You have 150 ml of a solution of benzoic acid in water estimated to contain about 5 g of acid. The distribution coefficient of benzoic acid in benzene and water is approx 10. Calculate the amount of acid that would be left in the water solution after three 50 ml extractions with benzene. Do the same calculation using one 150 ml extraction with benzene to determine which method is more efficient.

My ans:
k = (concentration of benzoic acid in benzene)/(benzoic acid conc in water) = 10

But I am not sure how to proceed from this. Any help appreciated

Kd = (concn organic phase)/(concn aqueous phase) = 10

Let x = g in aqueous phase.
Then 5-x = g in benzene phase.
10 = [(5-x)/50 mL]/(x/150 mL)
Solve for x which will give you the amount (in grams) in the aqueous phase for the first extraction.I have approximately 1.15 g. For the second extraction, the organic phase starts with 1.15 so it becomes 1.15-x and the water is x, go through it again. Etc.

I found approximately 0.062 g in the water phase after three extractions. I didn't do the one extraction with a larger amount BUT it should not extract as well as three smaller ones.

To calculate the amount of acid that would be left in the water solution after three 50 ml extractions with benzene, you can use the distribution coefficient (k) and the initial amount of acid.

1. Calculate the initial moles of benzoic acid in the water solution:
- The molar mass of benzoic acid is approximately 122.12 g/mol.
- So, the initial moles of benzoic acid in the water solution can be calculated as:
moles = mass / molar mass
moles = 5 g / 122.12 g/mol

2. Determine the initial moles of benzoic acid in the water solution:
- Since the distribution coefficient (k) is defined as the ratio of the concentration of benzoic acid in benzene to its concentration in water, we can calculate the initial concentration of benzoic acid in water solution:
concentration = mass / volume
concentration = 5 g / 150 ml
= (5 g / 150 ml) * (1000 ml / 1 L)
= 33.3 g/L

3. Determine the amount of acid remaining in the water solution after each extraction:
- In the first extraction, the concentration of benzoic acid in the organic phase (benzene) will be:
benzene concentration = k * water concentration
= 10 * 33.3 g/L
= 333.3 g/L

- The total moles of benzoic acid that moved to the benzene phase after the first extraction can be calculated as:
moles extracted = (benzene concentration) * (volume of benzene)
= 333.3 g/L * 50 ml
= (333.3 g/L) * (50 ml) * (1 L / 1000 ml)
= 16.665 g

- Therefore, the moles of benzoic acid remaining in the water solution after the first extraction can be calculated as:
remaining moles = initial moles - moles extracted
= (5 g / 122.12 g/mol) - (16.665 g / 122.12 g/mol)

- Repeat this calculation for the second and third extraction, using the remaining moles from the previous extraction.

To compare the efficiency of the three 50 ml extractions with a single 150 ml extraction, you can perform the same calculations for the 150 ml extraction using the same steps.

Hope that helps! Let me know if you have any further questions.

To calculate the amount of acid that would be left in the water solution after three 50 ml extractions with benzene, we need to consider the distribution of benzoic acid between benzene and water.

Let's assume that x grams of benzoic acid are extracted in each extraction step. After the first extraction, the distribution of acid between benzene and water can be calculated as follows:

Concentration of benzoic acid in the water solution before extraction:
= (5 g)/(150 ml) = 0.03333 g/ml

Concentration of benzoic acid in the water solution after the first extraction:
= (0.03333 g/ml) - (x g)/(50 ml)
= (1.5 g - x)/(50 ml)

Concentration of benzoic acid in benzene after the first extraction:
= (10)(x g)/(50 ml)
= (10x)/(50 ml) = (x/5) g/ml

Similarly, after the second extraction, the distribution of acid will be:

Concentration of benzoic acid in the water solution after the second extraction:
= (1.5 g - x)/(50 ml) - (x g)/(50 ml)
= (1.5 - 2x)/(50 ml)

Concentration of benzoic acid in benzene after the second extraction:
= (10)(x g)/(50 ml) + (x/5) g/ml
= (11x/5) g/ml

Finally, after the third extraction:

Concentration of benzoic acid in the water solution after the third extraction:
= (1.5 - 2x)/(50 ml) - (x g)/(50 ml)
= (1.5 - 3x)/(50 ml)

Concentration of benzoic acid in benzene after the third extraction:
= (10)(x g)/(50 ml) + (11x/5) g/ml
= (13x/5) g/ml

The amount of acid left in the water solution after three extractions can be determined by subtracting the total amount of acid extracted (3x) from the initial amount (5 g):

Amount of benzoic acid left in the water solution after three extractions
= 5 g - 3x

To determine the efficiency of the extraction methods, we can perform a similar calculation for a single 150 ml extraction with benzene:

Concentration of benzoic acid in the water solution after a single extraction:
= (5 g - x)/(150 ml)

Concentration of benzoic acid in benzene after a single extraction:
= (10x)/(150 ml) = (x/15) g/ml

The amount of acid left in the water solution after a single extraction can be determined by subtracting the amount of acid extracted (x) from the initial amount (5 g):

Amount of benzoic acid left in the water solution after a single extraction
= 5 g - x

Now, compare the amount of benzoic acid left in the water solution after three 50 ml extractions (5 g - 3x) with the amount left after a single 150 ml extraction (5 g - x). The method with the smaller amount left in the water solution is more efficient.

Please note that this calculation assumes ideal conditions and perfect extraction. In reality, there may be some loss of benzoic acid during the extraction process.