Chemistry - Partition Law

Problem Set

Partition Law Problem 1

What will the scientists do if they want to extract and help purify a desired product from a reaction mixture? This is when the Partition Law(or Distribution Law) saves the day!

First, we set some variables:

Let $A$ and $B$ be two immiscible liquids(cannot dissolve or react with each other), such as water and oil.

Let the solubility of $X$ in liquid $A$ be $s_A$ .

Let the solubility of $X$ in liquid $B$ be $s_B$ .

Now, we do the following steps:

1) Mix solute $X$ into the mixture of liquid $A$ and $B$ .

2) Stir them constantly and wait till liquid $A$ and liquid $B$ are divided into two layers(they will as they cannot dissolve each other).

3) Now, solute $X$ is dissolved in both solvent $A$ and solvent $B$ but the concentration is different.

This is what Partition Law wants to deliver to us.

Definition: At a fixed temperature, when a solute is added to a system consisting of two solvent, it will distribute itself between the two solvent in proportion to its solubility in each of the two solvent separately.

If we express the definition in formula, it will be

$K=\frac{s_A}{s_B}$

where $K$ is the partition coefficient.

If at $40^\circ C$ , the solubility of $X$ is $50g/100g$ water and $100g/100g$ ether, then the partition coefficient

$K=\frac{s_{water}}{s_{ether}}=\frac{50g/100g}{100g/100g}=\frac{1}{2}$ .

Now there is an example:

If $1L$ water can dissolve $100g$ organic compound $X$ and I use $1L$ ether to extract it, what is the mass in grams of $X$ are left in the ether layer?

Details:

$K=\frac{s_{H_2O}}{s_{ether}}=\frac{1}{2}$

Solution:

Let the amount extracted be $x$ $g$ .

The solubility of $X$ in water is $s_{H_2O}=\frac{(100-x)g}{1L}$

The solubility of $X$ in ether is $s_{ether}=\frac{xg}{1L}$

Then,

$K=\frac{s_{H_2O}}{s_{ether}}$

$\frac{1}{2}=\frac{\frac{(100-x)g}{1L}}{\frac{(x)g}{1L}}$

$x=66.7$ $g$

So, the mass of $X$ that left in the ether layer is $66.7$ $g$ .

**Note that I use $g/L$ as the unit of solubility instead of $g/g$ in my note before.

Solubility of $5g/1L$ water means $1L$ water can dissolve a maximum of $5g$ of that solute.

Solubility of $5g/100g$ water means $100g$ water can dissolve a maximum of $5g$ of that solute.

So, the unit does not make any changes to the answer as long as we use the same unit. As the example above, we cannot use a unit of solubility written as $g/L$ while the other written as $g/g$ .

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Chemistry - Partition Law

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