Distribution law lesson 1 of 2

describes distribution law of a solute between immiscible solvents and calculated examples . For A-level UACE Exams.



## Distribution Law: Understanding Solute Partitioning Between Immiscible Solvents

The **distribution law**, also known as **Nernst's partition law**, describes the **equilibrium distribution of a solute between two immiscible solvents**. In simpler terms, it explains how a solute will distribute itself between two non-mixing liquids at a constant temperature.

**Key Points:**

* **Immiscible solvents:** These are liquids that don't dissolve in each other, like oil and water.
* **Solute:** The substance that dissolves in both solvents.
* **Equilibrium:** A state where the concentration of the solute in each solvent remains constant over time, even though the molecules continue to move between the layers.

**The Law:**

The distribution law states that the **ratio of the equilibrium concentrations of a solute in two immiscible solvents is constant at a constant temperature**. This constant ratio is called the **distribution coefficient** or **partition coefficient**, denoted by **K_d**.

**Mathematically:**

```
K_d = C₁ / C₂
```

where:

* C₁ is the concentration of the solute in solvent 1
* C₂ is the concentration of the solute in solvent 2

**Factors Affecting Kd:**

* **Nature of the solute:** Solutes with more affinity for one solvent will have a higher concentration in that solvent, leading to a larger Kd value.
* **Nature of the solvents:** The polarity and ability of solvents to interact with the solute influence the distribution.
* **Temperature:** Kd can change slightly with temperature, although the change is often negligible for most A-Level applications.

**Applications:**

* **Extraction:** Separating a desired compound from a mixture by selectively dissolving it in one solvent and separating the layers.
* **Chromatography:** Utilizing the differing distribution behavior of components to separate them in a mixture.
* **Understanding drug action:** Predicting how drugs distribute between different compartments in the body based on their lipophilicity (affinity for fats).

**Calculated Examples:**

**Example 1:**

A solution of benzoic acid is shaken with water and benzene. After reaching equilibrium, the concentration of benzoic acid in the benzene layer (0.1 M) is found to be ten times higher than its concentration in the water layer (0.01 M). Calculate the distribution coefficient (Kd) for benzoic acid between benzene and water.

```
K_d = C_benzene / C_water = 0.1 M / 0.01 M = 10
```

**Example 2:**

A dye has a Kd value of 5 between chloroform and water. If 10 mg of the dye is dissolved in 10 mL of chloroform, how much dye will be present in 100 mL of water after reaching equilibrium?

* We can assume the initial concentration of the dye in chloroform (C_chloroform) is 1 mg/mL (10 mg / 10 mL).
* We need to find the equilibrium concentration in water (C_water).

```
K_d = C_chloroform / C_water
5 = 1 mg/mL / C_water
C_water = 0.2 mg/mL
```

Therefore, at equilibrium, 0.2 mg/mL x 100 mL = 20 mg of the dye will be present in the water layer.

**Remember:** These are just basic examples. A-Level UACE Exams might involve more complex scenarios and calculations related to distribution law, requiring a deeper understanding of the concepts and their applications.