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34 September 2020 Tablets & Capsules can be calculated with the fraction of component i (x i ) and the initial moisture content of component i (M i ) via: N M f = ∑ x i ∙ M i i = 1 (3) The expected final water activity can then be interpo- lated from the mixture's sorption isotherm. Examples of predicting water activity of mixtures The following example demonstrates how to calculate moisture migration for a blend of two components. The components include a coarse-grade lactose monohydrate (Pharmatose 80M) with low hygroscopicity and an API, which are blended together in an 80:20 ratio (% w/w). The initial water activity of the API is 0.1, and the initial water activity of the lactose is 0.3. The sorption isotherm of the lactose is measured using a ProUmid SPS-1µ advanced system operated at 25°C with 20 minutes between weighing cycles, equilibrium condition 0.003 percent w/w per 80 minutes, and boundary conditions of 8-12 hours. The isotherm of the API is imaginary. The component isotherms are combined using Equation (2) into the sorption isotherm of the proposed 80:20 blend, as shown in Figure 6. From the sorption isotherms, one can see that the API contains 0.060 percent moisture at 0.1 water activity, and the lactose contains 0.005 percent moisture at a water activity of 0.3. The total moisture content of the blend will therefore be: (80% ∙ 0.005%) + (20% ∙ 0.060%) = 0.016% and product [19], and they provide all the information needed to predict the final amount of moisture migration in a system. Figure 5 provides examples of sorption iso- therms for some commonly used excipients. You can determine a sample's final moisture status using sorption isotherms based on the following two facts: 1. Moisture will migrate from high water activity to low water activity, until achieving equilibrium. 2. The mass balance of the total system is zero. This means that the total amount of free moisture in a closed system (which can include air) is not changing upon moisture migration. Knowing the amount of free water at different water activities for all components is critical for predicting moisture migration. For simplification, the following description and examples use reversible sorption iso- therms, but similar calculations can be made for irrevers- ible sorption isotherms. For a mixture of two components, the mixture's sorp- tion isotherm is the linear combination of the individual components' sorption isotherms. The free moisture con- tent of a blend at a specific water activity (M Aw ) can be calculated with knowledge of the fraction of N different components i (x i ) and the moisture content of compo- nent i at that water activity value (M i,Aw ) via: N M A w = ∑ x i ∙ M i,A w i = 1 (2) The total amount of free water defines the sample's final moisture status. Upon preparation of a blend of N components, the final moisture content (M f ) of the blend Figure 6 Sorption isotherms of lactose, API, and an 80:20 (% w/w) blend of the two (Data points relevant for the example calculation are indicated in the graph.) Pharmatose 80M API Blend (20% API) Δm (%) 0.30 0.25 0.20 0.15 0.10 0.05 0.00 0.060 0.015 0.025 0.005 Relative humidity (%) 0 20 40 60 80 100 Figure 5 Sorption isotherms for five commonly used excipients (Measured using a ProUmid SPS-1µ advanced system, operated at 25°C, with 20 minutes between weighing cycles, equilibrium condition 0.003% w/w per 80 minutes, and boundary conditions of 8-12 hours.) Pharmacel 101 SuperTab 11SD SuperTab 21AN SuperTab 30GR Pharmatose 200M Δm (%) Relative humidity (%) 12 10 8 6 4 2 0 0 20 40 60 80 100