MECHANISTIC ANALYSIS OF CO-CRYSTAL DISSOLUTION, SURFACE pH, AND DISSOLUTION ADVANTAGE AS A GUIDE FOR RATIONAL SELECTION

Abstract

The dissolution behavior of a dibasic drug ketoconazole (KTZ) under the influence of pH has been evaluated and compared to its three 1:1 co-crystals with diacidic coformers, fumaric acid (FUM), succinic acid (SUC) and adipic acid (ADP). Mass transport models were developed by applying Fick’s Law of diffusion to dissolution with simultaneous chemical reactions in the hydrodynamic boundary layer adjacent to the dissolving surface to predict the interfacial pH and flux of the parent drug and cocrystals. All three co-crystals have the ability to modulate the interfacial pH to different extents compared to the parent drug due to the acidity of the coformers. Dissolution pH dependence of KTZ is significantly reduced by the co-crystallization with acidic coformers. Due to the different dissolution pH dependence, there exists a transition pH where the flux of the co-crystal is the same as the parent drug. Below this transition pH, the drug flux is higher, but above it, the co-crystal flux is higher. The development of these mass transport models provide a mechanistic understanding of the dissolution behavior and help identify co-crystalline solids with optimal dissolution characteristics.

Introduction

Co-crystals are multicomponent solids that contain two or more different molecular components in the same crystal lattice with well-defined stoichiometry and they have emerged as a promising solid state modification strategy to enhance the solubility, dissolution, and bioavailability of poorly water soluble compounds.^1–5 One of the advantages that co-crystal has to offer is the large diversity in formation; however, this can also complicate the process of selecting the proper solid form for development since each co-crystal form can have very distinct physicochemical properties.^6–8 The selection of a solid form with optimal physicochemical properties that can achieve the desired bioperformance is essential for successful oral drug product development. Among these properties, solubility and dissolution are important criteria for the selection process because they play significant roles in determining oral absorption.⁹ Depending on the properties of the co-crystals and solution conditions, co-crystals can display neither, either, or both solubility and/or dissolution advantages compared to the parent drug.¹⁰ A thorough understanding on the solution chemistry of co-crystals would help to rationalize the selection process.

Recent publications on co-crystal dissolution under the impact of pH and surfactant have provided in-depth evaluations of the dissolution mechanism.^{11, 12} These studies have demonstrated the importance of interfacial pH in determining the flux of co-crystals with ionizable components.^{11, 12} With approriate mass transport models, dissolution conditions where the co-crystal exhibits both thermodynamic stability and dissolution advantage can be mapped.¹¹ These publications have set a strong foundation for developing mass transport models for co-crystals with diverse ionization properties. Herein, the dissolution mechanism is applied to co-crystals containing diverse ionization properties, specifically for co-crystals with dibasic drugs with diacidic coformers.

The model drug studied here is ketoconazole (KTZ), an antifungal drug used primarily for fungal infections.^13–15 It is a weakly dibasic drug with poor intrinsic solubility, pH dependent dissolution and variable oral absorption.^13–15 The dissolution of KTZ below pH 3 is rapid, but the rate is significantly reduced above pH 5 as the drug becomes less ionized.^{15, 16} Many studies have demonstrated that the oral absorption of KTZ is impaired for patients with reduced gastric acid production.^17–19 Therefore, the prerequisite for adequate KTZ dissolution and oral absorption is sufficient gastric acidity. There are studies showing the oral absorption of KTZ can be improved by co-administering with stomach acid stimulant or acidic beverage.^{13, 15, 19} Knowing KTZ performs better under acidic conditions, it is of interest to identify whether co-crystallization with acidic coformers would help to improve its performance. The purpose of this study is to evaluate and compare the pH dependent dissolution of KTZ to its three co-crystals discovered by Martin et al.²⁰: ketoconazole fumaric acid (KTZ-FUM), ketoconazole succinic acid (KTZ-SUC) and ketoconazole adipic acid (KTZ-ADP) co-crystals.

Materials and methods

Materials

Ketoconazole (KTZ) was purchased from Bosche Scientific (New Brunswick, NJ) and used as received. Adipic acid (ADP), succinic acid (SUC), fumaric acid (FUM), were purchased from Sigma-Aldrich (St. Louis, MO) and used as received. Methanol, 2-propanol and hydrochloric acid were purchased from Fisher Scientific (Pittsburgh, PA). Acetone was purchased from Acros Organics (NJ). Sodium hydroxide pellets were purchased from J.T. Baker (Philipsburg, NJ). Trifluoroacetic acid was purchased from Aldrich Company (Milwaukee, WI). Water used in this study was filtered through a double deionized purification system (Milli Q Plus Water System) from Millipore Co. (Bedford, MA).

Co-crystal synthesis

Co-crystals were prepared by reaction crystallization method²¹ at room temperature. KTZ-SUC and KTZ-FUM were prepared by adding 1:1 molar ratio of KTZ and coformers in acetone solution. KTZ-ADP was prepared by adding 1:1 molar ratio of KTZ and ADP in 2-propanol solution. Solid phases were characterized by X-ray powder diffraction (XRPD) and differential scanning calorimetry (DSC).

Co-crystal dissolution measurements

Constant surface area dissolution rates of KTZ and its co-crystals were determined using a rotating disk apparatus. Drug or co-crystal powder (~150 mg) was compressed in a stainless steel rotating disk die with a tablet radius of 0.50 cm at approximately 85 MPa for 2 minutes using a hydraulic press. The die containing the compact was mounted onto a stainless steel shaft attached to an overhead, variable speed motor. The disk was exposed to 150 mL of dissolution medium in a water jacketed beaker with temperature controlled at 25°C and a rotation speed of 200 RPM was used. All dissolution experiments were performed in water with pH adjusted using HCl or NaOH. The bulk pH during dissolution was maintained constant by manually adding HCl or NaOH as necessary. Sink conditions were maintained throughout the experiments by ensuring the concentrations at the last time point of the dissolution were less than 10% of the solubility. Solution concentrations were measured using HPLC and solid phases after dissolution were analyzed by XRPD.

High Performance Liquid Chromatagraphic (HPLC)

Waters HPLC equipped with a photodiode array detector was used for all analysis. The mobile phase was composed of 60% methanol and 40% water with 0.1% trifluoroacetic acid and the flow rate of 1 mL/min was used. Separation was achieved using Waters, Atlantis, T3 column (5.0 μm, 100 Å) with dimensions of 4.6 × 250 mm. The sample injection volume was 20 μL. However, it was increased to 100 μL for the dissolution of KTZ at pH 5 and 6 due to the low concentration. The wavelengths for the analytes were as follows: 230 nm for KTZ, 220 nm for FUM and 210 nm for both SUC and ADP.

XRPD

XRPD diffractograms of solid phases were collected with a benchtop Rigaku Miniflex X-ray diffractometer using Cu-Kα radiation (λ = 1.54 Å), a tube voltage of 30 kV, and a tube current of 15 mA. Data was collected from 5 to 40° at a continuous scan rate of 2.5°/min.

DSC

Crystalline samples were analyzed by DSC using a TA instrument 2910 MDSC system equipped with a refrigerated cooling unit. All experiments were performed by heating the samples at a rate of 10 °C/min under a dry nitrogen atmosphere. Temperature and enthalpy of the instrument were calibrated using high purity indium standard.

Theoretical

The mass transport models presented in this study describe the dissolution mechanisms of a dibasic drug and its three 1:1 co-crystals with diacidic coformers in solution containing hydrogen ion, hydroxide ion and water as the reactive species. The mass transport analyses are based on the classic film theory that postulates the existence of a diffusion boundary layer adjacent to the dissolving surface.²² Due to the ionization properties, both drug and coformers can undergo chemical reactions with the reactive species from the bulk solution and thus alter the pH at the dissolving surface. The dissolution process is determined by the concentration gradient across the diffusion boundary layer and influenced by the simultaneous diffusion and chemical reactions occurring at the dissolving surface and in the adjacent boundary layer.^{22, 23}

The chemical equilibria and the equations for chemical equilibrium constants during the dissolution of a diabasic drug, B, can be described as follows:

$${\text{H}}_2\text{O}+{\text{BH}}_2^{2+}\leftrightharpoons {\text{H}}_3{\text{O}}^{+}+{\text{BH}}^{+}$$ (1)

$${\text{K}}_{\text{a}1}^{\text{B}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{BH}}^{+}\right]}{\left[{\text{BH}}_2^{2+}\right]}$$ (2)

$${\text{H}}_2\text{O}+{\text{BH}}^{+}\leftrightharpoons {\text{H}}_3{\text{O}}^{+}+\text{B}$$ (3)

$${\text{K}}_{\text{a}2}^{\text{B}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right][\text{B}]}{\left[{\text{BH}}^{+}\right]}$$ (4)

$${\text{H}}_3{\text{O}}^{+}+{\text{OH}}^{-}\leftrightharpoons 2{\text{H}}_2\text{O}$$ (5)

$${\text{K}}_{\text{w}}=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$$ (6)

$${\text{OH}}^{-}+{\text{BH}}_2^{2+}\leftrightharpoons {\text{H}}_2\text{O}+{\text{BH}}^{+}$$ (7)

$${\text{K}}_1=\frac{\left[{\text{BH}}^{+}\right]}{\left[{\text{BH}}_2^{2+}\right]\left[{\text{OH}}^{-}\right]}$$ (8)

$${\text{OH}}^{-}+{\text{BH}}^{+}\leftrightharpoons {\text{H}}_2\text{O}+\text{B}$$ (9)

$${\text{K}}_2=\frac{[\text{B}]}{\left[{\text{BH}}^{+}\right]\left[{\text{OH}}^{-}\right]}$$ (10)

where $K_{a1}^B$ and $K_{a2}^B$ are the ionization constants for the dibasic drug. Under aqueous conditions, the solubility of a dibasic drug as a function of pH can be described as follow:

$${\text{S}}_{\text{B}}={[\text{B}]}_0(1+\frac{{\text{H}}^{+}}{{\text{K}}_{\text{a}2}^{\text{B}}}+\frac{{\left({\text{H}}^{+}\right)}^2}{{\text{K}}_{\text{a}1}^{\text{B}}{\text{K}}_{\text{a}2}^{\text{B}}})$$ (11)

where [B]₀ is the intrinsic solubility of the drug.

The chemical equilibria and the equations for chemical equilibrium constants during the dissolution of a 1:1 co-crystal with dibasic drug, B and diacidic coformer, H₂A, can be described as follows:

$${\left({\text{BH}}_2\text{A}\right)}_{\text{solid}}\leftrightharpoons \text{B}+{\text{H}}_2\text{A}$$ (12)

$${\text{K}}_{\text{sp}}=[\text{B}]\left[{\text{H}}_2\text{A}\right]$$ (13)

$${\text{H}}_2\text{O}+{\text{BH}}_2^{2+}\leftrightharpoons {\text{H}}_3{\text{O}}^{+}+{\text{BH}}^{+}$$ (14)

$${\text{K}}_{\text{a}1}^{\text{B}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{BH}}^{+}\right]}{\left[{\text{BH}}_2^{2+}\right]}$$ (15)

$${\text{H}}_2\text{O}+{\text{BH}}^{+}\leftrightharpoons {\text{H}}_3{\text{O}}^{+}+\text{B}$$ (16)

$${\text{K}}_{\text{a}2}^{\text{B}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right][\text{B}]}{\left[{\text{BH}}^{+}\right]}$$ (17)

$${\text{H}}_2\text{O}+{\text{H}}_2\text{A}\leftrightharpoons {\text{H}}_3{\text{O}}^{+}+{\text{HA}}^{-}$$ (18)

$${\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2\text{A}\right]}$$ (19)

$${\text{H}}_2\text{O}+{\text{HA}}^{-}\leftrightharpoons {\text{H}}_3{\text{O}}^{+}+{\text{A}}^{2-}$$ (20)

$${\text{K}}_{\text{a}2}^{{\text{H}}_2\text{A}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right][{\text{A}}^{2-}]}{\left[{\text{HA}}^{-}\right]}$$ (21)

$${\text{H}}_3{\text{O}}^{+}+{\text{OH}}^{-}\leftrightharpoons 2{\text{H}}_2\text{O}$$ (22)

$${\text{K}}_{\text{w}}=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$$ (23)

$${\text{OH}}^{-}+{\text{BH}}_2^{2+}\leftrightharpoons {\text{H}}_2\text{O}+{\text{BH}}^{+}$$ (24)

$${\text{K}}_1=\frac{\left[{\text{BH}}^{+}\right]}{\left[{\text{BH}}_2^{2+}\right]\left[{\text{OH}}^{-}\right]}$$ (25)

$${\text{OH}}^{-}+{\text{BH}}^{+}\leftrightharpoons {\text{H}}_2\text{O}+\text{B}$$ (26)

$${\text{K}}_2=\frac{[\text{B}]}{\left[{\text{BH}}^{+}\right]\left[{\text{OH}}^{-}\right]}$$ (27)

$${\text{H}}_2\text{A}+{\text{OH}}^{-}\leftrightharpoons {\text{H}}_2\text{O}+{\text{HA}}^{-}$$ (28)

$${\text{K}}_3=\frac{\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2\text{A}\right]\left[{\text{OH}}^{-}\right]}$$ (29)

$${\text{OH}}^{-}+{\text{HA}}^{-}\leftrightharpoons {\text{H}}_2\text{O}+{\text{A}}^{2-}$$ (30)

$${\text{K}}_4=\frac{[{\text{A}}^{2-}]}{\left[{\text{HA}}^{-}\right]\left[{\text{OH}}^{-}\right]}$$ (31)

where K_sp is the solubility product of the co-crystal, $K_{a1}^{H_{2}A}$ and $K_{a2}^{H_{2}A}$ are the ionization constants of the coformer. In aqueous solutions, the stoichiometric solubility of the co-crystal as a function of pH can be described as:

$${\text{S}}_{\text{cc}}=\sqrt{{\text{K}}_{\text{sp}}(1+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}}{\left[{\text{H}}^{+}\right]}+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}{\text{K}}_{\text{a}2}^{{\text{H}}_2\text{A}}}{{\left[{\text{H}}^{+}\right]}^2})(1+\frac{\left[{\text{H}}^{+}\right]}{{\text{K}}_{\text{a}2}^{\text{B}}}+\frac{{\left[{\text{H}}^{+}\right]}^2}{{\text{K}}_{\text{a}1}^{\text{B}}{\text{K}}_{\text{a}2}^{\text{B}}})}$$ (32)

Previously developed mass transport models for co-crystal dissolution have demonstrated that the concentrations of the co-crystal components at the dissolving surface can be different if they have different diffusion coefficients.¹² Based on the solubility product behavior of co-crystals, two models, the interfacial equilibrium model and surface saturation model, have been developed to describe the dissolution process.¹² The interfacial equilibrium model maintains constant solubility product at all time during dissolution, while the surface saturation model only maintains the drug concentration at the stoichiometric solubility of the co-crystal.¹² The theoretical comparison between the two models has demonstrated the mass transport analysis based on the surface saturation model provides flux predictions that are more aligned with the experimental data.¹² The model drug, KTZ, is a larger molecule compared to the carboxylic acid coformers and thus has smaller diffusion coefficient compared to the coformers. According to the surface saturation model, the concentration of the slower diffusing component at the dissolving surface is maintained at the solubility of the co-crystal, while the concentration of the faster diffusing component is lower.¹² For the dissolution of 1:1 co-crystals with dibasic drugs and diacidic coformers, the concentrations of the components at the dissolving surface can be written as:

$${[\text{B}]}_{\text{T},0}=[\text{B}]+\left[{\text{BH}}^{+}\right]+\left[{\text{BH}}_2^{2+}\right]={\text{S}}_{\text{cc}}$$ (33)

$${\left[{\text{H}}_2\text{A}\right]}_{\text{T},0}=\left[{\text{H}}_2\text{A}\right]+\left[{\text{HA}}^{-}\right]+\left[{\text{A}}^{2-}\right]=(\frac{{\text{D}}_{\text{B}}}{{\text{D}}_{{\text{H}}_2\text{A}}}{)}^{2/3}{\text{S}}_{\text{cc}}$$ (34)

where D_B and ${\text{D}}_{{\text{H}}_2\text{A}}$ are the diffusion coefficients of the drug and coformer, respectively, and subscript T,0 denotes the total concentration of the ionized and nonionized species at the dissolving surface.

The mass transport analyses for both drug and co-crystals are based on the assumptions that all chemical reactions within the diffusion layer happen instantaneously and the aqueous diffusivities of the ionized and non-ionized forms are the same. Detail derivations of the mass transport models for the drug and co-crystals using the surface saturation model are provided in the Appendix.

Results

Interfacial pH and solubility advantage predictions

The important physicochemical properties required for accurate flux predictions include solubility products, ionization constants, and diffusion coefficients and these properties of the three model co-crystals are summarized in Table 1. The larger molecular size of KTZ results in about a two fold lower diffusion coefficient compared to the three coformers. It is important to capture the surface concentrations of the co-crystal components since they determine the pH at the dissolving surface and, thereby, the rate of dissolution. With a knowledge of the physicochemical properties, the interfacial pH of KTZ and its three co-crystals can be predicted using equation 54 and 86 respectively from the Appendix. The ability of KTZ to alter the pH microenvironment at the dissolving surface is compared to that of the three co-crystals in Figure 1. At low bulk pH, the interfacial pH of KTZ is higher than the bulk solution pH because it is a base and mostly ionized under these conditions and thus increases the pH at the dissolving surface. As bulk pH increases above the pK_a values, the ability of KTZ in increasing the interfacial pH is limited by its lower degree of ionization at the dissolving surface. Therefore, interfacial pH is approximately the same as the bulk pH. By co-crystallizing with acidic coformers, all three cocrystals still have the ability to increase the interfacial pH at low bulk pH. However, the increase is much smaller compared to drug because the ionization of KTZ is being suppressed by the acidic coformers. As bulk pH increases above the pK_a values of the coformers, interfacial pH of the co-crystals is dominated by the acidity of the coformers. The ionization of the coformers lowers the interfacial pH and results in a buffering effect at the dissolving surface, in which the interfacial pH does not change substantially with bulk pH. The interfacial pH in the buffering region for KTZ-FUM is 4.1, KTZ-SUC is 4.7 and KTZ-ADP is 4.8. Even though KTZ-FUM has the lowest K_sp among three co-crystals, it is still able to buffer the interfacial pH to the lowest because FUM is the most acidic among the three coformers.

Table 1.

Physicochemical properties of model cocrystals and their components.

Cocrystal (B-H2A)	Ksp (× 10−2 mM2)a	pKa values		Diffusion coefficients (× 10−6 cm2/sec)d
		Bb	H2Ac	$D_{B_{aq}}$	$D_{H{A}_{aq}}$
KTZ-FUM	0.15	2.94, 6.51	3.03, 4.38	3.56	8.67
KTZ-SUC	2.7		4.2, 5.6		8.38
KTZ-ADP	3.4		4.44, 5.44		7.07

^a)From reference ²⁹

^b)From reference ¹⁴

^c)From reference ²⁰

^d)Determined using Othmer Thaker equation.³⁰

Figure 1.

Interfacial pH of KTZ (), KTZ-ADP (), KTZ-SUC () and KTZ-FUM () as a function of bulk pH. Interfacial pH of both drug and cocrystals were calculated using equations 54 and 86, respectively, and with the physicochemical parameters shown in Table 1.

Co-crystals may be supersaturating drug delivery systems and precipitation of the parent drug can happen during dissolution. However, depending on solution conditions, co-crystals can exhibit higher or lower solubility compared to the parent drug. The solubility behavior of KTZ has been evaluated and compared to the three co-crystals by Y.M. Chen and et al.²⁴ The different solubility pH dependence between the drug and co-crystal has resulted in a pH_max value that serves as a transition point for determining the thermodynamic stability of co-crystal.²⁴ Above this pH, the co-crystal is more soluble than the drug and can convert to the stable drug form during dissolution. Below pH_max the drug is more soluble than co-crystal. Utilizing the predicted interfacial pH, the thermodynamic stability of the co-crystals at the dissolving surface can be assessed by comparing the ratio of the co-crystal solubility to that of the drug (S_cc/S_drug) defined as solubility advantage. As shown in Figure 2, all three cocrystals do not achieve a solubility advantage (S_cc/S_drug > 1) until bulk pH ≥ 4 because the interfacial pH has to be above their pH_max values, which range from 3.6 to 3.8.²⁴ Solubility advantage increases with bulk pH. However, it plateaus off at bulk pH 5 to 8 where change in interfacial pH is minimal due to the buffering ability of the coformers. The solubility advantage at the dissolving surface can indicate the driving force for drug precipitation during dissolution. For example, KTZ-ADP has the highest solubility advantage in the buffering region (bulk pH 5 to 8), as seen in Figure 2 and therefore likely has the greatest tendency to convert to the stable drug form during dissolution.

Figure 2.

Solubility advantage for KTZ-FUM (), KTZ-SUC () and KTZ-ADP () as a function of bulk pH. The solubility of the drug and cocrystals were calculated using equation 11 and 32 respectively and based on the interfacial pH predicted from Figure 1.

Effect of pH on KTZ dissolution

The effect of pH on the dissolution of KTZ was evaluated by performing rotating disk dissolution as a function of bulk pH. The dissolution concentration profiles of KTZ as a function of bulk pH are shown in Figure 3. The basicity of KTZ resulted in a significant decrease in dissolution rate from pH 2 to 6. Given its poor intrinsic solubility (4.7 × 10⁻⁶ M)²⁴, the KTZ concentration could not reach the detection limit until 5, 10 and 30 minutes after dissolution at bulk pH 4, 5 and 6, respectively, where ionization is minimal. Therefore, the dissolution at bulk pH 6 was extended to 60 minutes to determine the dissolution rate. The large error bars at bulk pH 6 are likely associated with the low concentration of KTZ.

Figure 3.

Dissolution concentration profiles of KTZ at bulk pH 2 to 5 (a) and bulk pH 6 (b).

The pH dependent dissolution behavior of KTZ can be predicted based on the knowledge of interfacial pH. As shown in Figure 4, the flux of KTZ is highly dependent on pH. It drops from 2.4×10⁻³ at pH 2 to 1.6×10⁻⁶ mmole/cm² min at pH 6, which is about a 2000 fold difference. This large pH effect on the dissolution rate of KTZ may be responsible for its variability in oral absorption observed in patients. The predictive power of the mass transport model was also evaluated in Figure 4 and there is quite good agreement between the experimental results and theoretical predictions. There are some deviations in flux predictions for KTZ at low bulk pH, which is possibly due to the assumption that the diffusion coefficients of the unionized and ionized species are the same. In multicomponent electrolyte mass transport system, the diffusion of the charged species can be significantly different from the neutral species due to the electrostatic interactions between the diffusing species in order to maintain charge neutrality.^25–27 In general, the fast diffusing ion is coupled with an ion of opposite charge to counteract the charge separation between the ions.^25–27.

Figure 4.

Theoretical (solid line) and experimental (solid circle) flux comparison of KTZ as a function of bulk pH. The flux of KTZ was calculated using equation 57 based on the interfacial pH predicted in Figure 1 and the physicochemical properties shown in Table 1.

Effect of pH on the dissolution of KTZ co-crystals

The pH effect on the dissolution of three model KTZ co-crystals was also evaluated and the dissolution concentration profiles as a function of bulk pH are shown in Figure 5, 6 and 7 for KTZ-FUM, KTZ-SUC and KTZ-ADP, respectively. Typically, the concentrations of both drug and coformer are measured during dissolution. Knowledge of the coformer concentration is important because it can help to confirm the stability of the co-crystal during dissolution in cases where XRPD is not sensitive enough to detect solid phase transformation. Among the three co-crystals, only FUM concentration was measured as the concentrations of ADP and SUC were both below the limit of detection. As shown in Figure 5, the coformer concentrations match those of the drug at all bulk pH conditions studied here, which indicates no solid phase transformation occurred during the dissolution of KTZ-FUM. Although the SUC concentration was not measured, the stability of KTZ-SUC was confirmed by the linear dissolution behavior shown in Figure 6 and also the XRPD data after dissolution. The XRPD analysis showed pure co-crystal phase after the dissolution of KTZ-ADP, however, the nonlinear dissolution behavior observed at bulk pH 4 to 7 is very likely a sign of solid phase transformation resulting in decreased dissolution at later timepoints. This observation is consistent with the solubility advantage prediction for KTZ-ADP at the dissolving surface; the driving force for solid phase transformation is the highest for KTZ-ADP (S_cc/S_drug = 11).

Figure 5.

Dissolution concentration profiles of KTZ-FUM in terms of KTZ (a) and FUM (b) concentrations as a function of bulk pH.

Figure 6.

Dissolution concentration profiles of KTZ-SUC in terms of KTZ concentrations as a function of bulk pH.

Figure 7.

Dissolution concentration profiles of KTZ-ADP in terms of KTZ concentrations at bulk pH 2 and 3 (a) and the nonlinear dissolution behavior at bulk pH 4 to 7 (b).

To evaluate the predictive power of the mass transport model, the flux of the three co-crystals were calculated from the dissolution rates and compared to the theoretical predictions in Figure 8. With the predicted interfacial pH shown in Figure 1, the mass transport model adequately describes the dissolution behavior of co-crystals under the influence of pH. There is an excellent agreement between the theoretical flux predictions and experimental data. Precipitation kinetics have not been considered for the current mass transport model, so co-crystal flux can be over predicted if drug crystallization happened during dissolution, which appears to be the case for KTZ-ADP.

Figure 8.

Theoretical (solid line) and experimental (solid circle) flux comparison of a) KTZFUM, b) KTZ-SUC and c) KTZ-ADP. The theoretical flux of cocrystals were calculated using equation 88 based on the interfacial pH predicted in Figure 1 and the physicochemical properties shown in Table 1.

The flux of all three co-crystals decrease with increasing bulk pH and reach essentially constant values in the buffering region since there is no significant change in interfacial pH. With the acidic coformers, the dissolution flux from co-crystals still decreases with pH at low bulk pH because the pH effect on dissolution is dominated by the basic drug. At bulk pH above the acidic pKa values, the pH effect is dominated by the coformers as ionization starts, while the drug ionization is suppressed. The acidity of the coformers dampens the pH effect on the dissolution of the co-crystals compared to the parent drug. There is about 5x difference in the flux of KTZ-FUM between pH 2 and 6, 11x for KTZ-SUC and 15x for KTZ-ADP. These differences are significantly smaller compared to the almost 2000x difference for KTZ; hense the co-crystals offer an opportunity to mitigate the effect of pH on dissolution in the physiologic pH range.

Comparison of dissolution behavior

The pH effect on the flux of KTZ is compared with KTZ-FUM in Figure 9, KTZ-SUC in Figure 10 and KTZ-ADP in Figure 11. These Figures demonstrate that the dissolution pH dependence of KTZ is significantly reduced by co-crystallizing with acidic coformers. Due to the different pH dependence, there exists a pH where the flux of the drug is the same as the co-crystal for each of the three co-crystals. This pH is an important transition point at which the co-crystal displays dissolution advantage above it and no advantage below it. Dissolution advantage of co-crystal is defined as the ratio of the co-crystal flux over that of the drug (J_cc/J_drug).¹¹ A co-crystal has no dissolution advantage if J_cc/J_drug ≤ 1, but it would exhibit a dissolution advantage if J_cc/J_drug > 1.¹¹ The transition pH for KTZ-FUM is ~ 3.5, KTZ-SUC and KTZ-ADP are both ~ 3. As shown in Figures 10 to 11, all three co-crystals display no dissolution advantage with J_cc/J_drug ≤ 1 at or around the transition pH, but display a significant dissolution advantage with J_cc/J_drug >> 1 above the transition pH. Among the three co-crystals, KTZ-FUM has the highest dissolution advantage, follows by KTZ-SUC and KTZ-ADP.

Figure 9.

a) Flux comparison between KTZ-FUM () and KTZ () as a function of bulk pH. b) Dissolution advantage of KTZ-FUM determined from the experimental flux values. The solid lines represent the theoretical predictions and the symbols represent the experimental flux for KTZ-FUM (circle) and KTZ (triangle).

Figure 10.

a) Flux comparison between KTZ-SUC () and KTZ () as a function of bulk pH. b) Dissolution advantage of KTZ-SUC determined from the experimental flux. The solid lines represent the theoretical predictions and the symbols represent the experimental flux for KTZ-FUM (circle) and KTZ (triangle).

Figure 11.

a) Flux comparison between KTZ-ADP() and KTZ () as a function of bulk pH. b) Dissolution advantage of KTZ-ADP determined from the experimental flux. The solid lines represent the theoretical predictions and the symbols represent the experimental flux for KTZ-FUM (circle) and KTZ (triangle).

Discussion

Each of the three co-crystals studied in this work demonstrate a different ability to modulate the dissolution behavior of the parent drug and this ability is dependent on the properties of the coformers and co-crystals. The substantial pH effect on the dissolution of KTZ is significantly reduced with acidic formers, which is due to their ability to lower the interfacial pH. The significant reduction in flux variation due to pH can potentially mitigate the pH effect on the oral absorption of KTZ.

One of the important questions to be addressed during the co-crystal selection process is what the most desired co-crystal solubility should be. Higher co-crystal solubility would not necessarily lead to better performance because it is more prone to conversion during dissolution. The ideal candidate would be a co-crystal with a solubility advantage that can sustain supersaturation for a period of time to maximize absorption in the GI tract. Understanding the solution behavior of co-crystals would help assist the selection process. Given that it has lowest solubility advantage at the dissolving surface, KTZFUM is the most stable during dissolution and it has the highest dissolution advantage. With high dissolution advantage and low risk of conversion, KTZ-FUM would appear to have the most potential for further development. It is important to point ou that this work has considered the pH effect on the dissolution of co-crystals; however, other factors such as solubilizing agents should also be considered for the selection process.

Conclusions

This work provides a mechanistic understanding of the dissolution behavior of co-crystals with dibasic drug and diacidic coformers. The acidity of the coformers lowers the interfacial pH and results in a significant reduction in dissolution pH dependence compared to the parent drug. Because of the different pH dependence, there exists a transition pH that can serve as a turning point for the co-crystal dissolution advantage. Mass transport analysis of the interfacial pH allows for the evaluation of cocrystal solubility advantage at the dissolving surface, which provides useful information regarding the thermodynamic stability of co-crystals during dissolution. Co-crystals can modulate the dissolution behavior of the parent drug by altering both interfacial pH and solubility. Having a thorough understanding on the thermodynamic and kinetic behavior of co-crystals can help to select the most suitable co-crystal for further development.

Appendix

Mass transport analysis for drug

The flux of all the species across the diffusion layer include both the diffusion and chemical reactions happening during dissolution. At steady state, the diffusion and simultaneous chemical reactions of the individual species within the diffusion layer can be written using Fick’s law²³ as follows:

$$\frac{\partial [\text{B}]}{\partial \text{t}}={\text{D}}_{\text{B}}\frac{{\partial }^2[\text{B}]}{\partial {\text{x}}^2}+{\mathrm{\varphi }}_1=0$$ (35)

$$\frac{\partial \left[{\text{BH}}^{+}\right]}{\partial \text{t}}={\text{D}}_{\text{B}}\frac{{\partial }^2\left[{\text{BH}}^{+}\right]}{\partial {\text{x}}^2}+{\mathrm{\varphi }}_2=0$$ (36)

$$\frac{\partial \left[{\text{BH}}_2^{2+}\right]}{\partial \text{t}}={\text{D}}_{\text{B}}\frac{{\partial }^2\left[{\text{BH}}_2^{2+}\right]}{\partial {\text{x}}^2}+{\mathrm{\varphi }}_3=0$$ (37)

$$\frac{\partial \left[{\text{OH}}^{-}\right]}{\partial \text{t}}={\text{D}}_{{\text{OH}}^{-}}\frac{{\partial }^2\left[{\text{OH}}^{-}\right]}{\partial {\text{x}}^2}+{\mathrm{\varphi }}_4=0$$ (38)

$$\frac{\partial \left[{\text{H}}^{+}\right]}{\partial \text{t}}={\text{D}}_{{\text{H}}^{+}}\frac{{\partial }^2\left[{\text{H}}^{+}\right]}{\partial {\text{x}}^2}+{\mathrm{\varphi }}_5=0$$ (39)

where φ_1–5 are the reaction rate functions. At equilibrium, the reaction rate of the reactant should be the opposite of the product:

$${\mathrm{\varphi }}_3=-{\mathrm{\varphi }}_1-{\mathrm{\varphi }}_2$$ (40)

The reaction rate of the acid and the base should be the same:

$${\mathrm{\varphi }}_3+{\mathrm{\varphi }}_5={\mathrm{\varphi }}_1+{\mathrm{\varphi }}_4$$ (41)

Based on equations 40 and 41, the following mass balance equations can be written:

$${\text{D}}_{\text{B}}\frac{{\text{d}}^2[\text{B}]}{{\text{dx}}^2}+{\text{D}}_{\text{B}}\frac{{\text{d}}^2\left[{\text{BH}}^{+}\right]}{{\text{dx}}^2}+{\text{D}}_{\text{B}}\frac{{\text{d}}^2\left[{\text{BH}}_2^{2+}\right]}{{\text{dx}}^2}=0$$ (42)

$${\text{D}}_{{\text{OH}}^{-}}\frac{{\text{d}}^2\left[{\text{OH}}^{-}\right]}{{\text{dx}}^2}+{\text{D}}_{\text{B}}\frac{{\text{d}}^2[\text{B}]}{{\text{dx}}^2}={\text{D}}_{{\text{H}}^{+}}\frac{{\text{d}}^2\left[{\text{H}}^{+}\right]}{{\text{dx}}^2}+{\text{D}}_{\text{B}}\frac{{\text{d}}^2\left[{\text{BH}}_2^{2+}\right]}{{\text{dx}}^2}$$ (43)

Integrating equations 42 and 43 once, gives:

$${\text{D}}_{\text{B}}\frac{\text{d}[\text{B}]}{\text{dx}}+{\text{D}}_{\text{B}}\frac{\text{d}\left[{\text{BH}}^{+}\right]}{\text{dx}}+{\text{D}}_{\text{B}}\frac{\text{d}\left[{\text{BH}}_2^{2+}\right]}{\text{dx}}={\text{C}}_1$$ (44)

$${\text{D}}_{{\text{OH}}^{-}}\frac{\text{d}\left[{\text{OH}}^{-}\right]}{\text{dx}}+{\text{D}}_{\text{B}}\frac{\text{d}[\text{B}]}{\text{dx}}={\text{D}}_{{\text{H}}^{+}}\frac{\text{d}\left[{\text{H}}^{+}\right]}{\text{dx}}+{\text{D}}_{\text{B}}\frac{\text{d}\left[{\text{BH}}_2^{2+}\right]}{\text{dx}}+{\text{C}}_2$$ (45)

By charge neutrality:

$${\text{D}}_{{\text{OH}}^{-}}\frac{\text{d}\left[{\text{OH}}^{-}\right]}{\text{dx}}={\text{D}}_{{\text{H}}^{+}}\frac{\text{d}\left[{\text{H}}^{+}\right]}{\text{dx}}+{\text{D}}_{\text{B}}\frac{\text{d}\left[{\text{BH}}^{+}\right]}{\text{dx}}+2{\text{D}}_{\text{B}}\frac{\text{d}\left[{\text{BH}}_2^{2+}\right]}{\text{dx}}$$ (46)

By combining equation 44, 45 and 46, it can be shown that:

$${\text{C}}_1=-{\text{C}}_2$$ (47)

Integrating equations 44 and 45,

$${\text{D}}_{\text{B}}[\text{B}]+{\text{D}}_{\text{B}}\left[{\text{BH}}^{+}\right]+{\text{D}}_{\text{B}}\left[{\text{BH}}_2^{2+}\right]={\text{C}}_1\text{x}+{\text{C}}_3$$ (48)

$${\text{D}}_{{\text{OH}}^{-}}\left[{\text{OH}}^{-}\right]+{\text{D}}_{\text{B}}[\text{B}]={\text{D}}_{{\text{H}}^{+}}\left[{\text{H}}^{+}\right]+{\text{D}}_{\text{B}}\left[{\text{BH}}_2^{2+}\right]+{\text{C}}_2\text{x}+{\text{C}}_4$$ (49)

Boundary conditions:

$$\begin{array}{ll}\text{At}\text{x}=0:& \text{At}\text{x}=\text{h}:\\ [B]=[B{]}_0\left(\text{intrinsic}\text{solubility}\text{of}\text{the}\text{drug}\right)& [B]=0\left(\text{sink}\text{condition}\right)\\ [B{H}^{+}]=unknown& [B{H}^{+}]=0\left(\text{sink}\text{condition}\right)\\ \left[B{H}_2^{2+}\right]=unknown& \left[B{H}_2^{2+}\right]=0\left(\text{sink}\text{condition}\right)\\ [H^{+}]=[H^{+}{]}_0& [H^{+}]=[H^{+}{]}_h\\ [O{H}^{-}]=[O{H}^{-}{]}_0& [O{H}^{-}]=[O{H}^{-}{]}_h\end{array}$$

Evaluation of interfacial pH

Applying the above boundary conditions to equation 48 and 49, at x = 0:

$${\text{D}}_{\text{B}}{\left[\text{B}\right]}_0+{\text{D}}_{\text{B}}\left[\text{B}{\text{H}}^{+}\right]+{\text{D}}_{\text{B}}\left[\text{B}{\text{H}}_2^{2+}\right]={\text{C}}_3$$ (50)

$${\text{D}}_{{\text{OH}}^{-}}{\left[{\text{OH}}^{-}\right]}_0+{\text{D}}_{\text{B}}{[\text{B}]}_0={\text{D}}_{{\text{H}}^{+}}{\left[{\text{H}}^{+}\right]}_0+{\text{D}}_{\text{B}}\left[{\text{BH}}_2^{2+}\right]+{\text{C}}_4$$ (51)

and at x = h, assuming sink conditions, equation 48 and 49 can be written as:

$$0={\text{C}}_1\text{h}+{\text{C}}_3$$ (52)

$${\text{D}}_{{\text{OH}}^{-}}{\left[{\text{OH}}^{-}\right]}_{\text{h}}={\text{D}}_{{\text{H}}^{+}}{\left[{\text{H}}^{+}\right]}_{\text{h}}+{\text{C}}_2\text{h}+{\text{C}}_4$$ (53)

Combining equations 50 to 53 and algebraically solving for interfacial pH, [H⁺]₀, yields the following equation:

$$2D_B\frac{{[B]}_0}{K_{a1}^B{K}_{a2}^B}{{\left[H^{+}\right]}_0}^3+\left(D_{H^{+}}+D_B\frac{{[B]}_0}{K_{a2}^B}\right){{\left[H^{+}\right]}_0}^2+\left(D_{O{H}^{-}}{\left[O{H}^{-}\right]}_h-D_{H^{+}}{\left[H^{+}\right]}_h\right){\left[H^{+}\right]}_0-D_{O{H}^{-}}{K}_w=0$$ (54)

Evaluation of flux

Combine equation 50 and 52, and solve for −C₁ for the total flux of the drug species across the diffusion layer:

$${\text{J}}_{\text{B}}=\frac{{\text{D}}_{\text{B}}}{\text{h}}{[\text{B}]}_0\left(1+\frac{{\left[{\text{H}}^{+}\right]}_0}{{\text{K}}_{\text{a}2}^{\text{B}}}+\frac{{{\left[{\text{H}}^{+}\right]}_0}^2}{{\text{K}}_{\text{a}1}^{\text{B}}{\text{K}}_{\text{a}2}^{\text{B}}}\right)$$ (55)

For rotating disk, the thickness of the hydrodynamic boundary layer can be defined according to Levich model²⁸:

$$\text{h}=1.612{\text{D}}^{\frac{1}{3}}{\text{v}}^{\frac{1}{6}}{\mathrm{\omega }}^{-\frac{1}{2}}$$ (56)

where ν is the kinematic viscosity and ω is the angular velocity in radians per unit time. Substitute equation 56 into 55, the flux of the drug becomes:

$${\text{J}}_{\text{B}}=0.62{\text{D}}_{\text{B}}^{2/3}{\mathrm{\omega }}^{1/2}{\text{v}}^{-1/6}{[\text{B}]}_0\left(1+\frac{{\left[{\text{H}}^{+}\right]}_0}{{\text{K}}_{\text{a}2}^{\text{B}}}+\frac{{{\left[{\text{H}}^{+}\right]}_0}^2}{{\text{K}}_{\text{a}1}^{\text{B}}{\text{K}}_{\text{a}2}^{\text{B}}}\right)$$ (57)

Mass transport analysis for co-crystal

The flux of all the species across the diffusion layer include both the diffusion and chemical reactions happening during dissolution. At steady state, the diffusion and simultaneous chemical reactions of the individual species within the diffusion layer can be written using Fick’s law²³ as follows:

$$\frac{\partial [\text{B}]}{\partial \text{t}}={\text{D}}_{\text{B}}\frac{{\partial }^2[\text{B}]}{\partial {\text{x}}^2}+{\mathrm{\varphi }}_1=0$$ (58)

$$\frac{\partial \left[{\text{BH}}^{+}\right]}{\partial \text{t}}={\text{D}}_{\text{B}}\frac{{\partial }^2\left[{\text{BH}}^{+}\right]}{\partial {\text{x}}^2}+{\mathrm{\varphi }}_2=0$$ (59)

$$\frac{\partial \left[{\text{BH}}_2^{2+}\right]}{\partial \text{t}}={\text{D}}_{\text{B}}\frac{{\partial }^2\left[{\text{BH}}_2^{2+}\right]}{\partial {\text{x}}^2}+{\mathrm{\varphi }}_3=0$$ (60)

$$\frac{\partial \left[{\text{OH}}^{-}\right]}{\partial \text{t}}={\text{D}}_{{\text{OH}}^{-}}\frac{{\partial }^2\left[{\text{OH}}^{-}\right]}{\partial {\text{x}}^2}+{\mathrm{\varphi }}_4=0$$ (61)

$$\frac{\partial \left[{\text{H}}^{+}\right]}{\partial \text{t}}={\text{D}}_{{\text{H}}^{+}}\frac{{\partial }^2\left[{\text{H}}^{+}\right]}{\partial {\text{x}}^2}+{\mathrm{\varphi }}_5=0$$ (62)

$$\frac{\partial \left[{\text{H}}_2\text{A}\right]}{\partial \text{t}}={\text{D}}_{{\text{H}}_2\text{A}}\frac{{\partial }^2\left[{\text{H}}_2\text{A}\right]}{\partial {\text{x}}^2}+{\mathrm{\varphi }}_6=0$$ (63)

$$\frac{\partial \left[{\text{HA}}^{-}\right]}{\partial \text{t}}={\text{D}}_{{\text{H}}_2\text{A}}\frac{{\partial }^2\left[{\text{HA}}^{-}\right]}{\partial {\text{x}}^2}+{\mathrm{\varphi }}_7=0$$ (64)

$$\frac{\partial [{\text{A}}^{2-}]}{\partial \text{t}}={\text{D}}_{{\text{H}}_2\text{A}}\frac{{\partial }^2[{\text{A}}^{2-}]}{\partial {\text{x}}^2}+{\mathrm{\varphi }}_8=0$$ (65)

At equilibrium, the reaction rate of the reactant should be the opposite of the product. Based on the chemical equilibria, the followings can be written:

$${\mathrm{\varphi }}_3=-{\mathrm{\varphi }}_1-{\mathrm{\varphi }}_2$$ (66)

$${\mathrm{\varphi }}_6=-{\mathrm{\varphi }}_7-{\mathrm{\varphi }}_8$$ (67)

The reaction rate of the acid and the base should be the same:

$${\mathrm{\varphi }}_3+{\mathrm{\varphi }}_5+{\mathrm{\varphi }}_6={\mathrm{\varphi }}_1+{\mathrm{\varphi }}_4+{\mathrm{\varphi }}_8$$ (68)

Based on equation 66, 67 and 68, the following mass balance equations can be written:

$${\text{D}}_{\text{B}}\frac{{\text{d}}^2[\text{B}]}{{\text{dx}}^2}+{\text{D}}_{\text{B}}\frac{{\text{d}}^2\left[{\text{BH}}^{+}\right]}{{\text{dx}}^2}+{\text{D}}_{\text{B}}\frac{{\text{d}}^2\left[{\text{BH}}_2^{2+}\right]}{{\text{dx}}^2}=0$$ (69)

$${\text{D}}_{{\text{H}}_2\text{A}}\frac{{\text{d}}^2\left[{\text{H}}_2\text{A}\right]}{{\text{dx}}^2}+{\text{D}}_{{\text{H}}_2\text{A}}\frac{{\text{d}}^2\left[{\text{HA}}^{-}\right]}{{\text{dx}}^2}+{\text{D}}_{{\text{H}}_2\text{A}}\frac{{\text{d}}^2[{\text{A}}^{2-}]}{{\text{dx}}^2}=0$$ (70)

$${\text{D}}_{{\text{OH}}^{-}}\frac{{\text{d}}^2\left[{\text{OH}}^{-}\right]}{{\text{dx}}^2}+{\text{D}}_{\text{B}}\frac{{\text{d}}^2[\text{B}]}{{\text{dx}}^2}+{\text{D}}_{{\text{H}}_2\text{A}}\frac{{\text{d}}^2[{\text{A}}^{2-}]}{{\text{dx}}^2}={\text{D}}_{{\text{H}}^{+}}\frac{{\text{d}}^2\left[{\text{H}}^{+}\right]}{{\text{dx}}^2}+{\text{D}}_{\text{B}}\frac{{\text{d}}^2\left[{\text{BH}}_2^{2+}\right]}{{\text{dx}}^2}{\text{D}}_{{\text{H}}_2\text{A}}\frac{{\text{d}}^2\left[{\text{H}}_2\text{A}\right]}{{\text{dx}}^2}$$ (71)

Integrate equation 69 to 71 once, gives:

$${\text{D}}_{\text{B}}\frac{\text{d}[\text{B}]}{\text{dx}}+{\text{D}}_{\text{B}}\frac{\text{d}\left[{\text{BH}}^{+}\right]}{\text{dx}}+{\text{D}}_{\text{B}}\frac{\text{d}\left[{\text{BH}}_2^{2+}\right]}{\text{dx}}={\text{C}}_1$$ (72)

$${\text{D}}_{{\text{H}}_2\text{A}}\frac{\text{d}\left[{\text{H}}_2\text{A}\right]}{\text{dx}}+{\text{D}}_{{\text{H}}_2\text{A}}\frac{\text{d}\left[{\text{HA}}^{-}\right]}{\text{dx}}+{\text{D}}_{{\text{H}}_2\text{A}}\frac{\text{d}[{\text{A}}^{2-}]}{\text{dx}}={\text{C}}_2$$ (73)

$${\text{D}}_{{\text{OH}}^{-}}\frac{\text{d}\left[{\text{OH}}^{-}\right]}{\text{dx}}+{\text{D}}_{\text{B}}\frac{\text{d}[\text{B}]}{\text{dx}}+{\text{D}}_{{\text{H}}_2\text{A}}\frac{\text{d}[{\text{A}}^{2-}]}{\text{dx}}={\text{D}}_{{\text{H}}^{+}}\frac{\text{d}\left[{\text{H}}^{+}\right]}{\text{dx}}+{\text{D}}_{\text{B}}\frac{\text{d}\left[\text{B}{\text{H}}_2^{2+}\right]}{\text{dx}}+{\text{D}}_{{\text{H}}_2\text{A}}\frac{\text{d}\left[{\text{H}}_2\text{A}\right]}{\text{dx}}+{\text{C}}_3$$ (74)

By charge neutrality:

$${\text{D}}_{{\text{OH}}^{-}}\frac{\text{d}\left[{\text{OH}}^{-}\right]}{\text{dx}}+{\text{D}}_{{\text{H}}_2\text{A}}\frac{\text{d}\left[{\text{HA}}^{-}\right]}{\text{dx}}+2{\text{D}}_{{\text{H}}_2\text{A}}\frac{\text{d}[{\text{A}}^{2-}]}{\text{dx}}={\text{D}}_{{\text{H}}^{+}}\frac{\text{d}\left[{\text{H}}^{+}\right]}{\text{dx}}+{\text{D}}_{\text{B}}\frac{\text{d}\left[{\text{BH}}^{+}\right]}{\text{dx}}+2{\text{D}}_{\text{B}}\frac{\text{d}\left[{\text{BH}}_2^{2+}\right]}{\text{dx}}$$ (75)

By combining equation 72 to 75, it can be shown that:

$${\text{C}}_3={\text{C}}_2-{\text{C}}_1$$ (76)

Integrating equations 72 to 74, gives:

$${\text{D}}_{\text{B}}\left[\text{B}\right]+{\text{D}}_{\text{B}}\left[\text{B}{\text{H}}^{+}\right]+{\text{D}}_{\text{B}}\left[\text{B}{\text{H}}_2^{2+}\right]={\text{C}}_1\text{x}+{\text{C}}_4$$ (77)

$${\text{D}}_{{\text{H}}_2\text{A}}\left[{\text{H}}_2\text{A}\right]+{\text{D}}_{{\text{H}}_2\text{A}}\left[{\text{HA}}^{-}\right]+{\text{D}}_{{\text{H}}_2\text{A}}\left[{\text{A}}^{2-}\right]={\text{C}}_2\text{x}+{\text{C}}_5$$ (78)

$${\text{D}}_{{\text{OH}}^{-}}\left[{\text{OH}}^{-}\right]+{\text{D}}_{\text{B}}[\text{B}]+{\text{D}}_{{\text{H}}_2\text{A}}\left[{\text{A}}^{2-}\right]={\text{D}}_{{\text{H}}^{+}}\left[{\text{H}}^{+}\right]+{\text{D}}_{\text{B}}\left[{\text{BH}}_2^{2+}\right]+{\text{D}}_{{\text{H}}_2\text{A}}\left[{\text{H}}_2\text{A}\right]+{\text{C}}_3\text{x}+{\text{C}}_6$$ (79)

Based on equations 33 and 34, the concentrations of the nonionized drug and coformer at the dissolving surface can be written to include in the following boundary conditions:

$$\begin{array}{ll}\text{At}\text{x}=0:& \mathrm{at}\mathrm{x}=\mathrm{h}:\\ [\text{B}]=\frac{\sqrt{K_{sp}(1+\frac{K_{a1}^{H_{2}A}}{{\left[H^{+}\right]}_0}+\frac{K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}}{{{\left[H^{+}\right]}_0}^2})(1+\frac{{\left[H^{+}\right]}_0}{K_{a2}^B}+\frac{{{\left[H^{+}\right]}_0}^2}{K_{a1}^B{K}_{a2}^B})}}{(1+\frac{{\left[H^{+}\right]}_0}{K_{a2}^B}+\frac{{{\left[H^{+}\right]}_0}^2}{K_{a1}^B{K}_{a2}^B})}& [\text{B}]=0(\text{sink}\text{condition})\\ \left[H_{2}A\right]={(\frac{D_B}{D_{H_{2}A}})}^{2/3}\frac{\sqrt{K_{sp}(1+\frac{K_{a1}^{H_{2}A}}{{\left[H^{+}\right]}_0}+\frac{K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}}{{{\left[H^{+}\right]}_0}^2})(1+\frac{{\left[H^{+}\right]}_0}{K_{a2}^B}+\frac{{{\left[H^{+}\right]}_0}^2}{K_{a1}^B{K}_{a2}^B})}}{(1+\frac{K_{a1}^{H_{2}A}}{{\left[H^{+}\right]}_0}+\frac{K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}}{{{\left[H^{+}\right]}_0}^2})}& \left[H_{2}A\right]=0(\text{sink}\text{condition})\\ \left[B{H}^{+}\right]=unknown& \left[B{H}^{+}\right]=0(\text{sink}\text{condition})\\ \left[B{H_2}^{2+}\right]=unknown& \left[B{H_2}^{2+}\right]=0(\text{sink}\text{condition})\\ [H{A}^{-}]=unknown& \left[H{A}^{-}\right]=0(\text{sink}\text{condition})\\ \left[A^{2-}\right]=unknown& \left[A^{2-}\right]=0(\text{sink}\text{condition})\\ \left[H^{+}\right]={\left[H^{+}\right]}_0& \left[H^{+}\right]={\left[H^{+}\right]}_h\\ \left[O{H}^{-}\right]={\left[O{H}^{-}\right]}_0& \left[O{H}^{-}\right]={\left[O{H}^{-}\right]}_h\end{array}$$

Evaluation of interfacial pH

Applying the above boundary conditions to equations 77 to 79, at x = 0:

$${\text{D}}_{\text{B}}\frac{\sqrt{{\text{K}}_{\text{sp}}(1+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}}{{\left[{\text{H}}^{+}\right]}_0}+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}{\text{K}}_{\text{a}2}^{{\text{H}}_2\text{A}}}{{{\left[{\text{H}}^{+}\right]}_0}^2})(1+\frac{{\left[{\text{H}}^{+}\right]}_0}{{\text{K}}_{\text{a}2}^{\text{B}}}+\frac{{{\left[{\text{H}}^{+}\right]}_0}^2}{{\text{K}}_{\text{a}1}^{\text{B}}{\text{K}}_{\text{a}2}^{\text{B}}})}}{(1+\frac{{\left[{\text{H}}^{+}\right]}_0}{{\text{K}}_{\text{a}2}^{\text{B}}}+\frac{{{\left[{\text{H}}^{+}\right]}_0}^2}{{\text{K}}_{\text{a}1}^{\text{B}}{\text{K}}_{\text{a}2}^{\text{B}}})}+{\text{D}}_{\text{B}}\left[\text{B}{\text{H}}^{+}\right]+{\text{D}}_{\text{B}}\left[\text{B}{\text{H}}_2^{2+}\right]={\text{C}}_4$$ (80)

$${\text{D}}_{{\text{H}}_2\text{A}}{(\frac{{\text{D}}_{\text{B}}}{{\text{D}}_{{\text{H}}_2\text{A}}})}^{2/3}\frac{\sqrt{{\text{K}}_{\text{sp}}(1+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}}{{\left[{\text{H}}^{+}\right]}_0}+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}{\text{K}}_{\text{a}2}^{{\text{H}}_2\text{A}}}{{{\left[{\text{H}}^{+}\right]}_0}^2})(1+\frac{{\left[{\text{H}}^{+}\right]}_0}{{\text{K}}_{\text{a}2}^{\text{B}}}+\frac{{{\left[{\text{H}}^{+}\right]}_0}^2}{{\text{K}}_{\text{a}1}^{\text{B}}{\text{K}}_{\text{a}2}^{\text{B}}})}}{(1+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}}{{\left[{\text{H}}^{+}\right]}_0}+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}{\text{K}}_{\text{a}2}^{{\text{H}}_2\text{A}}}{{{\left[{\text{H}}^{+}\right]}_0}^2})}+{\text{D}}_{{\text{H}}_2\text{A}}\left[{\text{HA}}^{-}\right]+{\text{D}}_{{\text{H}}_2\text{A}}\left[{\text{A}}^{2-}\right]={\text{C}}_5$$ (81)

$$\begin{gathered} {\text{D}}_{{\text{OH}}^{-}}{\left[{\text{OH}}^{-}\right]}_0+{\text{D}}_{\text{B}}\frac{\sqrt{{\text{K}}_{\text{sp}}(1+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}}{{\left[{\text{H}}^{+}\right]}_0}+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}{\text{K}}_{\text{a}2}^{{\text{H}}_2\text{A}}}{{{\left[{\text{H}}^{+}\right]}_0}^2})(1+\frac{{\left[{\text{H}}^{+}\right]}_0}{{\text{K}}_{\text{a}2}^{\text{B}}}+\frac{{{\left[{\text{H}}^{+}\right]}_0}^2}{{\text{K}}_{\text{a}1}^{\text{B}}{\text{K}}_{\text{a}2}^{\text{B}}})}}{(1+\frac{{\left[{\text{H}}^{+}\right]}_0}{{\text{K}}_{\text{a}2}^{\text{B}}}+\frac{{{\left[{\text{H}}^{+}\right]}_0}^2}{{\text{K}}_{\text{a}1}^{\text{B}}{\text{K}}_{\text{a}2}^{\text{B}}})}+{\text{D}}_{{\text{H}}_2\text{A}}\left[{\text{A}}^{2-}\right]={\text{D}}_{{\text{H}}^{+}}{\left[{\text{H}}^{+}\right]}_0+{\text{D}}_{\text{B}}\left[{\text{BH}}_2^{2+}\right]+ \\ {\text{D}}_{{\text{H}}_2\text{A}}{(\frac{{\text{D}}_{\text{B}}}{{\text{D}}_{{\text{H}}_2\text{A}}})}^{2/3}\frac{\sqrt{{\text{K}}_{\text{sp}}(1+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}}{{\left[{\text{H}}^{+}\right]}_0}+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}{\text{K}}_{\text{a}2}^{{\text{H}}_2\text{A}}}{{{\left[{\text{H}}^{+}\right]}_0}^2})(1+\frac{{\left[{\text{H}}^{+}\right]}_0}{{\text{K}}_{\text{a}2}^{\text{B}}}+\frac{{{\left[{\text{H}}^{+}\right]}_0}^2}{{\text{K}}_{\text{a}1}^{\text{B}}{\text{K}}_{\text{a}2}^{\text{B}}})}}{(1+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}}{{\left[{\text{H}}^{+}\right]}_0}+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}{\text{K}}_{\text{a}2}^{{\text{H}}_2\text{A}}}{{{\left[{\text{H}}^{+}\right]}_0}^2})}+{\text{C}}_6 \end{gathered}$$ (82)

and at x = h, assuming sink conditions, equations 77 to 79 can be written as:

$$0={\text{C}}_1\text{h}+{\text{C}}_4$$ (83)

$$0={\text{C}}_2\text{h}+{\text{C}}_5$$ (84)

$${\text{D}}_{{\text{OH}}^{-}}{\left[{\text{OH}}^{-}\right]}_{\text{h}}={\text{D}}_{{\text{H}}^{+}}{\left[{\text{H}}^{+}\right]}_{\text{h}}+{\text{C}}_3\text{h}+{\text{C}}_6$$ (85)

Combining equations 80 to 85 and algebraically solving for interfacial pH, [H⁺]₀, yields the following equation:

$${\text{Ax}}^8+{\text{Bx}}^7+{\text{Cx}}^6+{\text{Dx}}^5+{\text{Ex}}^4+{\text{Fx}}^3+{\text{Gx}}^2+\text{Hx}+\text{I}=0$$ (86)

where:

$$A=4{D_B}^2{K}_{sp}-{D_{H^{+}}}^2{K}_{a1}^B{K}_{a2}^B;$$

$$\begin{gathered} B= \\ 4D_B{K}_{sp}\left(D_B{K}_{a1}^B+2D_B{K}_{a1}^{H_{2}A}-{D_B}^{\frac{2}{3}}{D_{H_{2}A}}^{\frac{1}{3}}{K}_{a1}^{H_{2}A}\right)-{D_{H^{+}}}^2{K}_{a1}^B{K}_{a_2}^B\left(K_{a1}^{H_{2}A}+K_{a1}^B\right)- \\ 2D_{H^{+}}{K}_{a1}^B{K}_{a2}^B\left(D_{O{H}^{-}}{\left[O{H}^{-}\right]}_h-D_{H^{+}}{\left[H^{+}\right]}_h\right); \end{gathered}$$

$$\begin{gathered} C={D_B}^2{K}_{sp}\left({K_{a1}^B}^2+8K_{a1}^{H_{2}A}{K}_{a1}^B+8K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}+4K_{a1}^{H_2{A}^2}\right)-2{D_B}^{\frac{5}{3}}{D_{H_{2}A}}^{\frac{1}{3}}{K}_{sp}(2K_{a1}^{H_2{A}^2}+3K_{a1}^{H_{2}A}{K}_{a1}^B+ \\ 4K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A})-2D_{H^{+}}{K}_{a1}^B{K}_{a2}^B\left(D_{O{H}^{-}}{\left[O{H}^{-}\right]}_h-D_{H^{+}}{\left[H^{+}\right]}_h\right)\left(K_{a1}^{H_{2}A}+K_{a1}^B\right)-K_{a1}^B{K}_{a2}^B(D_{O{H}^{-}}{\left[O{H}^{-}\right]}_h- \\ {D}_{H^{+}}{\left[H^{+}\right]}_h{)}^2-{D_{H^{+}}}^2{K}_{a1}^B{K}_{a2}^B\left(K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}+K_{a1}^{H_{2}A}{K}_{a1}^B+K_{a1}^B{K}_{a2}^B\right)+2D_{O{H}^{-}}{D}_{H^{+}}{K}_w{K}_{a1}^B{K}_{a2}^B+ \\ {D_B}^{\frac{4}{3}}{D_{H_{2}A}}^{\frac{2}{3}}{K}_{sp}{K}_{a1}^{H_2{A}^2}; \end{gathered}$$

$$\begin{gathered} D=2{D_B}^2{K}_{sp}{K}_{a1}^{H_{2}A}{K}_{a1}^B\left(K_{a1}^B+4K_{a2}^{H_{2}A}+2K_{a1}^{H_{2}A}\right)-2{D_B}^{\frac{5}{3}}{D_{H_{2}A}}^{\frac{1}{3}}{K}_{sp}{K}_{a1}^{H_{2}A}(2K_{a2}^{H_{2}A}+2K_{a1}^{H_{2}A}{K}_{a1}^B+ \\ 2K_{a1}^B{K}_{a2}^B+K_{a1}^{H_{2}A}{K}_{a1}^B+{K_{a1}^B}^2+4K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}+6K_{a2}^{H_{2}A}{K}_{a1}^B)+2{D_B}^{\frac{4}{3}}{D_{H_{2}A}}^{\frac{2}{3}}{K}_{sp}{K}_{a1}^{H_2{A}^2}\left(2K_{a2}^{H_{2}A}+K_{a1}^B\right)+ \\ 8{D_B}^2{K}_{sp}{K}_{a1}^{H_2{A}^2}{K}_{a2}^{H_{2}A}+2D_{O{H}^{-}}{D}_{H^{+}}{K}_w{K}_{a1}^B{K}_{a2}^B\left(K_{a1}^{H_{2}A}+K_{a1}^B\right)- \\ 2K_{a1}^B{K}_{a2}^B\left(D_{O{H}^{-}}{\left[O{H}^{-}\right]}_h-D_{H^{+}}{\left[H^{+}\right]}_h\right)\left(D_{H^{+}}{K}_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}+D_{H^{+}}{K}_{a_1}^{H_{2}A}{K}_{a_1}^B+D_{H^{+}}{K}_{a_1}^B{K}_{a2}^B-D_{O{H}^{-}}{K}_w\right)- \\ {D_{H^{+}}}^2{K}_{a1}^B{K}_{a2}^B\left(K_{a1}^{H_{2}A}{K}_{a1}^B+K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}\right)-K_{a1}^B{\left(D_{O{H}^{-}}{\left[O{H}^{-}\right]}_h-D_{H^{+}}{\left[H^{+}\right]}_h\right)}^2\left(K_{a1}^{H_{2}A}{K}_{a2}^B+K_{a1}^B{K}_{a2}^B\right); \end{gathered}$$

$$\begin{gathered} E={D_B}^2{K}_{sp}{K_{a1}^B}^2\left(2K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}+K_{a1}^{H_2{A}^2}\right)-2{D_B}^{\frac{5}{3}}{D_{H_{2}A}}^{\frac{1}{3}}{K}_{sp}{K}_{a1}^{H_{2}A}(9K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}{K}_{a1}^B+2K_{a1}^{H_{2}A}{K}_{a1}^B{K}_{a2}^B+ \\ 2K_{a2}^{H_{2}A}{K_{a1}^B}^2+4K_{a1}^{H_{2}A}{K_{a2}^{H_{2}A}}^2+4K_{a2}^{H_{2}A}{K}_{a1}^B{K}_{a2}^B+K_{a1}^{H_{2}A}{K_{a1}^B}^2+{K_{a1}^B}^2{K}_{a2}^B)+{D_B}^{\frac{4}{3}}{D_{H_{2}A}}^{\frac{2}{3}}{K}_{sp}{K_{a1}^{H_{2}A}}^2{(K_{a1}^B}^2+ \\ 2K_{a1}^B{K}_{a2}^B+4{K_{a2}^{H_{2}A}}^2+8K_{a2}^{H_{2}A}{K}_{a1}^B)-{\left(D_{O{H}^{-}}{K}_w\right)}^2{K}_{a1}^B{K}_{a2}^B+ \\ 2K_{a1}^B{K}_{a2}^B\left(D_{O{H}^{-}}{\left[O{H}^{-}\right]}_h-D_{H^{+}}{\left[H^{+}\right]}_h\right)(D_{O{H}^{-}}{K}_w{K}_{a1}^{H_{2}A}+D_{O{H}^{-}}{K}_w{K}_{a1}^B-D_{H^{+}}{K}_{a1}^B{K}_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}- \\ {D}_{H^{+}}{K}_{a1}^{H_{2}A}{K}_{a1}^B{K}_{a2}^B)+2D_{O{H}^{-}}{D}_{H^{+}}{K}_w{K}_{a1}^B{K}_{a2}^B\left(K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}+K_{a1}^{H_{2}A}{K}_{a1}^B+K_{a1}^B{K}_{a2}^B\right)- \\ {K}_{a1}^B{K}_{a2}^B{\left(D_{O{H}^{-}}{\left[O{H}^{-}\right]}_h-D_{H^{+}}{\left[H^{+}\right]}_h\right)}^2\left(K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}+K_{a1}^{H_{2}A}{K}_{a1}^B+K_{a1}^B{K}_{a2}^B\right)+ \\ 4{D_B}^2{K}_{sp}{K}_{a1}^{H_2{A}^2}\left(2K_{a2}^{H_{2}A}{K}_{a1}^B+K_{a2}^{H_2{A}^2}\right)-{D_{H^{+}}}^2{\left(K_{a1}^B{K}_{a2}^B\right)}^2{K}_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}; \end{gathered}$$

$$\begin{gathered} F= \\ 2{D_B}^{\frac{4}{3}}{D_{H_{2}A}}^{\frac{2}{3}}{K}_{sp}{K}_{a1}^{H_2{A}^2}{K}_{a1}^B\left(2K_{a2}^{H_{2}A}{K}_{a1}^B+4K_{a2}^{H_{2}A}{K}_{a2}^B+4K_{a2}^{H_2{A}^2}+K_{a1}^B{K}_{a2}^B\right)- \\ 2{D_B}^{\frac{5}{3}}{D_{H_{2}A}}^{\frac{1}{3}}{K}_{sp}{K}_{a1}^{H_2{A}^2}{K}_{a1}^B\left(6K_{a2}^{H_2{A}^2}+6K_{a2}^{H_{2}A}{K}_{a2}^B+K_{a1}^B{K}_{a2}^B+3K_{a2}^{H_{2}A}{K}_{a1}^B+2\frac{K_{a2}^{H_{2}A}}{K_{a1}^{H_{2}A}}{K}_{a1}^B{K}_{a2}^B\right)- \\ {\left(D_{O{H}^{-}}{K}_w\right)}^2{K}_{a1}^B{K}_{a2}^B\left(K_{a1}^{H_{2}A}+K_{a1}^B\right)+2D_{O{H}^{-}}{K}_w{K}_{a1}^B{K}_{a2}^B\left(D_{O{H}^{-}}{\left[O{H}^{-}\right]}_h-D_{H^{+}}{\left[H^{+}\right]}_h\right)(K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}+ \\ {K}_{a1}^{H_{2}A}{K}_{a1}^B+K_{a1}^B{K}_{a2}^B)+2D_{O{H}^{-}}{D}_{H^{+}}{K}_w{K_{a1}^B}^2{K}_{a1}^{H_{2}A}\left({K_{a2}^{H_{2}A}{K}_{a2}^B+K_{a2}^B}^2\right)-{K_{a1}^B}^2{K}_{a1}^{H_{2}A}(D_{O{H}^{-}}{\left[O{H}^{-}\right]}_h- \\ {D}_{H^{+}}{\left[H^{+}\right]}_h{)}^2\left(K_{a2}^{H_{2}A}{K}_{a2}^B+{K_{a2}^B}^2\right)-2D_{H^{+}}\left(D_{O{H}^{-}}{\left[O{H}^{-}\right]}_h-D_{H^{+}}{\left[H^{+}\right]}_h\right)K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}{\left(K_{a1}^B{K}_{a2}^B\right)}^2+ \\ 2{D_B}^2{K}_{sp}{K_{a1}^{H_{2}A}}^2{K}_{a2}^{H_{2}A}{K}_{a1}^B\left(K_{a1}^B+2K_{a2}^{H_{2}A}\right); \end{gathered}$$

$$\begin{gathered} G= \\ {D_B}^{\frac{4}{3}}{D_{H_{2}A}}^{\frac{2}{3}}{K}_{sp}{K}_{a1}^{H_2{A}^2}{K}_{a1}^B\left(K_{a1}^B{K_{a2}^B}^2+8K_{a2}^{H_2{A}^2}{K}_{a2}^B+8K_{a2}^{H_{2}A}{K}_{a1}^B{K}_{a2}^B+4K_{a2}^{H_2{A}^2}{K}_{a1}^B\right)- \\ 2{D_B}^{\frac{5}{3}}{D_{H_{2}A}}^{\frac{1}{3}}{K}_{sp}{K}_{a1}^{H_2{A}^2}{K}_{a2}^{H_{2}A}{K}_{a1}^B\left(4K_{a2}^{H_{2}A}{K}_{a2}^B+3K_{a1}^B{K}_{a2}^B+2K_{a2}^{H_{2}A}{K}_{a1}^B\right)- \\ {\left(D_{O{H}^{-}}{K}_w\right)}^2{K}_{a1}^B{K}_{a2}^B\left(K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}+K_{a1}^{H_{2}A}{K}_{a1}^B+K_{a1}^B{K}_{a2}^B\right)+2D_{O{H}^{-}}{K}_w{K}_{a1}^{H_{2}A}{K_{a1}^B}^2{K}_{a2}^B(D_{O{H}^{-}}{\left[O{H}^{-}\right]}_h- \\ {D}_{H^{+}}{\left[H^{+}\right]}_h)\left(K_{a2}^{H_{2}A}+K_{a2}^B\right)+2D_{O{H}^{-}}{D}_{H^{+}}{K}_w{K}_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}{\left(K_{a1}^B{K}_{a2}^B\right)}^2- \\ {\left(D_{O{H}^{-}}{\left[O{H}^{-}\right]}_h-D_{H^{+}}{\left[H^{+}\right]}_h\right)}^2{K}_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}{\left(K_{a1}^B{K}_{a2}^B\right)}^2+{D_B}^2{K}_{sp}{(K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}{K}_{a1}^B)}^2; \end{gathered}$$

$$\begin{gathered} H=4{D_B}^{\frac{4}{3}}{D_{H_{2}A}}^{\frac{2}{3}}{K}_{sp}{K_{a1}^{H_{2}A}}^2{K}_{a2}^{H_{2}A}{K}_{a1}^B{K}_{a2}^B\left(K_{a1}^B{K}_{a2}^B+2K_{a2}^{H_{2}A}{K}_{a1}^B\right)-{\left(D_{O{H}^{-}}{K}_w\right)}^2{K}_{a1}^{H_{2}A}{K_{a1}^B}^2{K}_{a2}^B(K_{a2}^{H_{2}A}+ \\ {K}_{a2}^B)+2D_{O{H}^{-}}{K}_w{K}_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}{\left(K_{a1}^B{K}_{a2}^B\right)}^2\left(D_{O{H}^{-}}{\left[O{H}^{-}\right]}_h-D_{H^{+}}{\left[H^{+}\right]}_h\right)- \\ 4{D_B}^{\frac{5}{3}}{D_{H_{2}A}}^{\frac{1}{3}}{K}_{sp}{(K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}{K}_{a1}^B)}^2{K}_{a2}^B; \end{gathered}$$

$$I=4{D_B}^{\frac{4}{3}}{D_{H_{2}A}}^{\frac{2}{3}}{K}_{sp}{\left(K_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}{K}_{a1}^B{K}_{a2}^B\right)}^2-{\left(D_{OH}-K_w\right)}^2{K}_{a1}^{H_{2}A}{K}_{a2}^{H_{2}A}{\left(K_{a1}^B{K}_{a2}^B\right)}^2.$$

Evaluation of flux of the co-crystal components

Combine equations 80 and 83, and solve for −C₁ for the total flux of co-crystal in terms of drug species across the diffusion layer:

$${\text{J}}_{\text{cc}}=\frac{{\text{D}}_{\text{B}}}{\text{h}}\sqrt{{\text{K}}_{\text{sp}}(1+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}}{{\left[{\text{H}}^{+}\right]}_0}+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}{\text{K}}_{\text{a}2}^{{\text{H}}_2\text{A}}}{{{\left[{\text{H}}^{+}\right]}_0}^2})(1+\frac{{\left[{\text{H}}^{+}\right]}_0}{{\text{K}}_{\text{a}2}^{\text{B}}}+\frac{{{\left[{\text{H}}^{+}\right]}_0}^2}{{\text{K}}_{\text{a}1}^{\text{B}}{\text{K}}_{\text{a}2}^{\text{B}}})}$$ (87)

Substitute equations 56 into equation 87:

$${\text{J}}_{\text{cc}}=0.62{\text{D}}_{\text{B}}^{2/3}{\mathrm{\omega }}^{1/2}{\text{v}}^{-1/6}\sqrt{{\text{K}}_{\text{sp}}(1+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}}{{\left[{\text{H}}^{+}\right]}_0}+\frac{{\text{K}}_{\text{a}1}^{{\text{H}}_2\text{A}}{\text{K}}_{\text{a}2}^{{\text{H}}_2\text{A}}}{{{\left[{\text{H}}^{+}\right]}_0}^2})(1+\frac{{\left[{\text{H}}^{+}\right]}_0}{{\text{K}}_{\text{a}2}^{\text{B}}}+\frac{{{\left[{\text{H}}^{+}\right]}_0}^2}{{\text{K}}_{\text{a}1}^{\text{B}}{\text{K}}_{\text{a}2}^{\text{B}}})}$$ (88)