Extent, Rate, Models, and Manipulation of Dissolution

Learning Outcomes for Solubility and Dissolution

Describe the significance and importance of solubility and dissolution in the formulation of medicinal products.
Understand the role of solutions as effective drug delivery vehicles.
Explain the various factors that influence solubility and dissolution behavior.
Identify specific strategies used to optimize the design of solution-based formulations.
Perform quantitative calculations to characterize solubility and dissolution processes.

The Dissolution Process

Definition: Dissolution is the transfer of molecules or ions from the solid state into a solution.
The Three-Step Process:
1. Removal of a solute molecule from the solid structure.
2. Creation of a cavity within the solvent to accommodate the solute.
3. Insertion of the solute molecule into the prepared cavity.
Molecular Interactions:
- Solute-Solute molecules: Interactions involving $W_{a}$ (solute molecule cohesive forces).
- Solvent-Solvent molecules: Interactions involving $W_{b}$ (solvent molecule cohesive forces).
- Solute-Solvent interactions: Interactions involving $W_{c}$ (adhesive forces).
Energy Exchange: Dissolution requires overcoming cohesive forces ($W_{a}$ and $W_{b}$), which are replaced by adhesive forces ($W_{c}$).
Ideal Solutions: In an ideal solution, the cohesive forces are exactly equal to the adhesive forces, meaning the energy exchange is zero.

Heat of Mixing (\Delta H_{mix}$)\n\n* If cohesive forces are not equal to adhesive forces, the solution is non-ideal.\n* Endothermic Reaction: If \Delta H_{mix} is positive, energy must be supplied to the system to facilitate the formation of a solution. This occurs when cohesive forces are stronger than adhesive forces.\n* Exothermic Reaction: If \Delta H_{mix} is negative, energy is released during the formation of the solution. This occurs when adhesive forces are greater than the cohesive forces.\n\n# Dissolution Mechanisms and Diffusion\n\n* Boundary Layers: These develop at the interface between the solid (e.g., a tablet) and the solvent.\n* Concentration Gradient:\n * C_s: The concentration representing the solubility at the solid/solvent boundary. This concentration is high and potentially supersaturated.\n * C: The concentration of the drug in the bulk solution.\n * The difference between C_s $and$ C $at the boundary limits the rate of dissolution. Drug diffusion occurs in the direction of high concentration ($ C_s $) to low concentration ($ C).\n* Rate Determining Step (RDS):\n * The release of solute molecules from the solid into the boundary layer is virtually instantaneous.\n * The diffusion of solute molecules across the static boundary layer (the stagnant layer) is the slowest process.\n * Consequently, diffusion is the rate-determining step for the overall dissolution process.\n\n# Noyes-Whitney Equation\n\n* This equation describes the rate of dissolution of solids when the process is diffusion-controlled and involves no chemical reaction.\n\n\frac{dm}{dt} = \frac{D \times A \times (C_s - C)}{h}\n\n* Variables:\n * \frac{dm}{dt} $: Rate of dissolution ($ mg \, s^{-1}).\n * D $: Diffusion coefficient ($ cm^{2} \, s^{-1}).\n * A $: Surface area of the solid ($ cm^{2}).\n * C_s $: Saturation solubility of the drug in the dissolution medium ($ mg \, cm^{-3}).\n * C $: Concentration of the drug in the bulk dissolution medium at time$ t $($ mg \, cm^{-3}).\n * h $: Thickness of the diffusion layer ($ cm).\n * \frac{(C_s - C)}{h}: The concentration gradient.\n\n# Sink Conditions\n\n* Definition: Sink conditions refer to the ability of a solvent to dissolve between 5 to 10 times the amount of drug being solubilized.\n* Because a drug has a fixed solubility in a specific solvent, achieving sink conditions requires increasing the volume of the solvent.\n* Mathematical Simplification: If the volume of solution is so large that the bulk concentration (C $) does not exceed 10% of the saturation solubility ($ C_s $), then$ (C_s - C) \approx C_s.\n* Modified Noyes-Whitney Equation for Sink Conditions:\n\n\frac{dm}{dt} = \frac{D \times A \times C_s}{h}\n\n* Practical Applications:\n * In Vivo: Sink conditions often apply in the body because drug absorption into the bloodstream is frequently faster than the rate of dissolution.\n * In Vitro Dissolution Studies: Researchers must consider media choice, media volume, and sample replacement to maintain sink conditions.\n* Non-Sink Conditions: If C $exceeds 10% of$ C_s $. If$ C = C_s, the solution is saturated, and the overall dissolution rate becomes zero.\n\n# Factors Influencing Dissolution Rate Variables\n\n* D (Diffusion Coefficient): May be decreased if the viscosity of the medium is increased (e.g., by adding viscosity modifiers).\n* A (Surface Area): Increased through micronization (reducing particle size) or by using amorphous forms of materials.\n* h (Diffusion Layer Thickness): Decreased by increasing agitation (e.g., stirring or shaking).\n* Cs (Saturation Solubility): May be altered by changing the pH of the medium, using drug salts, or employing buffers.\n* C (Bulk Concentration): Decreased by increasing the fluid volume or through the removal of the drug (e.g., by drinking and gastric emptying or by systemic absorption).\n* Concentration Progression: As dissolution proceeds, C $increases, causing$ (C_s - C) to decrease; therefore, the dissolution rate naturally decreases over time as the bulk concentration rises.\n\n# Intrinsic Dissolution Rate (IDR)\n\n* Overview: While the rate of dissolution depends on variables like agitation and surface area, the Intrinsic Dissolution Rate (IDR) is a measure designed to be independent of these experimental variables.\n* Definition: IDR is the rate of dissolution of a pure pharmaceutical active when conditions such as surface area, agitation/stirring speed, pH, and ionic strength of the dissolution medium are kept strictly constant.\n* IDR Equation:\n\n\frac{dm}{dt \times A} = k(C_s - C)\n\n* In this formula, k $is the intrinsic dissolution rate constant, with units of$ cm \, s^{-1} $.$ k = \frac{D}{h}.\n* Measurement of IDR: Usually measured using the Static Disc Method.\n * A dissolution vessel contains fluid at 37 \, ^{\circ}C.\n * A rotating paddle operates at a constant speed (e.g., 100 \, rpm).\n * A metal disk contains a compressed pellet of the drug with a known, constant surface area.\n* Purpose of IDR: It is used to depict the dissolution rate of a drug, determine batch-to-batch chemical equivalency, screen drug candidates, and understand solution behavior.\n\n# Example Problem: IDR Calculation\n\nScenario: A 5 \, g $sample of drug granules with a total surface area of$ 0.28 \, m^2 $is placed into$ 500 \, mL $water. After$ 1 \, min $,$ 0.6 \, g $of the drug had dissolved. The solubility of the drug is$ 15 \, mg \, mL^{-1}.\n\nPart A: Calculate the intrinsic dissolution rate constant (k).\n1. Check for Sink Conditions:\n * Drug dissolved in 500 \, mL = 0.6 \, g = 600 \, mg.\n * Bulk concentration (C $) =$ \frac{600 \, mg}{500 \, mL} = 1.2 \, mg \, mL^{-1}.\n * Saturation solubility (C_s $) =$ 15 \, mg \, mL^{-1}.\n * 10\% $of$ C_s = 1.5 \, mg \, mL^{-1}.\n * Since 1.2 < 1.5 $, sink conditions apply. We use the modified IDR equation:$ \frac{dm}{dt \times A} = k \times C_s.\n2. Convert Units:\n * m = 600 \, mg.\n * t = 1 \, min = 60 \, s.\n * A = 0.28 \, m^2 = 2800 \, cm^2.\n * C_s = 15 \, mg \, mL^{-1} = 15 \, mg \, cm^{-3}.\n3. Calculate k:\n * \frac{600}{60 \times 2800} = k \times 15\n * 3.57 \times 10^{-3} = k \times 15\n * k = 2.38 \times 10^{-4} \, cm \, s^{-1}.\n\nPart B: Calculate the diffusion coefficient (D $) if the boundary layer thickness ($ h $) is$ 50 \, \mu m.\n1. Convert h:\n * h = 50 \, \mu m = 5 \times 10^{-3} \, cm.\n2. Calculate D:\n * k = \frac{D}{h} \implies D = k \times h\n * D = (2.38 \times 10^{-4} \, cm \, s^{-1}) \times (5 \times 10^{-3} \, cm)\n * D = 1.19 \times 10^{-6} \, cm^{2} \, s^{-1}.\n\n# Dissolution of Powders and Hixson-Cromwell Cube Root Law\n\n* The Noyes-Whitney equation assumes a constant surface area, which is true for static/rotating disc measurements but false for powders because the surface area decreases as particles dissolve.\n* Hixson-Cromwell Cube Root Law: Used for materials where surface area changes during the process.\n* Assumptions:\n * The powder consists of uniformly sized particles.\n * The dissolution rate is based on the cube root of the weight of the particles.\n * Particle radius is not constant during dissolution.\n* Equation:\n\n\sqrt[3]{M_0} - \sqrt[3]{M_t} = Kt\n\n* Variables:\n * M_0 $: Mass of the particles at the start ($ t = 0).\n * M_t $: Mass of undissolved particles at time$ t.\n * K $: The cube root dissolution rate constant (Note: This is different from the intrinsic dissolution rate constant$ k).\n* Linear Plot: Plotting the cube root of the mass undissolved (\sqrt[3]{M_t} $) against time ($ t) yields a linear plot where:\n * Y \text{-intercept} = \sqrt[3]{M_0}\n * \text{Gradient} = -K\n\n# Example Problem: Hixson-Cromwell Law\n\nScenario: 0.5 \, g $of drug powder is added to$ 500 \, mL $water. Dissolution is monitored over$ 40 \, mins. \n\nData Table:\n| Time (min) | Bulk Conc (mg \, mL^{-1} $) | Mass Dissolved ($ g $) | Mass Undissolved ($ M_t $in$ g $) |$ \sqrt[3]{M_t} $($ g^{1/3}) |\n| :--- | :--- | :--- | :--- | :--- |\n| 10 | 0.490 | 0.245 | 0.255 | 0.634 |\n| 20 | 0.784 | 0.392 | 0.108 | 0.476 |\n| 30 | 0.936 | 0.468 | 0.032 | 0.317 |\n| 40 | 0.992 | 0.496 | 0.004 | 0.159 |\n\nCalculations:\n* Mass Dissolved: \text{Concentration} \times \text{Volume (500 mL)} / 1000.\n* Mass Undissolved (M_t $):$ 0.5 \, g - \text{Mass Dissolved}.\n* Using Excel to plot \sqrt[3]{M_t} vs Time:\n * Regression equation: y = -0.0158x + 0.7929.\n * K = 0.0158 \, g^{1/3} \, min^{-1}.\n * \text{Intercept} = 0.7929 $. Calculate$ M_0 $:$ 0.7929^{3} = 0.498 \, g $(close to the actual starting mass of$ 0.5 \, g).\n\n# Directed Learning Exercises\n\nExercise 1:\nA sample of drug granules is added to 250 \, mL $water. After$ 10 \, mins $,$ 0.15 \, g $of drug has dissolved. The solubility of the drug is$ 8 \, mg \, mL^{-1}.\n* a. If the intrinsic dissolution rate constant (k $) is$ 2.23 \times 10^{-5} \, cm \, s^{-1} $, what is the total surface area of the granules (in$ m^2)?\n* b. If the thickness of the boundary layer is 100 \, \mu m $, what is the diffusion coefficient ($ D)?\n\nExercise 2:\n2.5 \, g $of a second drug with a surface area of$ 0.21 \, m^2 $is added to$ 500 \, mL $water. After$ 10 \, mins $,$ 1.2 \, g $of drug has dissolved. The solubility of the drug is$ 20 \, mg \, mL^{-1}.\n* a. What is the value of the intrinsic dissolution rate constant, k?\n* b. If the diffusion coefficient is 4.06 \times 10^{-7} \, cm^{2} \, s^{-1} $, what is the thickness of the boundary layer (in$ \mu m$$)?

Questions & Discussion

Specimen Exam Question: Explain how the following methods, used for the measurement of dissolution rate, differ from the specialized methods used to measure intrinsic dissolution rate:

Beaker method
Rotating basket method
Flask-stirrer method
Paddle method", "title": "Extent, Rate, Models, and Manipulation of Dissolution"}