3.5 The Partition Coefficient and Distribution Law

3.5 The Partition Coefficient

When a solute is introduced to a system containing two immiscible liquids, it distributes itself between the two solvents. This distribution reaches a dynamic equilibrium, where the rate at which the solute moves from the first solvent to the second equals the rate at which it moves from the second to the first.

Consider two immiscible liquids, such as ether (less dense, forms the upper layer) and water. If a solute $X$ that dissolves in both liquids is added to this system and shaken vigorously, the solute will distribute itself between the two phases. At equilibrium, the concentration of solute $X$ in each layer becomes constant.

The distribution of solute $X$ between water and ether can be represented by the following equilibrium:

$\mathrm{X}\text{ (in water)}\rightleftharpoons \mathrm{X}\text{ (in ether)}$

The equilibrium constant for this distribution is known as the partition coefficient and is denoted by $K_{pc}$ or $K_p$. It is defined as the ratio of the concentration of the solute in the organic (ether) layer to its concentration in the aqueous (water) layer.

${\mathrm{K}}_{\mathrm{PC}}=\frac{[\mathrm{X}\text{ in ether}]}{[\mathrm{X}\text{ in water}]}$

Nernst Distribution Law

The concept governing the distribution of a solute between two immiscible solvents is known as the distribution law or Nernst distribution law. This law states that at a constant temperature, a solute distributes itself between two immiscible solvents in contact with each other in such a way that the ratio of its concentrations in the two solvents is constant, provided the solute is in the same molecular state in both solvents.

The partition coefficient ($K_{pc}$) is therefore defined as: The ratio of the concentrations of a solute in two different immiscible solvents in contact with each other when equilibrium has been established at a particular temperature.

Since the partition coefficient is a ratio of two concentrations (e.g., ${\mathrm{mol}\mathrm{dm}}^{-3}/{\mathrm{mol}\mathrm{dm}}^{-3}$), it is a dimensionless quantity and thus has no units. Its value is constant at a constant temperature.

3.5.1 Calculation of Partition Coefficient

The partition coefficient can be calculated using the equilibrium expression when the solute exists in the same physical state in both solvents. This principle is widely used in solvent extraction, a technique often used to separate organic products from aqueous reaction mixtures.

Worked Examples

ACTIVITY 3.1: Calculating the partition coefficient

Procedure:

1. Measure $100{\mathrm{cm}}^3$ of a $0.150\mathrm{M}$ solution of aqueous methylamine ($\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$) and add it into a separatory funnel.
2. Add $75{\mathrm{cm}}^3$ of an organic solvent into the separatory funnel.
3. Shake the separatory funnel gently but thoroughly to allow the solute to distribute between the two solvents. Allow the system to reach equilibrium (about 5-10 minutes).
4. Let the layers settle and separate the two solvents into different beakers.
5. Measure the concentration of the solute in each solvent.

Calculations:

$100{\mathrm{cm}}^3$ of a $0.150{\mathrm{mol}\mathrm{dm}}^{-3}$ solution of aqueous methylamine ($\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$) is shaken with $75.0{\mathrm{cm}}^3$ of an organic solvent. After equilibrium, $25{\mathrm{cm}}^3$ of the aqueous layer is run off and titrated against $0.225\mathrm{M}\mathrm{HCl}$. $7.05{\mathrm{cm}}^3$ of $\mathrm{HCl}$ was used.

Data for calculations:

1. Volume of aqueous methylamine solution: $100{\mathrm{cm}}^3$
2. Initial concentration of methylamine: $0.150{\mathrm{mol}\mathrm{dm}}^{-3}$
3. Volume of organic solvent: $75.0{\mathrm{cm}}^3$
4. Volume of aqueous layer titrated: $25.0{\mathrm{cm}}^3$
5. Volume of $\mathrm{HCl}$ used for titration: $7.05{\mathrm{cm}}^3$
6. Concentration of $\mathrm{HCl}$: $0.225{\mathrm{mol}\mathrm{dm}}^{-3}$

Goal: Calculate the partition coefficient of methylamine in the organic solvent and water.

Solution:

Step 1: Write down the equilibrium equation.

$\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_{2(\mathrm{aq})}\rightleftharpoons \mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2\text{ (organic solvent)}$

Step 2: Write down the ${\mathrm{K}}_{\mathrm{pc}}$ expression.

${\mathrm{K}}_{\mathrm{pc}}=\frac{[\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2\text{ (organic layer)}]}{[\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_{2(\mathrm{aq})}]}$

Step 3: Determine the total moles of methylamine in the original solution.

1. Given Values:

   • Initial concentration of $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2=0.150\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$
   • Initial volume of $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$ solution $=100\mathrm{c}{\mathrm{m}}^3=0.100\mathrm{d}{\mathrm{m}}^3$

2. Apply Formula: Moles ($n$) = Concentration ($M$) $\times$ Volume ($V_{d{m}^3}$)

3. Calculation: $\text{Total moles }\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2=0.150\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}\times 0.100\mathrm{d}{\mathrm{m}}^3=0.015\mathrm{mol}$

Step 4: Determine the number of moles of methylamine in the aqueous layer after equilibrium.

The reaction during titration is: $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_{2(\mathrm{aq})}+\mathrm{HC}{\mathrm{l}}_{(\mathrm{aq})}\to \mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_3\mathrm{C}{\mathrm{l}}_{(\mathrm{aq})}$ The stoichiometry is $1\mathrm{mol}$ of $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$ reacts with $1\mathrm{mol}$ of $\mathrm{HCl}$.

1. Calculation:
   • Moles of $\mathrm{HCl}$ reacted = $0.225\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}\times 0.00705\mathrm{d}{\mathrm{m}}^3=0.00158625\mathrm{mol}$
   • Moles of $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$ in $25\mathrm{c}{\mathrm{m}}^3$ aqueous layer $=0.00158625\mathrm{mol}$.
   • Moles of $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$ in the entire $100\mathrm{c}{\mathrm{m}}^3$ aqueous layer (after equilibrium) = $\frac{0.00158625\mathrm{mol}}{25\mathrm{c}{\mathrm{m}}^3}\times 100\mathrm{c}{\mathrm{m}}^3=0.006345\mathrm{mol}$

Step 5: Determine the number of moles of $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$ present in the organic layer.

$\text{Moles }\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2\text{ (organic layer)}=0.015\mathrm{mol}-0.006345\mathrm{mol}=0.008655\mathrm{mol}$

Step 6: Change the number of moles into concentrations.

• Concentration ($\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$ in aqueous layer) = $\frac{0.006345\mathrm{mol}}{0.100\mathrm{d}{\mathrm{m}}^3}=0.06345\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$
• Concentration ($\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$ in organic layer) = $\frac{0.008655\mathrm{mol}}{0.075\mathrm{d}{\mathrm{m}}^3}=0.1154\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$

Step 7: Substitute the values into the ${\mathrm{K}}_{\mathrm{pc}}$ expression.

$K_{PC}=\frac{0.1154\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}}{0.06345\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}}\approx 1.82$ Since the value of $K_{pc}$ is larger than 1, methylamine is more soluble in the organic solvent than in water.

3.5.2 Factors Affecting the Numerical Value of Partition Coefficient

The partition coefficient ($K_{pc}$) quantifies how a solute distributes between two immiscible phases. Several factors can influence its numerical value:

1. The Polarity of the Solute:

   • Polar solutes generally have a greater affinity for polar phases (e.g., water) due to favorable dipole-dipole interactions or hydrogen bonding. Conversely, non-polar solutes prefer non-polar phases.

2. The Solvent Polarity:

   • The nature of the two immiscible solvents significantly determines the partitioning. This is explained by the "like dissolves like" principle.

3. Temperature:

   • Temperature influences the solubility and kinetic energy of the solute in both solvents. Generally, changing temperature alters the $K_{pc}$ value.

4. Molecular Structure and Size:

   • The size and specific functional groups within a solute molecule affect its interactions with solvents. Larger molecules or those with complex structures can experience increased steric hindrance or different solvation energies.

Possible Questions/Answers

• Q: When 2 grams of a solute is shaken with a mixture of $150{\mathrm{cm}}^3$ of water and $20{\mathrm{cm}}^3$ of chloroform. After shaking, $1.5$ grams of the solute is found in the chloroform layer. Calculate the partition coefficient.
  • A:
    1. Mass in water: $2\mathrm{g}-1.5\mathrm{g}=0.5\mathrm{g}$
    2. Concentration in chloroform: $\frac{1.5\mathrm{g}}{20{\mathrm{cm}}^3}=0.075\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$
    3. Concentration in water: $\frac{0.5\mathrm{g}}{150{\mathrm{cm}}^3}=0.00333\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$
    4. $K_{PC}$: $\frac{0.075}{0.00333}\approx 22.5$

Significance of Partition Coefficient

Partition coefficients are crucial in various chemical and biological processes, including:

• Liquid-liquid extraction: For separating compounds in the lab.
• Drug distribution: Predicting how pharmaceuticals will be absorbed and distributed in the body (e.g., crossing cell membranes).
• Chromatography: In paper or column chromatography, the distribution of a solute between the stationary and mobile phases is governed by $K_{pc}$.