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  • These solutions are unstable and will revert back to a saturated solution if a “seed crystal” is added or if there is a change in temperature or pressure. (E.g., honey is a supersaturated solution of sugar). These solutions are unstable and will revert back to a saturated solution if a “seed crystal” is added or if there is a change in temperature or pressure. (E.g., honey is a supersaturated solution of sugar). These solutions are unstable and will revert back to a saturated solution if a “seed crystal” is added or if there is a change in temperature or pressure. (E.g., honey is a supersaturated solution of sugar).
  • These solutions are unstable and will revert back to a saturated solution if a “seed crystal” is added or if there is a change in temperature or pressure. (E.g., honey is a supersaturated solution of sugar). These solutions are unstable and will revert back to a saturated solution if a “seed crystal” is added or if there is a change in temperature or pressure. (E.g., honey is a supersaturated solution of sugar). These solutions are unstable and will revert back to a saturated solution if a “seed crystal” is added or if there is a change in temperature or pressure. (E.g., honey is a supersaturated solution of sugar).
  • These solutions are unstable and will revert back to a saturated solution if a “seed crystal” is added or if there is a change in temperature or pressure. (E.g., honey is a supersaturated solution of sugar). These solutions are unstable and will revert back to a saturated solution if a “seed crystal” is added or if there is a change in temperature or pressure. (E.g., honey is a supersaturated solution of sugar). These solutions are unstable and will revert back to a saturated solution if a “seed crystal” is added or if there is a change in temperature or pressure. (E.g., honey is a supersaturated solution of sugar).

Transcript

  • 1. Chemistry 30S Unit 4 - Solutions
  • 2.  
  • 3. Solutions
    • A solution is defined as a homogeneous mixture of two or more substances where all substances are distributed as individual molecules or ions.
    Chemistry 30S Unit 4 - Solutions
  • 4. Some Terms to Know Chemistry 30S Unit 4 - Solutions solute SOLVENT miscible immiscible soluble insoluble solubility dissolve
  • 5. What makes up a solution?
    • A solution has two components: the solute and the solvent.
    • The solvent is the substance in greater amount (it does the dissolving).
    • The solute is considered to be the dissolved substance.
    Chemistry 30S Unit 4 - Solutions
  • 6. What are the various types of solutions?
    • A solution can be a solid, liquid, or gas.
    • Gas Solutions
      • Consist of gases or vapors dissolved in one another.
    • Liquid Solutions
      • Consists of a liquid solvent in which a gas, solid, or liquid is dissolved.
    • Solid Solution
      • Are mixtures of solids that are uniformly spread throughout at the atomic or molecular level.
    Chemistry 30S Unit 4 - Solutions
  • 7. Types of Solutions
    • We can describe the types of solutions in terms of the state or phase of the solute and solvent.
    Chemistry 30S Unit 4 - Solutions
  • 8.  
  • 9. What is electronegativity?
    • It is a measure of an atom's attraction for electrons in the covalent bond.
    • When two different atoms are covalently bonded, the atom with the higher electronegativity will attract the shared electrons stronger than the other atom can. This bond is known as a polar bond .
    Chemistry 30S Unit 4 - Solutions
  • 10. What is electronegativity? Chemistry 30S Unit 4 - Solutions
  • 11. What is a Polar Molecule?
    • A good example is HCl, hydrogen chloride gas. The difference in electronegativities is about 1.0, suggesting polar character.
    Chemistry 30S Unit 4 - Solutions
  • 12. Water - A Polar Molecule
    • If we look at a water molecule, we can see that there is a lack of symmetrical electron distribution around the molecule. Because of this, water has been shown to be polar.
    Chemistry 30S Unit 4 - Solutions Bending Water Demo
  • 13. Polarity
    • The Polarity of the solute and solvent molecules will affect the solubility.
    • Generally polar solute molecules will dissolve in polar solvents and non-polar solute molecules will dissolve in non-polar solvents.
    • The polar solute molecules have a positive and a negative end to the molecule. If the solvent molecule is also polar, then positive ends of solvent molecules will attract negative ends of solute molecules.
    • This is a type of intermolecular force known as dipole-dipole interaction.
    Chemistry 30S Unit 4 - Solutions
  • 14. The Solution Process Chemistry 30S Unit 4 - Solutions
  • 15. The Solution Process
    • When a solute dissolves in a solvent, the individual particles of the solute separate from the other particles of the solute and move between the spaces of the solvent particles. The solvent particles collide with the solute particles and forces of attraction between solute and solvent particles "hold" the solute particles in the spaces.
    • For a solute to be dissolved in a solvent, the attractive forces between the solute and solvent molecules must be greater than the forces of attraction between the solute molecules.
    Chemistry 30S Unit 4 - Solutions
  • 16. The Solution Process
    • There are 3 steps to the dissolving process:
    • The solvent particles must move apart to make room for solute particles. This process requires energy to overcome forces of attraction between solvent particles. The first step in the dissolving process is endothermic.
    • The solute particles must separate form the other solute particles. This process also requires energy to overcome the forces of attraction between the solute particles. The second step in the dissolving process is endothermic.
    • When the solute particles move between the solvent particles the forces of attraction between solute and solvent take hold and the particles "snap" back and move closer. This process releases energy. The final step in the dissolving process is exothermic.
    Chemistry 30S Unit 4 - Solutions
  • 17. Energy Changes During Dissolving
    • The total heat change in the dissolving process is the sum of the three heat changes. If the sum of the heat absorbed in the first two steps of the dissolving process is greater than the heat released in the last step, the dissolving of that substance will be endothermic . If the dissolving process for a substance is endothermic, the container will feel cooler as the substance dissolves.
    • In some cases, the exothermic process is larger than the sum of the two endothermic processes. In these cases, as the solid dissolves large amounts of heat are evolved.
    Chemistry 30S Unit 4 - Solutions
  • 18. How ionic compounds dissolve in water
    • polar waters stick to ions on the surface of an ionic crystal, partially neutralizing the ion's charge
    • attraction of surface ions to the rest of the crystal is weakened
    • water molecules surround the surface ions and carry them away from the crystal
    • result: soluble ionic compounds dissociate into ions in water NaCl(s)  Na + (aq) + Cl - (aq) K 2 SO 4 (s)  2 K + (aq) + SO 4 2- (aq)
    Chemistry 30S Unit 4 - Solutions
  • 19. How ionic compounds dissolve in water
    • The separation of ions is called dissociation
    • Whereas the process of surrounding the solute ions with solvent molecules is called solvation
    • If the solvent used to dissolve the solute is water, the process is called hydration
    Chemistry 30S Unit 4 - Solutions
  • 20. How covalent compounds dissolve in water
    • When a covalent compound is dissolved in a solvent, the entire molecule is pulled away from the solid structure as it forms a solution.
    • It forms an aqueous solution.
    • Example:
    • C 12 H 22 O 11(s) + H 2 0 (l)  C 12 H 22 O 11(aq)
    Chemistry 30S Unit 4 - Solutions
  • 21. Solubility Curves
    • Recall, solubility refers to the amount of solute that can dissolve in a given amount of solvent at given conditions (like temperature).
    • A Solubility curve is a graph of the relationship between solubility and temperature.
    Chemistry 30S Unit 4 - Solutions
  • 22. Solubility Curves
    • Solubility curves , like the one shown here, tell us what mass of solute will dissolve in 100g (or 100mL) of water over a range of temperatures.
    Chemistry 30S Unit 4 - Solutions
  • 23. Unsaturated, Saturated, & Supersaturated Solutions
    • A saturated solution contains the maximum amount of solute that can be dissolved under the conditions at which the solution exists.
    • An unsaturated solution is a solution where less solute than the maximum amount possible is dissolved in the solution.
    • A supersaturated solution holds more dissolved solute than is possible in a saturated solution under the same conditions.
    Chemistry 30S Unit 4 - Solutions
  • 24. Unsaturated, Saturated, & Supersaturated Solutions Chemistry 30S Unit 4 - Solutions
  • 25. Temperature & Solubility
    • The solubility of a solute(solid) in a liquid usually increases with an increase in temperature. However, there are some exceptions.
    • The solubility of a gas in a liquid always decreases with increasing temperature.
    Chemistry 30S Unit 4 - Solutions
  • 26. Temperature & Solubility
    • The reason that the solubility of solids is so variable is that much of the energy for solvation is required to separate the solid particles from each other in either the crystal lattice or from the molecular solid structure.
    • This energy absorption is not required for a gas because all the particles are already separated.
    • Consequently, the overall net process becomes exothermic with the result that solubility is inversely proportional to temperature. The higher kinetic energy of gas particles allows them to escape from a solution more readily. As a result, the solubility of gases decreases with an increase in temperature.
    Chemistry 30S Unit 4 - Solutions
  • 27. How does pressure affect solubility?
    • Increasing pressure increases the solubility of a gas in a liquid because increasing the pressure forces the gas particles into contact with the liquid. As the gas particles contact the liquid, forces of attraction from the liquid cause the gas to condense and dissolve.
    Chemistry 30S Unit 4 - Solutions                           
  • 28. Boiling Point Elevation
    • If we add solute to a solvent, the vapour pressure of the solution is lowered.
    • This relationship can be explained by the following considerations:
      • at the surface of the solution where evaporation occurs, there are fewer solvent particles due to the presence of solute particles: reduced vapour pressure
      • the solute particles absorb energy and therefore reduces the energy available to evaporate the solvent particles: reduced vapour pressure
      • energy is required to overcome the intermolecular forces between the solute and the solvent particles: reduced vapour pressure.
      • The more solute you add, the less vaporizing that will occur .
    Chemistry 30S Unit 4 - Solutions
  • 29. Boiling Point Elevation
    • Since the definition of boiling is the temperature at which the vapour pressure equals the pressure above the liquid, it can be readily seen that if the vapour pressure is lowered, it will require additional energy to raise the temperature to where the vapour pressure equals that of the pressure above the solution. Hence the boiling point elevation .
    Chemistry 30S Unit 4 - Solutions
  • 30. Freezing Point Depression
    • In order for a liquid to freeze, it must achieve a very ordered state that results in the formation of a crystal. If there are impurities in the liquid, i.e. solute, then the liquid in inherently less ordered. Therefore a solution is more difficult to freeze than the pure solvent and a lower temperature is required to freeze the solution.
    Chemistry 30S Unit 4 - Solutions
  • 31. Freezing Point Depression
    • Another way to explain this is to say, as a solution is cooled, solvent molecules lose average kinetic energy to enable them to settle into the crystal structure of the pure solvent. As the crystal grows, solute molecules interfere with the growth of the solvent crystals. To compensate, more kinetic energy must be taken from the solution thus depressing the freezing point.
    Chemistry 30S Unit 4 - Solutions
  • 32. Chemistry 30S Unit 4 - Solutions
  • 33. Chemistry 30S Unit 4 - Solutions Why does ROAD SALT melt ice?
  • 34. Chemistry 30S Unit 4 - Solutions
  • 35. Concentrations of Solutions
    • It is important for chemists to know how much solute is in a certain volume of solution. For example, chemists provide the data for safe concentrations of various substances in drinking water. When chemists speak about concentration they are referring to the amount of solute per unit of solution, unlike solubility that is the amount of solute per unit of solvent.
    Chemistry 30S Unit 4 - Solutions
  • 36. Concentrations of Solutions
    • There are several different units for concentration:
      • parts per million (ppm) means grams of solute for every million grams of solution.
      • parts per billion (ppb) means grams of solute in every billion grams of solution.
      • percent solutions (%) means grams of solute in 100 g of solution or mL of solute in 100 mL of solution for solutions of liquids.
      • grams of solute in a Litre of solution (g/L)
      • grams of solute in every dm 3 (cubic decimetre)
    Chemistry 30S Unit 4 - Solutions
  • 37. Concentrations of Solutions
    • The units of parts per million and billion are often used in describing concentrations of solutes in water and air quality. The percent solute is often used when describing alcoholic beverages (for example, 5% alcohol by volume). The other two units are often used in the medical field to describe the quantities of solutes, like cholesterol, in the blood.
    Chemistry 30S Unit 4 - Solutions
  • 38. Molar Concentration or Molarity
    • The units most commonly used for concentration in chemistry is moles of solute in one Litre of solution, mol/L. This is also known as molar concentration, or molarity , M . The M for molarity should not be con fused with the M for molar mass.
    Chemistry 30S Unit 4 - Solutions
  • 39. Molar Concentration or Molarity
    • Concentration can be calculated according to the formula below:
      • Where:
        • C is concentration in mol/L,
        • n is moles of solute, and
        • V is volume of solution in Litres
    Chemistry 30S Unit 4 - Solutions
  • 40. Molar Concentration or Molarity
    • Example 1. Calculate the concentration of a sodium chloride solution if 0.200 moles is dissolved in 250. mL of solution.
    Chemistry 30S Unit 4 - Solutions
  • 41. Molar Concentration or Molarity
    • Example 2. What volume of a 1.25 mol/L solution contains 5.00 moles of solute?
    Chemistry 30S Unit 4 - Solutions
  • 42. Molar Concentration or Molarity
    • Example 3. How many moles of solute are needed to make 400. mL of a 0.225 mol./L solution?
    Chemistry 30S Unit 4 - Solutions
  • 43. Preparing A Solution
    • In making a solution with a solid solute and liquid solvent, chemists follow a few basic steps:
      • Determine the mass of the solute needed.
      • Add the solid to a volumetric flask of the desired volume needed. Volumetric flasks come in various volumes, from as small as 10 mL to more than 2 L.
      • Fill the volumetric flask about half-full of solvent and swirl until the solid dissolves.
    Chemistry 30S Unit 4 - Solutions
  • 44. Preparing A Solution
      • 4. When the solute is dissolved, add solvent until the bottom of the meniscus reaches the mark on the neck of the flask.
      • 5. Cover the top of the flask and invert several times to mix the solution
    Chemistry 30S Unit 4 - Solutions
  • 45. Preparing A Solution
    • Example 1. Describe the steps needed to make 200. mL of a 0.100 mol/L solution of Cu(NO 3 ) 2 .
    Chemistry 30S Unit 4 - Solutions
  • 46. Concentration of Ions
    • Recall from a previous lesson that ionic compounds dissociate into positive and negative ions when dissolved in water. For example, the dissociation equation of aluminum chloride in water is
    • AlCl 3 ( s )  Al 3+ ( aq ) + 3 Cl¯( aq )
    Chemistry 30S Unit 4 - Solutions
  • 47. Concentration of Ions
    • AlCl 3 ( s )  Al 3+ ( aq ) + 3 Cl¯( aq )
    • The equation indicates that when aluminum chloride dissolves it dissociates into one aluminum ion and 3 chloride ions in solution. We can calculate the concentration of each kind of ion by using the stoichiometry of this equation.
    Chemistry 30S Unit 4 - Solutions
  • 48. Concentration of Ions
    • Example 1. What is the concentration of chloride ions in a 0.200 mol/L solution of aluminum chloride?
    Chemistry 30S Unit 4 - Solutions
  • 49. Dilutions
    • Chemists will often store solutions in a more concentrated form to save on storage space and the ease of making new solutions. These types of solutions are called stock solutions. Virtually an infinite number of solutions of varying concentrations can be made from a single stock solution, simply by adding more solvent. To dilute a solution means to add more solvent without adding more solute. Since the number of moles of solute remains the same (since no more solute is added), calculations of new concentrations becomes easier.
    Chemistry 30S Unit 4 - Solutions
  • 50. Preparing Dilutions
    • The concentration calculations that we have done essentially involved preparing a solution from scratch. We started with separate solvent and solute and figured out how much of each you would need to use.
    • Quite often, however, solutions are prepared by diluting a more concentrated solution. For example, if you needed a one molar solution you could start with a six molar solution and dilute it.
    Chemistry 30S Unit 4 - Solutions
  • 51. Diluting A Solution
    • Preparing a diluted solution is very similar to preparing a solution, except there is no solute to mass. To prepare a diluted solution:
      • i. Determine the volume of stock solution needed. ii. Add the stock solution to a volumetric flask. iii. Add water to the volumetric flask up to the mark and mix.:
    Chemistry 30S Unit 4 - Solutions
  • 52. Diluting A Solution
    • M 1 V 1 = M 2 V 2
      • M 1 = molarity of initial solution
      • V 1 = volume of initial solution
      • M 2 = molarity of diluted solution
      • V 2 = volume of diluted solution
    Chemistry 30S Unit 4 - Solutions
  • 53. Diluting A Solution
    • Example 1. Describe how to make 1.00 L of a 1.00 mol/L solution of HCl from a concentrated solution of 11.8 mol/L.
    Chemistry 30S Unit 4 - Solutions
  • 54. Diluting A Solution
    • Example 2. 450. mL of a 0.150 mol/L NaCl solution is mixed with 125 mL of a 0.220 mol/L NaCl solution. What is the new NaCl concentration?
    Chemistry 30S Unit 4 - Solutions
  • 55. Diluting A Solution
    • Example 3. How much water must be added to 25.0 mL of a 1.00 mol/L stock solution of NaOH to make a 0.100 mol/L solution?
    Chemistry 30S Unit 4 - Solutions
  • 56. Diluting A Solution
    • Example 4. If 500. mL of water is added to 300. mL of a 0.100 mol/L solution of NaCl, what is the new concentration?
    Chemistry 30S Unit 4 - Solutions
  • 57. Diluting A Solution
    • Example 5. What is the concentration of the stock solution if 50.0 mL of the solution is diluted to make 220. mL of a 0.400 mol/L solution?
    Chemistry 30S Unit 4 - Solutions
  • 58. Unsaturated, Saturated, & Supersaturated Solutions Chemistry 30S Unit 4 - Solutions saturated unsaturated supersaturated