**Problem Set**

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 \(g/g\).

## Comments

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TopNewestWhy is \(x = 66.7g\)? – Daniel Lim · 2 years, 8 months ago

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\(x=200-2x\)

\(3x=200\)

\(x=66.7\) – Christopher Boo · 2 years, 8 months ago

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– Daniel Lim · 2 years, 8 months ago

Oh... ThanksLog in to reply

Thx for the information – Benjamin Setiawan · 6 months, 1 week ago

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What does "extract" mean? – Daniel Lim · 2 years, 8 months ago

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– Daniel Lim · 2 years, 8 months ago

ThanksLog in to reply

– Christopher Boo · 2 years, 8 months ago

To obtain a substance by chemical or mechanical action, as by pressure, distillation, or evaporation. (Google)Log in to reply

Literal Definition: to remove/separate a certain substance(useful) from another substance through various Physical and Chemical methods. – Anish Puthuraya · 2 years, 8 months agoLog in to reply

Thanks for the post! Never learned this back in secondary school. – Marissa Chong · 2 years, 8 months ago

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Thanks Christopher for this post which is completely new to me. – Soham Dibyachintan · 2 years, 9 months ago

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Peter Taylor-

I like the way you added the problems into the notes. However, I don't know how to add a line between the problem and the note. Can you teach me, please? – Christopher Boo · 2 years, 9 months ago

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Imgur

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– Christopher Boo · 2 years, 9 months ago

Thanks!Log in to reply

– Peter Taylor Staff · 2 years, 9 months ago

I also took the liberty of making you a picture. It's not the best, and it is a little bit of stretch, seeing as the post is mostly about partitioning and not about miscibility. If you don't like it, I can take the picture away.Log in to reply

– Christopher Boo · 2 years, 9 months ago

The picture is very nice, thanks!Log in to reply

nice.. :) – Joyan Florenz Delos Verges · 2 years, 9 months ago

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good post – Fauline St · 2 years, 9 months ago

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Good Post... – Anish Puthuraya · 2 years, 9 months ago

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