Problem 22 - Analytical Chemistry

Question

A solute, \(S\), has a distribution ratio between water and ether of \(7.5 .\) Calculate the extraction efficiency if we extract a 50.0 -mL aqueous sample of \(S\) using \(50.0 \mathrm{~mL}\) of ether as (a) a single portion of \(50.0 \mathrm{~mL} ;\) (b) two portions, each of \(25.0 \mathrm{~mL} ;(\mathrm{c})\) four portions, each of \(12.5 \mathrm{~mL} ;\) and \((\mathrm{d})\) five portions, each of \(10.0 \mathrm{~mL}\). Assume the solute is not involved in any secondary equilibria.

Step-by-Step Solution

Answer

(a) 88.24%; (b) 95.57%; (c) 98.54%; (d) 98.98%

Step 1: Understanding Distribution Ratio

The distribution ratio (\( D \)) is defined as the ratio of the concentration of a solute in the organic phase (ether) to its concentration in the aqueous phase (water). Here, \( D = 7.5 \). This implies that for every 1 part of solute in water, there are 7.5 parts in ether.

Step 2: Extraction Efficiency for Single Extraction

For a single extraction, the efficiency (\( E \)) is given by the formula \[ E = \frac{VD}{V_D + V_A} \times 100 \] where \( V_D \) and \( V_A \) are the volumes of the organic and aqueous phases, respectively. Both are 50.0 mL, so \[ E = \frac{50.0 \times 7.5}{50.0 \times 7.5 + 50.0} \times 100 = \frac{375}{425} \times 100 \approx 88.24\% \].

Step 3: Understanding Multiple Extraction

For multiple batch extractions, the efficiency involves considering sequential extractions. The remaining fraction after each extraction is fewer solutes staying in the aqueous phase: \( x = \left( \frac{V_A}{V_D + V_A} \right) \). The remaining fraction after \( n \) extractions is \( x^n \).

Step 4: Calculating Efficiency for Two Extractions

For two extractions of 25.0 mL each, \( V_D = 25.0 \text{ mL} \) for each extraction. The remaining fraction after one extraction is \( x = \frac{50.0}{25.0 \times 7.5 + 50.0} = \frac{50.0}{237.5} = 0.2105 \). The remaining fraction after two extractions is \( (0.2105)^2 \approx 0.0443 \). Thus, the efficiency \( E = (1 - 0.0443) \times 100 \approx 95.57\% \).

Step 5: Calculating Efficiency for Four Extractions

For four extractions of 12.5 mL each, \( V_D = 12.5 \text{ mL} \) for each extraction. \( x = \frac{50.0}{12.5 \times 7.5 + 50.0} = \frac{50.0}{143.75} \approx 0.3478 \). The remaining fraction after four extractions is \( (0.3478)^4 \approx 0.0146 \). Thus, the efficiency \( E = (1 - 0.0146) \times 100 \approx 98.54\% \).

Step 6: Calculating Efficiency for Five Extractions

For five extractions of 10.0 mL each, \( V_D = 10.0 \text{ mL} \) for each extraction. \( x = \frac{50.0}{10.0 \times 7.5 + 50.0} = \frac{50.0}{125.0} = 0.4 \). The remaining fraction after five extractions is \( (0.4)^5 \approx 0.0102 \). Thus, the efficiency \( E = (1 - 0.0102) \times 100 \approx 98.98\% \).

Key Concepts

Distribution Ratio

The distribution ratio, often symbolized as \( D \), is a fundamental concept in liquid-liquid extraction. It describes how a solute distributes itself between two immiscible phases. Typically, these phases are an organic solvent (like ether) and water. The calculation of \( D \) is straightforward: it's the concentration of the solute in the organic phase divided by its concentration in the aqueous phase.
When \( D = 7.5 \), it indicates for every 1 part of solute in water, there are 7.5 parts in ether. This ratio is crucial because it informs us about the solute's preference for the organic phase, which directly impacts the efficiency of the extraction.
Understanding \( D \) helps predict how much of the solute will stay in the organic phase after extraction. In extraction problems, knowing the distribution ratio allows us to determine how efficient the extraction will be, a key factor in designing and optimizing extraction processes.

Extraction Efficiency

Extraction efficiency measures how effectively a solute is removed from the aqueous phase into the organic phase in a liquid-liquid extraction process. It's crucial for evaluating the success of an extraction.
Efficiency is calculated using the formula:

• For a single extraction: \[ E = \frac{VD}{V_D + V_A} \times 100 \]

where \( V_D \) is the volume of the organic phase, \( V_A \) is the volume of the aqueous phase, and \( D \) is the distribution ratio.
In our example, if both volumes are 50 mL and \( D = 7.5 \), the extraction efficiency for a single extraction is approximately 88.24%. This value tells us the percentage of the solute that moves into the ether in one extraction step.
Higher efficiency in a single step means a significant amount of the solute transfers into the organic solvent, reducing the need for multiple extractions.

Multiple Extractions

Multiple extractions are used to increase overall extraction efficiency by performing several smaller sequential extractions rather than one large one. This technique takes advantage of the fact that each extraction reduces the concentration of the solute in the aqueous phase.
For each extraction step, the fraction remaining in the aqueous phase is calculated as follows:

• Remaining fraction after one extraction: \( x = \frac{V_A}{V_D \times D + V_A} \)

This formula shows the portion of the solute that is not transferred to the organic phase in a single step.
The remaining fraction after multiple extractions can be determined by raising this value to the power of the number of extractions \( n \). Thus, the formula becomes \( x^n \).
Each additional extraction exponentially increases the efficiency. For example, when using two 25 mL extractions, the efficiency rises to about 95.57%. For four or five smaller extractions, this increases to approximately 98.54% and 98.98%, respectively.
This method is particularly advantageous when the distribution ratio is moderate, allowing a more thorough extraction while using less solvent overall for each step.