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I. The Nature of Solutions

Solutions. I. The Nature of Solutions. A. Definitions. Solution - homogeneous mixture. Solute - substance being dissolved. Solvent - present in greater amount. A. Definitions. Solvent - H 2 O. Solute - KMnO 4. B. Solvation. Solvation – the process of dissolving.

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I. The Nature of Solutions

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  1. Solutions I. The Nature of Solutions

  2. A. Definitions • Solution - homogeneous mixture Solute - substance being dissolved Solvent - present in greater amount

  3. A. Definitions Solvent - H2O Solute - KMnO4

  4. B. Solvation • Solvation – the process of dissolving solute particles are surrounded by solvent particles First... solute particles are separated and pulled into solution Then...

  5. - + - - + + acetic acid salt sugar B. Solvation Non- Electrolyte Weak Electrolyte Strong Electrolyte solute exists as ions and molecules solute exists as ions only solute exists as molecules only DISSOCIATION IONIZATION

  6. B. Solvation • Dissociation • separation of an ionic solid into aqueous ions NaCl(s)  Na+(aq) + Cl–(aq)

  7. B. Solvation • Ionization • breaking apart of some polar molecules into aqueous ions HNO3(aq) + H2O(l)  H3O+(aq) + NO3–(aq)

  8. B. Solvation • Molecular Solvation • molecules stay intact C6H12O6(s)  C6H12O6(aq)

  9. NONPOLAR NONPOLAR POLAR POLAR B. Solvation “Like Dissolves Like”

  10. B. Solvation • Soap/Detergent • polar “head” with long nonpolar “tail” • dissolves nonpolar grease in polar water

  11. UNSATURATED SOLUTION more solute dissolves SATURATED SOLUTION no more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form C. Solubility concentration

  12. C. Solubility • Solubility • maximum grams of solute that will dissolve in 100 g of solvent at a given temperature • varies with temp • based on a saturated soln

  13. C. Solubility • Solubility Curve • shows the dependence of solubility on temperature

  14. C. Solubility • Solids are more soluble at... • high temperatures. • Gases are more soluble at... • low temperatures & • high pressures (Henry’s Law). • EX: nitrogen narcosis, the “bends,” soda

  15. Solutions • Solutions WS

  16. Solutions II. Concentration

  17. A. Concentration • The amount of solute in a solution. • Describing Concentration • % by mass - medicated creams • % by volume - rubbing alcohol • ppm, ppb - water contaminants • molarity - used by chemists • molality - used by chemists

  18. A. Concentration Water Quality Report

  19. mass of solvent only 1 kg water = 1 L water B. Molality

  20. B. Molality • Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water. 75 g MgCl2 1 mol MgCl2 95.21 g MgCl2 0.25 kg water = 3.2m MgCl2

  21. B. Molality • How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water? 0.500 kg water 1.54 mol NaCl 1 kg water 58.44 g NaCl 1 mol NaCl = 45.0 g NaCl

  22. C. Dilution • Preparation of a desired solution by adding water to a concentrate. • Moles of solute remain the same.

  23. C. Dilution • What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 (15.8M)V1 = (6.0M)(250mL) V1 = 95 mL of 15.8M HNO3

  24. 500 mL of 1.54M NaCl 500 mLwater 500 mL volumetric flask 500 mL mark 45.0 gNaCl D. Preparing Solutions • 1.54m NaCl in 0.500 kg of water • mass 45.0 g of NaCl • add water until total volume is 500 mL • mass 45.0 g of NaCl • add 0.500 kg of water

  25. 95 mL of15.8M HNO3 250 mL mark water for safety D. Preparing Solutions • 250 mL of 6.0M HNO3by dilution • measure 95 mL of 15.8M HNO3 • combine with water until total volume is 250 mL • Safety: “Do as you oughtta, add the acid to the watta!”

  26. Turn in one paper per team. Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. For each of the following solutions: 1) 100.0 mL of 0.50M NaCl 2) 0.25m NaCl in 100.0 mL of water 3) 100.0 mL of 3.0M HCl from 12.1M concentrate. Solution Preparation Activity

  27. 100.0 mL of 0.50 M NaCl .100 L(0.50 mol NaCl)(58g NaCl) = 2.9 g NaCl (1.0L)(1 mol NaCl) Dissolve 2.9 g of NaCl in water and fill to 100.0 mL Solution Preparation Activity –Solution 1

  28. 0.25 m NaCl in 100.0 mL of water 0.1000 L(0.25 mol)(58 g NaCl) = 1.5 g NaCl (1.0 L) (1 mol NaCl) Dissolve 1.5 g of NaCl in 100.0 mL of water Solution Preparation Activity –Solution 2

  29. 100.0 mL of 3.0M HCl from 12.1M conc M1V1 = M2V2 V1= M2V2 M1 V1 = (0.1000 L)(3.0 mol/L HCl) = 0.025 L of conc (12.1 mol/L HCl) Add 25 mL of concentrate and fill to 100.0 mL Solution Preparation Activity –Solution 3

  30. Practice Time • Do Solution Concentration WS

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