can provide information about the solubility and partitioning behavior of a substance, which can be useful in predicting its behavior in different environments and designing experiments.

How do you calculate the distribution coefficient?

The distribution coefficient, denoted as Kow, is calculated by dividing the concentration of the substance in the organic phase (ligroin) by the concentration in the aqueous phase (water). This can be determined experimentally or by using mathematical models.

What affects the distribution coefficient?

The distribution coefficient can be affected by various factors, including pH, temperature, and the nature of the substance and the two phases. For example, a substance may have different distribution coefficients in different solvents or at different pH levels.

How is the distribution coefficient measured experimentally?

The distribution coefficient can be measured experimentally by conducting a two-phase extraction, where the substance is added to the two immiscible phases and allowed to partition between them. The concentrations of the substance in each phase can then be measured and used to calculate the distribution coefficient.

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