1 / 57

Solutions and Solubility: Understanding and Managing Lake Nyos Tragedy

Explore the causes of the Lake Nyos tragedy in Cameroon and learn about the possible solutions and solubility factors in preventing such disasters. Discover the science behind solubility and how it relates to the dissolution of different substances in a solvent.

tbernal
Télécharger la présentation

Solutions and Solubility: Understanding and Managing Lake Nyos Tragedy

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 13 Solutions Cameroon: location of Lake Nyos 50 miles underneath lake Nyos CO2 is produced by molten volcanic rock (magma) and held in solution by the pressure of the water above it.

  2. CHAPTER OUTLINE

  3. Tragedy in Cameroon • Lake Nyos • lake in Cameroon, West Africa • on August 22, 1986, > 1700 people & 3000 cattle died • Burped Carbon Dioxide Cloud Appeared From Nyos • CO2 seeps in from underground and dissolves in lake water to levels above normal saturation • though not toxic, CO2 is heavier than air – the people died from asphyxiation

  4. Possible Resolution • scientists have studied Lake Nyos and similar lakes in the region to try and keep such tragedies from reoccurring • currently, they are trying to keep the CO2 levels in the lake water from reaching very high supersaturation levels by pumping air into the water to agitate it By understanding solutions we are able to divert natural tragedies

  5. TYPE OFSOLUTIONS • A solution is a homogeneous mixture of two substances: Solute: substance being dissolved present in smaller amount Solvent: substance doing the dissolving present in larger amount Solutes and solvents may be of any form of matter: solid, liquid or gas.

  6. Common Types of Solution

  7. e.g. Brass (varies in composition)

  8. SOLUBILITY • Solutions form between solute and solvent molecules because of similarities between them. Like dissolves Like Ionic solids dissolve in water because the charged ions (polar) are attracted to the polar water molecules. Non-polar molecules such as oil and grease dissolve in non-polar solvents such as kerosene.

  9. SOLUBILITY • solutions that contain metal solutes and a metal solvent are calledalloys • when one substance (solute) dissolves in another (solvent) it is said to be soluble • salt is soluble in water, • bromine is soluble in methylene chloride chlorocarbon that is not miscible with water, but will dissolve in most organic solvents • when one substance does not dissolve in another it is said to be insoluble • oil is insoluble in water

  10. Salt Dissolving in Water partial + surround the anion partial – surround the cation Solvation is the process of attraction and association of molecules of a solvent with molecules or ions of a solute. As ions dissolve in a solvent they spread out and become surrounded by solvent molecules.

  11. SOLUBILITY • there is usually a limit to the solubility of one substance in another • gases are alwayssoluble in each other • two liquids that are mutually soluble are said to be miscible • alcohol and water are miscible • oil and water are immiscible

  12. Descriptions of Solubility • saturated solutions have the maximum amount of solute that will dissolve in that solvent at that temperature • unsaturated solutions can dissolve more solute • supersaturated solutions are holding more solute than they should be able to at that temperature • unstable

  13. Adding Solute to various Solutions unsaturated saturated supersaturated

  14. Supersaturated Solution A supersaturated solution has more dissolved solute than the solvent can hold. When disturbed, all the solute above the saturation level comes out of solution.

  15. Electrolytes • electrolytes are substances whose aqueous solution is a conductor of electricity • strong electrolytes, all the electrolyte molecules are dissociated into ions, SALTS • nonelectrolytes, none of the molecules are dissociated into ions, SUGARS • weak electrolytes, a small percentageof the molecules are dissociated into ions

  16. SOLUBILITY • Solubility refers to the maximum amount of solute that can be dissolved in a given amount of solvent. • Many factors affect the solubility of a solute in a solution. Type of solute Type of solvent Temperature Solubility is measured in grams of solute per 100 grams of solvent at a given temperature.

  17. SOLUBILITY • Solubility of most solids in water increases as temperature increases. • Using a solubility chart, the solubility of a solute at a given temperature can be determined. • For example, KNO3 has a solubility of 80 g/100 g H2O (80%) at 40 C.

  18. SOLUBILITY OF GASES • Solubility of gases in water decreases as temperature increases. • At higher temperatures more gas molecules have the energy to escape from solution. • Henry’s law states that the solubility of a gas is directly proportional to the pressure above the liquid. • For example, a can of soda is carbonated at high pressures in order to increase the solubility of CO2. Once the can is opened, the pressure is reduced and the excess gas escapes from the solution.

  19. Solubility and Pressure • the solubility of gases in water depends on the pressure of the gas • higher pressure = higher solubility higher pressure above the soln lower pressure above the soln less gas in soln more gas in soln

  20. Solubility and Pressure When soda pop is sealed, the CO2 is under pressure. Opening the container lowers the pressure, which decreases the solubility of CO2 and causes bubbles to form.

  21. Solution Concentration • dilute solutions have low solute concentrations • concentrated solutions have high solute concentrations

  22. CONCENTRATIONUNITS • The amount of solute dissolved in a certain amount of solution (occasionally amount of solvent) is called concentration. amount of solute Concentration = amount of solution • Three types of concentration units will be studied in this class: Mass Percent: (m/m) and (m/v) Molarity

  23. mass of solute + mass of solvent MASS PERCENT • Mass percent (% m/m) is defined as the mass of solute divided by the mass of solution.

  24. MASS/VOLUMEPERCENT • Mass/Volume percent (% m/v) is defined as the mass of solute divided by the volume of solution.

  25. Example 1: What is the mass % (m/m) of a NaOH solution that is made by dissolving 30.0 g of NaOH in 120.0 g of water? Mass of solution = 30.0 g + 120.0 g = 150.0 g

  26. Example 2: What is the mass % (m/v) of a solution prepared by dissolving 5.0 g of KI to give a final volume of 250 mL?

  27. USING PERCENTCONCENTRATION • Some examples of percent compositions, their meanings, and possible conversion factors are shown in the table below: • In the preparation of solutions, one often needs to calculate the amount of solute or solution. • To achieve this, percent composition can be used as a conversion factor.

  28. Example 1: A topical antibiotic solution is 1.0% (m/v) Clindamycin. How many grams of Clindamycin are in 65 mL of this solution? 1.0 g Clindamycin 65 mL solution x 100 mL solution = 0.65 g

  29. Example 2: How many grams of solute are needed to prepare 150 mL of a 40.0% (m/v) solution of LiNO3? 40.0 g LiNO3 150 mL solution x 100 mL solution = 60. g LiNO3

  30. MOLARITY • The most common unit of concentration used in the laboratory is molarity (M). • Molarity is defined as: moles of solute Molarity = Liter of solution

  31. Example 1: What is the molarity of a solution containing 1.4 mol of acetic acid in 250 mL of solution? Vol. of solution = = 5.6 M Molarity =

  32. Example 2: What is the molarity of a solution that contains 75 g of KNO3 in 350 mL of solution? Mol of solute = = 0.74 mol Vol of solvent =

  33. Weigh out 1 mole (58.45 g) of NaCl and add it to a 1.00 L volumetric flask. Add water to dissolve the NaCl, then add water to the mark. Swirl to Mix Step 1 Step 2 Step 3 Preparing a 1.00 M NaCl Solution

  34. USINGMOLARITY • Molarity relationship can be used to calculate: Amount of solute: Moles solute = Molarity x volume Volume of solution:

  35. 16 1 Example 1: How many moles of nitric acid are in 325 mL of 16 M HNO3 solution? Vol. of solution = mol of solute = = 5.2 mol

  36. 4.50 1 Example 2: How many grams of NaHCO3 are in 325 mL of 4.50 M solution of NaHCO3? Vol. of solution = mol of solute = = 1.46 mol = 123 g mass of solute =

  37. 1 2.0 Example 3: What volume (mL) of 2.0 M NaOH solution contains 20.0 g of NaOH? mol of solute = = 0.500 mol Vol. In L = = 0.25 L Vol. In mL = = 250 mL

  38. 1 0.300 Example 4: How many mL of a 0.300 M glucose (C6H12O6) IV solution is needed to deliver 10.0 g of glucose to the patient? mol of solute = = 0.0555 mol Vol. In L = = 0.185 L Vol. In mL = = 185 mL

  39. DILUTION Amount of solute remains constant • When more water is added to a solution, • Solutions are often prepared from more concentrated ones by adding water. This process is called dilution. Volume and concentration are inversely proportional Volume increases Concentration decreases Frozen juice Diluted juice Water

  40. DILUTION • The amount of solute depends on the concentration and the volume of the solution. Therefore, M1 x V1 = M2 x V2 Concentrated solution Dilute solution

  41. Example 1: Concentration decreases What is the molarity of the final solution when 75 mL of 6.0 M KCl solution is diluted to 150 mL? Volume increases M1 x V1 = M2 x V2 M1 = 6.0 M V1 = 75 mL M2 = ??? V2 = 150 mL M2 = 3.0 M

  42. Example 2: Volume increases What volume (mL) of 0.20 M HCl solution can be prepared by diluting 50.0 mL of 1.0 M HCl? Concentration decreases M1 x V1 = M2 x V2 M1 = 1.0 M V1 = 50.0 mL M2 = 0.20 M V2 = ??? V2 = 250 mL

  43. Making a Solution by Dilution M1 x V1 = M2 x V2 M1 = 12.0 M V1 = ? L M2 = 1.50 M V2 = 5.00 L dilute 0.625 L of 12.0 M solution to 5.00 L

  44. Solution Stoichiometry • we know that the balanced chemical equation tells us the relationship between moles of reactants and products in a reaction • 2 H2(g) + O2(g) → 2 H2O(l) implies for every 2 moles of H2 used, you need 1 mole of O2 and to make 2 moles of H2O • molarity is the relationship between moles of solute and liters of solution, thus we can measure the moles of a material in a reaction within a solution by knowing its molarity and volume

  45. Example 1: • How much 0.115 M KI solution, in liters, is required to completely precipitate all the Pb2+ in 0.104 L of 0.225 M Pb(NO3)2? • 2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq) Identify what the question is looking for: volume of KI solution, L 0.115 M KI  0.115 mol KI  1 L solution 0.225 M Pb(NO3)2 0.225 mol Pb(NO3)2  1 L solution Chem. Eq’n  2 mol KI  1 mol Pb(NO3)2

  46. = 0.40696 L = 0.407 L

  47. OSMOLARITY • Many important properties of solutions depend on the number of particles formed in solution. • Recall that when ionic substances (strong electrolytes) dissolve in water they form several particles for each formula unit. • For example: NaCl (s) Na+ (aq) + Cl (aq) 1 formula unit 2 particles

  48. OSMOLARITY CaCl2 (s) Ca2+ (aq) + 2 Cl (aq) 3 particles 1 formula unit

  49. OSMOLARITY • When covalent substances (non- or weak electrolytes) dissolve in water they form only one particle for each formula unit. • For example: C12H22O11 (s) C12H22O11 (aq) 1 formula unit 1 particle

  50. OSMOLARITY • Osmolarity of a solution is its molarity multiplied by the number of particles formed in solution. Osmolarity = i x Molarity Number of particles in solution

More Related