1. SOLUTIONS AND SOLUBILITY

2. Solutions and Solubility Solution: can be defined as a system in which molecules of a solute are dissolved in a given solvent vehicle. It is a mixture of 2 or more components forming a Single Phase. Solution: can be defined as a system in which molecules of a solute are dissolved in a given solvent vehicle. It is a mixture of 2 or more components forming a Single Phase. Solution Solute Small proportion e.g. ions- molecules Solvent Largest proportion

3. "" consists of (w/w) of sucrose in water. "Simple syrup" consists of 66.67% (w/w) of sucrose in water. Mwt of sucrose=342 Mwt of sucrose=342 Mwt of water=18 Mwt of water=18 The moles of sucrose=0.19 moles, and the moles of water=1.85 moles The moles of sucrose=0.19 moles, and the moles of water=1.85 moles EXAMPLE

4. Amount of sucrose = 66.7 g Amount of water = 100 - 66.7 = 33.3 g No. of moles of sucrose = 66.7/342 = 0.195 mole No. of moles of water = 33.3/ 18 = 1.85 mole CALCULATION HENCE, solute and solvent are determined depending on molar basis not weight basis

5. Classification of solutions according to the amount of solute Unsaturated(Subsaturated) Saturated (Stable) Supersaturated (Metastable) (Metastable) Dissolved solute < concentration needed for complete saturation at a definite temperature. Dissolved solute > concentration needed for complete saturation at a definite temperature. e.g. sodium thiosulphate- sodium acetate Equilibrium of dissolved with excess undissolved solid at a definite temperature.

6. 1- Seeding with a small crystal of the solute. 2- Vigorous shaking. 3- Scratching the wall of container in contact to solution. Supersaturated solutions Stable solutions ?

7. The maximum equilibrium solubility (solubility-saturated solubility) of a drug in a given medium is important because it affects the rate of solution (dissolution=amount dissolved/time). The maximum equilibrium solubility (solubility-saturated solubility) of a drug in a given medium is important because it affects the rate of solution (dissolution=amount dissolved/time). The higher the solubility, the more rapid is the rate of solution when no chemical reaction is involved. The higher the solubility, the more rapid is the rate of solution when no chemical reaction is involved.

8. Solubility IT is the concentration of solute in a saturated solution at a certain temperature. (Quantitative Terms)OR IT is the spontaneous interaction of 2 or more substances to form a homogeneous molecular dispersion (single phase). (Qualitative Terms)

9. Expression of solubility I) Quantitative USP expression Molar solution Molal solution Mole fraction Percent by weight Percent weight/volume Percent by volume II) Qualitative Very soluble Freely soluble Soluble Sparingly soluble Slightly soluble Very slightly soluble Practically insoluble

10. Quantitative USP expression Molar solution Molal solution Mole fraction Percent by weight Percent w/v Percent by volume Number of mls of solvent dissolving 1 gm of solute (gm/ml) Number of moles of solute/1 liter of solution (mol/L) Number of moles of solute/1000gm of solvent Mole fraction of (A)=Moles (A)/sum of moles of (A)+(B) Number of gms of solute/100 gms solution (%w/w) Number of gms of solute/100 ml solution (%w/v) Number of mls of solute/100 ml solution (%v/v)

11. Qualitative Very soluble Freely soluble Soluble Sparingly soluble Slightly soluble Very slightly soluble Practically insoluble < than 1 part of solvent ~1 - 10 ~10 - 30 ~30 – 100 ~100 – 1000 ~1000 – 10.000 > than 10.000

12. Example: Example: The solubility of a drug in water is 41.5 g per 1000 mL of solution at 18 o C. The density of the solution was 1.0375 g/mL and the molecular weight of the drug is 151.9. Express the solubility of the drug in terms of molarity, molality and percent by weight, calculate the mole fraction and mole percent of both components of the solution. a. Molarity: a. Molarity: ( Number of moles of solute/1 liter of solution (mol/L)) No. of moles of the drug = 41.5 / 151.9 = 0.2732 Molarity of the drug = 0.2732 / 1liter of solution = Molarity of the drug = 0.2732 / 1liter of solution = 0.2732 M b. Molality: (Number of moles of solute/1000gm of solvent) Grams of solution = volume x density = 1000 x 1.0375 = 1037.5 g = 1037.5 g Grams of solvent = 1037.5 - 41.5 = 996.0 g Molality of the drug = 0.2732 x 1000 / 996.0 = Molality of the drug = 0.2732 x 1000 / 996.0 = 0.2743 m

13. c. Percent by weight: (Number of gms of solute/100 gms solution (%w/w) (g of drug / g of solution) x 100 = (41.5/ ) x 100 = 4.00 % w/w (g of drug / g of solution) x 100 = (41.5/ 1037.5) x 100 = 4.00 % w/w d. Mole fraction and Mole percent: (Mole fraction of (A)=Moles (A)/sum of moles of (A)+(B)) No. of moles of water = 996 / 18.02 = 55.25 moles No. of moles of water = 996 / 18.02 = 55.25 moles Mole fraction of the drug (A) = 0.2732/(55.27 + 0.2732) = 0.0049 Mole fraction of water (B) = 55.27 / (55.27 + 0.2732) = 0.9951 Note that A drug + B water = 1 Mole percent of the drug = 0.0049 x 100 = 0.49% Mole percent of water = 0.9951 x 100 = 99.51%

14. Steps of Determination of Solubility: Steps of Determination of Solubility: I) Preparation of a of substance in a given solvent at a particular temperature by: I) Preparation of a saturated solution of substance in a given solvent at a particular temperature by: Addition of excess powder in the solvent and stirring till equilibrium is reached. Addition of excess powder in the solvent and stirring till equilibrium is reached. Filtration is done to separate the undissolved solid. Filtration is done to separate the undissolved solid. II) Analysis of the saturated solution.

15. Phase Solubility Analysis: Importance: 1- Determination of solubility of pure substance from non pure sample. 2- Determination of purity of sample. 3- Differentiate between optical isomers (unless 1:1)

16. Phase Solubility Analysis: (cont.) Steps of determination: 1gm 2gm 3gm 4gm shaken at constant (T,P) 1gm 2gm 3gm 4gm Equilibrium Separate solid from solution   Determine amount dissolved   Plot Y axis (solution conc)   Plot X axis (system conc)

17. Gibbs phase Rule: F= C – P + 2 F= Degree of freedom (T, P, C) C= Number of components P= number of phases At constant T and P (F= C- P)

18. Phase solubility curves: Phase solubility curves: Phase solubility diagram for a pure substance F=1 F=0  A- B: Conc is below saturation  B- C: Conc is above saturation  B – C: has no slope, indicating purity  Point D: Solubility of pure substance. A)For pure substance: Saturation

19. B) For non- pure substance: (one impurity) Phase solubility curve for substance contain one impurity  A- B: Conc is below saturation for both (1 phase)  At B: Saturation with major component  From B to C: Conc is above saturation with major component and below saturation for minor one (2 phases)  Section C – D: Saturation with both components (3 phases)  Value of AE: Solubility of major component.  Value of EF: Solubility of minor component  At BC: Pure solid major F=2 F= 1 F=0 F=1