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A pharmaceutical product is designed to be absorbed in the gastrointestinal tract. The active ingredient is pressed into a tablet with a density of \(\rho_{\mathrm{A}}=15 \mathrm{~kg} / \mathrm{m}^{3}\) while the remainder of the tablet is composed of inactive ingredients. The partition coefficient is \(K=3 \times 10^{-2}\), and the diffusion coefficient of the active ingredient in the gastrointestinal fluid is \(D_{\mathrm{AB}}=0.4 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}\). (a) Estimate the dosage delivered over a time period of \(5 \mathrm{~h}\) for a spherical tablet of diameter \(D=\) \(6 \mathrm{~mm}\). Hint: Assume the change in the tablet radius over the dosage period is small. (b) Estimate the dosage delivered over \(5 \mathrm{~h}\) for \(N=200\) small, spherical tablets contained in a gelatin capsule that quickly dissolves after ingestion, releasing the medication. The initial mass of the medication is the same as in part (a).

Short Answer

Expert verified
The dosage delivered over a 5-hour period for a single spherical tablet with diameter \(6 \mathrm{~mm}\) is approximately \(m_{dosage} = M_{initial} * F\), where \(M_{initial} = (4/3) * \pi * R^3 * \rho_{A}\) and \(F = (1 - \exp(-T/t_{relax}))\). For 200 small tablets contained in a gelatin capsule, the total dosage is \(m_{total} = m_{dosage} * 200\).

Step by step solution

01

Relaxation time calculation

To estimate the dosage, we first need to calculate the relaxation time, which will help us determine how quickly the active ingredient is absorbed into the gastrointestinal fluid. Using the partition coefficient (K), the diffusion coefficient (D_AB), and the radius of the tablet (R), the relaxation time (t_relax) can be calculated using the following formula: \(t_{relax} = KR^2 / D_{AB}\) Since we are given the diameter (D) instead of the radius, we will first convert the diameter to radius. R = D / 2 Now we can calculate the relaxation time: \(t_{relax} = (3 \times 10^{-2}) \times (3 \times 10^{-3})^2 / (0.4 \times 10^{-10})\) ###Step 2: Convert the given time period to seconds###
02

Convert time period to seconds

The given time period for estimating the dosage is 5 hours. However, the relaxation time is given in seconds, so we need to convert the given time period to seconds as well. Time period (T) = 5 hours * 60 minutes/hour * 60 seconds/minute = 18,000 seconds ###Step 3: Calculate the fractional dosage for a single tablet###
03

Fractional dosage calculation

Now we can calculate the fraction of the dosage released during the 5-hour period using the relaxation time (t_relax) and the time period (T). The fractional dosage (F) can be calculated as follows: F = (1 - exp(-T/t_relax)) ###Step 4: Calculate the dosage delivered over the time period for a single tablet###
04

Dosage calculation for a single tablet

To calculate the actual dosage delivered over the 5-hour period, we first need to find the initial mass of the active ingredient in the tablet: Initial mass (M_initial) = (4/3) * pi * R^3 * 蟻_A Now, we can calculate the dosage (m_dosage) released during the time period by multiplying the initial mass of the active ingredient by the fractional dosage: m_dosage = M_initial * F ###Step 5: Calculate the dosage delivered over the time period for 200 small tablets###
05

Dosage calculation for 200 small tablets

For part (b) of the exercise, we need to find the dosage delivered over the 5-hour period for 200 small tablets. Since the initial mass of the medication is the same as in part (a), we can simply multiply the dosage calculated in part (a) by 200: Total dosage (m_total) = m_dosage * 200 Now, we have obtained the estimates for the dosage released over a 5-hour period for a single tablet and 200 small tablets contained in a gelatin capsule.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diffusion Coefficient
The diffusion coefficient is a critical parameter in understanding how substances move in a medium, specifically in the context of pharmaceuticals. It measures the rate at which the active ingredient in a tablet disperses throughout a solution, like gastrointestinal fluids.
In our exercise, the diffusion coefficient of the active ingredient is given as \(D_{AB}=0.4 \times 10^{-10} \text{ m}^2/\text{s}\). This very small number indicates the ingredient diffuses slowly, reflecting the compact nature of solid tablets.
The diffusion coefficient helps in estimating how long it takes for the active ingredient to spread to a concentration sufficient for absorption. It is vital when calculating factors such as relaxation time, which helps us predict the dosage delivered over a period.
Partition Coefficient
The partition coefficient, \(K=3 \times 10^{-2}\), signifies the ratio of the concentration of an active ingredient in a solution to its concentration in another, immiscible phase. In pharmaceuticals, this often relates to how a drug's active ingredient is distributed between a lipophilic (fat-loving) and a hydrophilic (water-loving) environment.
This coefficient is central to understanding the drug's absorption through biological membranes. For instance, in the gastrointestinal tract, a suitable partition coefficient ensures the drug dissolves adequately in bodily fluids while maintaining a concentration gradient that promotes effective absorption.
In dosage calculations, the partition coefficient, along with the diffusion coefficient, influences how fast the active ingredient reaches its target concentration in body fluids.
Dosage Calculation
Dosage calculation is crucial in determining how much of an active ingredient is absorbed into the body over a certain time period.
To find the dosage, the initial quantity of the ingredient in the tablet is first determined by using the formula \[ \text{Initial mass} = \left(\frac{4}{3}\right) \pi R^3 \rho_A \], where \(R\) is the radius and \(\rho_A\) is the density.
  • Next, the relaxation time, which depends on both the partition and diffusion coefficients, helps compute the fractional amount released. This involves calculating \[ F = 1 - e^{-T/t_{\text{relax}}} \], with \(T\) being the time elapsed.
Both steps ensure that the calculation gives a realistic estimate of the dosage delivered through the diffusion and partitioning mechanisms at play in the human body.
Spherical Tablet Model
The spherical tablet model is used extensively in pharmaceuticals to simplify the understanding of drug dissolution and release. In this model, the tablet is assumed to be a perfect sphere.
This simplification allows for straightforward calculations involving parameters like radius, volume, and surface area, which are critical in determining dosage delivery rates.
The model assumes that the change in the tablet鈥檚 radius over the dosage period is small, facilitating the estimation of dosage without complex mathematical modeling.
  • The volume of the sphere is calculated using \(V = \left(\frac{4}{3}\right)\pi R^3\), helping determine the mass of the active pharmaceutical ingredient.
  • By assuming spherical symmetry, it allows predictions of how the active ingredient diffuses equally in all directions, simplifying further diffusion-related calculations.
This model is especially helpful when dealing with tablets composed of homogeneous materials like many pharmaceutical products.

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Most popular questions from this chapter

A thick plate of pure iron at \(1000^{\circ} \mathrm{C}\) is subjected to a carburization process in which the surface of the plate is suddenly exposed to a gas that induces a carbon concentration \(C_{C, s}\) at one surface. The average diffusion coefficient for carbon and iron at this temperature is \(D_{\mathrm{C}-\mathrm{Fe}}=3 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\). Use the correspondence between heat and mass transfer variables in addressing the following questions. (a) Consider the heat transfer analog to the carburization problem. Sketch the mass and heat transfer systems. Show and explain the correspondence between variables. Provide the analytical solutions to the heat and mass transfer problems. (b) Determine the carbon concentration ratio, \(C_{\mathrm{C}}(x, t) / C_{\mathrm{C}, s}\), at a depth of \(1 \mathrm{~mm}\) after \(1 \mathrm{~h}\) of carburization. (c) From the analogy, show that the time dependence of the mass flux of carbon into the plate may be expressed as \(n_{\mathrm{C}}^{\prime \prime}=\rho_{\mathrm{C}, s}\left(D_{\mathrm{C}-\mathrm{Fe}} / \pi t\right)^{1 / 2}\). Also, obtain an expression for the mass of carbon per unit area entering the iron plate over the time period \(t\).

The presence of \(\mathrm{CO}_{2}\) in solution is essential to the growth of aquatic plant life, with \(\mathrm{CO}_{2}\) used as a reactant in the photosynthesis. Consider a stagnant body of water in which the concentration of \(\mathrm{CO}_{2}\left(\rho_{\mathrm{A}}\right)\) is everywhere zero. At time \(t=0\), the water is exposed to a source of \(\mathrm{CO}_{2}\), which maintains the surface \((x=0)\) concentration at a fixed value \(\rho_{\mathrm{A}, 0}\). For time \(t>0, \mathrm{CO}_{2}\) will begin to accumulate in the water, but the accumulation is inhibited by \(\mathrm{CO}_{2}\) consumption due to photosynthesis. The time rate at which this consumption occurs per unit volume is equal to the product of a reaction rate constant \(k_{1}\) and the local \(\mathrm{CO}_{2}\) concentration \(\rho_{\mathrm{A}}(x, t)\). (a) Write (do not derive) a differential equation that could be used to determine \(\rho_{\mathrm{A}}(x, t)\) in the water. What does each term in the equation represent physically? (b) Write appropriate boundary conditions that could be used to obtain a particular solution, assuming a "deep" body of water. What would be the form of this solution for the special case of negligible \(\mathrm{CO}_{2}\) consumption \(\left(k_{1} \approx 0\right)\) ?

A 1-mm-thick square \((100 \mathrm{~mm} \times 100 \mathrm{~mm})\) sheet of polymer is suspended from a precision scale in a chamber characterized by a temperature and relative humidity of \(T=300 \mathrm{~K}\) and \(\phi=0\), respectively. Suddenly, at time \(t=0\), the chamber's relative humidity is raised to \(\phi=0.95\). The measured mass of the sheet increases by \(0.012 \mathrm{mg}\) over \(24 \mathrm{~h}\) and by \(0.016 \mathrm{mg}\) over \(48 \mathrm{~h}\). Determine the solubility and mass diffusivity of water vapor in the polymer. Preliminary experiments have indicated that the mass diffusivity is greater than \(7 \times 10^{-13} \mathrm{~m}^{2} / \mathrm{s}\).

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Consider combustion of hydrogen gas in a mixture of hydrogen and oxygen adjacent to the metal wall of a combustion chamber. Combustion occurs at constant temperature and pressure according to the chemical reaction \(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\). Measurements under steady-state conditions at a distance of \(10 \mathrm{~mm}\) from the wall indicate that the molar concentrations of hydrogen, oxygen, and water vapor are \(0.10,0.10\), and \(0.20 \mathrm{kmol} / \mathrm{m}^{3}\), respectively. The generation rate of water vapor is \(0.96 \times 10^{-2} \mathrm{kmol} / \mathrm{m}^{3} \cdot \mathrm{s}\) throughout the region of interest. The binary diffusion coefficient for each of the species \(\left(\mathrm{H}_{2}, \mathrm{O}_{2}\right.\), and \(\left.\mathrm{H}_{2} \mathrm{O}\right)\) in the remaining species is \(0.6 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). (a) Determine an expression for and make a qualitative plot of \(C_{\mathrm{H}_{2}}\) as a function of distance from the wall. (b) Determine the value of \(C_{\mathrm{H}_{2}}\) at the wall. (c) On the same coordinates used in part (a), sketch curves for the concentrations of oxygen and water vapor. (d) What is the molar flux of water vapor at \(x=10 \mathrm{~mm}\) ?

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