Problem 16 - Analytical Chemistry

Question

For a solute with a distribution ratio of \(25.0,\) show by calculation which is more effective, extraction of \(10 \mathrm{~mL}\) of an aqueous solution with \(10 \mathrm{~mL}\) organic solvent or extraction with two separate 5.0 -mL portions of organic solvent.

Step-by-Step Solution

Answer

Two separate 5.0-mL extractions are more effective.

Step 1: Understanding Distribution Ratio

The distribution ratio (\(D\)) is the ratio of the concentration of the solute in the organic solvent to its concentration in the aqueous phase. Here, \(D = 25.0\). This means that for every unit of solute in the aqueous phase, 25 units will be in the organic phase.

Step 2: Single Extraction Calculation

Calculate the amount of solute extracted in one step with \(10 \, \text{mL}\) of organic solvent. If \(V_a\) and \(V_o\) are the volumes of the aqueous and organic phases respectively, we use: \[ E_1 = \frac{D \, V_o}{D \, V_o + V_a}\]\(V_a = 10 \, \text{mL}\), \(V_o = 10 \, \text{mL}\), so:\[E_1 = \frac{25 \, \times \, 10}{25 \, \times \, 10 + 10} = \frac{250}{260} = \frac{25}{26}\]This is the fraction of solute extracted in a single 10 mL extraction.

Step 3: Two-Step Extraction Calculations

For two extractions with separate portions of \(5.0 \, \text{mL}\) organic solvent, compute extraction yield for each step. First step: \[E_1 = \frac{25 \, \times \, 5}{25 \, \times \, 5 + 10} = \frac{125}{135} = \frac{25}{27}\]Second step uses the remaining solute: \[E_2 = \frac{25 \, \times \, 5}{25 \, \times \, 5 + 10} = \frac{125}{135} = \frac{25}{27}\]Remaining solute after step one (fraction \( \frac{2}{27} \)) gets extracted: \(E_{total} = 1 - (1 - E_1)^2 = 1 - (\frac{2}{27})^2 = 1 - \frac{4}{729}\) or approximately \(\frac{725}{729}\).

Step 4: Comparing Extraction Yields

Compare the yields: a single extraction retrieves \(\frac{25}{26}\) = \(0.9615\), and two consecutive extractions retrieve \(\frac{725}{729}\) \(\approx 0.995\). Two separate extractions are more effective as more solute is extracted.

Key Concepts

Distribution Ratio

The distribution ratio, often abbreviated as "D," is a key concept in liquid-liquid extraction, an important technique in analytical chemistry. It represents the ratio of the concentration of a solute in the organic phase to its concentration in the aqueous phase. For instance, if the distribution ratio ( D ) is 25, this indicates that the solute prefers the organic phase highly compared to the aqueous phase.

In essence, the distribution ratio helps in determining how much of the solute will transfer into the organic phase during an extraction process. A higher value of D (like 25.0) signifies that a larger portion of the solute is in the organic layer.

• This concept is crucial for designing extraction procedures.
• It helps predict the efficiency and success of the extraction.

It simplifies the process by reducing the need for trial and error: when you know D , you can calculate the expected outcome of the extraction process.

Extraction Efficiency

Extraction efficiency is a measure of how effective an extraction process is at removing a solute from one liquid phase to another. In the context of liquid-liquid extraction, it is crucial to understand which conditions improve this efficiency.

The calculation of efficiency involves the distribution ratio as seen in the equation: \( E = \frac{D \times V_o}{D \times V_o + V_a} \)\( \). Here, \( V_o \) is the volume of the organic phase and \( V_a \) is the volume of the aqueous phase.

• A single step with 10 mL of organic solvent shows an efficiency of \( \frac{25}{26} \) or about 96%.
• Two steps with 5 mL portions enhance efficiency, reaching 99.5%.

This demonstrates that multiple extractions with smaller volumes can significantly increase the efficiency. With better use of solvent, more of the solute is extracted from the aqueous phase.

Analytical Chemistry

Analytical chemistry is a branch of chemistry focused on the composition of materials. This field employs various techniques, like liquid-liquid extraction, to separate, identify, and quantify different chemical components in a sample.

In the exercise example, knowing the distribution ratio and calculating extraction efficiency allows chemists to effectively separate components in mixtures by maximizing the amount of solute that can be extracted:

• It helps in designing optimal procedures for separating substances.
• This includes adjusting volumes and steps to achieve the best results.

This application reflects the broader goals of analytical chemistry — ensuring precision, accuracy, and efficiency in chemical analysis.

Solvent Extraction

Solvent extraction is a pivotal method in both industrial and laboratory settings used to separate components based on their solubility in different immiscible liquids. The process often involves partitioning a solute between an aqueous phase and an organic phase.

The choice of solvent and understanding the distribution ratio plays a vital role in its success.

• It relies on fine-tuning solvent type and quantity.
• Effective phase separation yields purified products.

In laboratory settings as presented in the exercise, solvent extraction allows for efficient retrieval of desired materials from complex mixtures. Understanding how factors like distribution ratio impact the process aids in optimizing the conditions for maximum yield.