Distribution Coefficient Calculation

Reference ID: MET-CED7 | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

The distribution coefficient calculation is a fundamental procedure in Process Engineering, specifically within liquid-liquid extraction (LLE) unit operations. It quantifies the equilibrium partitioning of a solute between two immiscible liquid phases: an aqueous raffinate and an organic extractant. This calculation is critical for designing mixer-settler stages, determining solvent requirements, and optimizing the recovery of high-value products such as pharmaceuticals (e.g., Penicillin G) or specialty chemicals.

In industrial practice, this model assumes a single-stage, continuous, co-current system operating at steady-state equilibrium. It is primarily used to predict the final concentration of a solute in both phases and to evaluate the efficiency of the extraction process based on the solvent-to-feed ratio.

Methodology & Formulas

The calculation relies on the Nernst partition law, adjusted for the chemical state of the solute; for ionizable species, the apparent distribution ratio D must be calculated to account for pH‑dependent dissociation, a factor explored in detail in our discussion of the pH effect on extraction efficiency.

The apparent distribution ratio D is determined by the intrinsic distribution coefficient K and the dissociation constant pK_a:

\[ D = K \cdot \left( \frac{1}{1 + 10^{(pH - pK_{a})}} \right) \]

The mass balance for the system, assuming no solute in the incoming solvent (y_in = 0), is defined as:

\[ \dot{m}_{F} \cdot x_{f} = \dot{m}_{F} \cdot x_{out} + \dot{m}_{S} \cdot y_{out} \]

Applying the equilibrium relation y_out = D \cdot x_out, the raffinate concentration x_out is derived as:

\[ x_{out} = \frac{\dot{m}_{F} \cdot x_{f}}{\dot{m}_{F} + (\dot{m}_{S} \cdot D)} \]

The extract concentration y_out and the fractional recovery η are subsequently calculated as:

\[ y_{out} = D \cdot x_{out} \] \[ \eta = \left( \frac{\dot{m}_{S} \cdot y_{out}}{\dot{m}_{F} \cdot x_{f}} \right) \cdot 100\% \]

Parameter	Constraint/Regime
pH Range	2.0 ≤ pH ≤ 2.5
Temperature	0 °C ≤ T ≤ 10 °C
Solute Concentration	x_f ≤ 0.05 (wt fraction)
Solvent-to-Feed Ratio	0.1 ≤ (\(\dot{m}_{S} / \dot{m}_{F}\)) ≤ 0.8

The distribution coefficient, often denoted as K_d, represents the ratio of the equilibrium concentration of a solute in the organic phase to its equilibrium concentration in the aqueous phase. For process engineers, this value is critical for determining the efficiency of separation processes. The calculation typically follows these principles:

Identify the concentration of the solute in the organic phase (C_o).
Identify the concentration of the solute in the aqueous phase (C_a).
Calculate the ratio using the formula K_d = C_o / C_a.

Several operational variables can introduce deviations from theoretical K_d values. Engineers should monitor the following:

Temperature fluctuations, as K_d is highly temperature-dependent.
The presence of competing solutes or impurities that alter activity coefficients.
System pH, which significantly impacts the ionization state of the solute.
Phase equilibrium time, ensuring the system has reached a true steady state before sampling.

The separation factor (α) is a measure of the relative selectivity of a solvent for two different solutes. It is calculated by taking the ratio of the individual distribution coefficients for the two components. If you are evaluating the performance of a solvent, use the following logic:

Determine K_d for solute A.
Determine K_d for solute B.
Calculate α = K_{d_A} / K_{d_B}.
A higher α value indicates a more effective separation process.

Worked Example: Distribution Coefficient for Penicillin G Extraction

Process: A single-stage mixer-settler extracts Penicillin G from an aqueous broth (raffinate) into n-Butanol (extract) at 5 °C and pH 2.0. Equilibrium is assumed, and the Nernst partition law applies for dilute solutions.

Knowns:

Feed rate, \( \dot{m}_F = 100.0 \) kg/h
Solvent rate, \( \dot{m}_S = 20.0 \) kg/h
Feed concentration, \( x_f = 0.001 \) kg solute per kg aqueous
Solvent inlet concentration, \( y_{in} = 0.0 \) kg solute per kg organic
Intrinsic distribution coefficient, \( K = 40.0 \)
pH = 2.0
pK_a of Penicillin G = 2.75
Temperature = 5.0 °C

Step-by-Step Calculation:

Calculate the apparent distribution coefficient \( D \) using pH correction for the weak acid solute: \[ D = K \cdot \frac{1}{1 + 10^{(pH - pK_a)}} = 40.0 \cdot \frac{1}{1 + 10^{(2.0 - 2.75)}} = 40.0 \cdot \frac{1}{1 + 10^{-0.75}} \] Since \(10^{-0.75} \approx 0.17783\), we have \[ D = \frac{40.0}{1.17783} \approx 33.96. \] For subsequent calculations, we retain more precision: \(D = 33.9608\).
Compute total solute entering the system: \[ T_{\text{solute,in}} = \dot{m}_F \cdot x_f = 100.0 \cdot 0.001 = 0.1 \; \text{kg/h} \]
Set up the material balance with equilibrium relation \( y_{\text{out}} = D \cdot x_{\text{out}} \): \[ \dot{m}_F \cdot x_f + \dot{m}_S \cdot y_{\text{in}} = \dot{m}_F \cdot x_{\text{out}} + \dot{m}_S \cdot y_{\text{out}} \] Since \( y_{\text{in}} = 0 \) and \( y_{\text{out}} = D \cdot x_{\text{out}} \): \[ 0.1 = 100.0 \cdot x_{\text{out}} + 20.0 \cdot (33.9608 \cdot x_{\text{out}}) = x_{\text{out}} \cdot (100.0 + 20.0 \cdot 33.9608) \] \[ 20.0 \cdot 33.9608 = 679.216 \quad \Rightarrow \quad \text{Denominator} = 100.0 + 679.216 = 779.216 \]
Solve for the raffinate concentration \( x_{\text{out}} \): \[ x_{\text{out}} = \frac{0.1}{779.216} = 1.2833 \times 10^{-4} \; \text{kg solute per kg aqueous} \approx 0.000128 \; \text{kg/kg} \]
Calculate the extract concentration \( y_{\text{out}} \): \[ y_{\text{out}} = D \cdot x_{\text{out}} = 33.9608 \cdot 1.2833 \times 10^{-4} = 4.358 \times 10^{-3} \; \text{kg solute per kg organic} \approx 0.00436 \; \text{kg/kg} \]
Compute the fractional recovery \( \eta \): \[ \eta = \frac{\dot{m}_S \cdot y_{\text{out}}}{\dot{m}_F \cdot x_f} \times 100\% = \frac{20.0 \cdot 4.358 \times 10^{-3}}{0.1} \times 100\% = 87.16\% \approx 87.2\% \]

Final Answer: The equilibrium raffinate concentration is \(1.28 \times 10^{-4}\) kg/kg (0.000128 kg/kg), the extract concentration is \(4.36 \times 10^{-3}\) kg/kg (0.00436 kg/kg), and the recovery is 87.2%.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle

Distribution Coefficient Calculation

Introduction & Context

Methodology & Formulas

Worked Example: Distribution Coefficient for Penicillin G Extraction

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