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SOLUTIONS AND SOLUBILITY

SOLUTIONS AND SOLUBILITY. 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. EXAMPLE.

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SOLUTIONS AND SOLUBILITY

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  1. SOLUTIONSANDSOLUBILITY

  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.

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

  4. CALCULATION 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 HENCE, solute and solvent are determined depending on molar basis not weight basis

  5. Dissolved solute < concentration needed for complete saturation at a definite temperature. Equilibrium of dissolved with excess undissolved solid at a definite temperature. Dissolved solute > concentration needed for complete saturation at a definite temperature. e.g. sodium thiosulphate- sodium acetate

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

  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=amountdissolved/time). • The higher the solubility, the more rapid is the rate of solution when no chemical reaction is involved.

  8. 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) Solubility

  9. 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)

  10. < than 1 part of solvent ~1 - 10 ~10 - 30 ~30 – 100 ~100 – 1000 ~1000 – 10.000 > than 10.000

  11. Example: The solubility of a drug in water is 41.5 g per 1000 mL of solution at 18oC. 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, molalityand percent by weight, calculate the mole fraction and mole percent of both components of the solution. 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 = 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 Grams of solvent = 1037.5 - 41.5 = 996.0 g Molality of the drug = 0.2732 x 1000 / 996.0 = 0.2743 m

  12. c. Percent by weight:(Number of gms of solute/100 gms solution (%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 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%

  13. Steps of Determination of Solubility: 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. • Filtration is done to separate the undissolved solid. II) Analysis of the saturated solution.

  14. 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) Phase Solubility Analysis:

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

  16. Gibbs phase Rule:F= C – P + 2F= Degree of freedom (T, P, C)C= Number of componentsP= number of phasesAt constant T and P (F= C- P)

  17. Phase solubility diagram for a pure substance A)For pure substance: F=0 • Phase solubility curves: Saturation F=1 • 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.

  18. Phase solubility curve for substance contain one impurity B) For non- pure substance: (one impurity) F=0 F= 1 F=1 F=1 F=2 • 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 • AtBC:Pure solid major

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